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MEASURE THEORY
T.K.SUBRAHMONIAN MOOTHATHU
Contents
1. Introduction: Riemann vs Lebesgue 2
2. Small subsets of Rd 2
3. About functions behaving nicely outside a small set 9
4. σ-algebras and measurable spaces 17
5. Measures and outer measures 22
6. Lebesgue measure on Rd 24
7. Pre-measure → outer measure → measure 30
8. Measurable functions 36
9. Approximating measurable functions with simple functions 40
10. Limit theorems for measurable functions 42
11. Integration with respect to a measure 46
12. Properties of the integral, and limit theorems 48
13. Lp spaces 53
14. Product spaces and product measures 56
15. Fundamental Theorem of Calculus (FTC) 59
16. Measures on metric spaces 61
17. Measure Theory and Functional Analysis 65
Prerequisites: Real Analysis and basic Metric Space Theory.
Convention: Axiom of choice and continuum hypothesis will be assumed throughout.
Topics for self-reading: Banach-Tarski paradox, Littlewood’s three principles.
Our approach: We will do measure theory first, and then will do integration theory. We will also
discuss the interaction of measure theory with topology in Rd.
Some textbooks for further reading:
1. R.G. Bartle, The elements of integration and Lebesgue measure, 1995.
2. D.L. Cohn, Measure Theory, 2nd Edn., 2013.1
2 T.K.SUBRAHMONIAN MOOTHATHU
3. H.L. Royden, Real Analysis, 3rd Edn., 1988.
4. E.M. Stein and R. Shakarchi, Real Analysis (Princeton Lectures in Analysis, 3), 2005.
1. Introduction: Riemann vs Lebesgue
Measure theory helps us to assign numbers to certain sets and functions - to a measurable set
we may assign its measure, and to an integrable function we may assign the value of its integral.
Lebesgue integration theory is a generalization and completion of Riemann integration theory. In
Lebesgue’s theory, we can assign numbers to more sets and more functions than what is possible in
Riemann’s theory. If we are asked to distinguish between Riemann integration theory and Lebesgue
integration theory by pointing out an essential feature, the answer is perhaps the following.
Riemann integration theory 7→ finiteness.
Lebesgue integration theory 7→ countable infiniteness.
Riemann integration theory is developed through approximations of a finite nature (eg: one
tries to approximate the area of a bounded subset of R2 by the sum of the areas of finitely many
rectangles), and this theory works well with respect to finite operations - if we can assign numbers to
finitely many sets A1, . . . , An and finitely many functions f1, . . . , fn, then we can assign numbers to
A1∪· · ·∪An, f1+· · ·+fn, maxf1, . . . , fn, etc. The disadvantage of Riemann integration theory is
that it does not behave well with respect to operations of a countably infinite nature - there may not
be any consistent way to assign numbers to∪∞
n=1An,∑∞
n=1 fn, limn→∞ fn, supfn : n ∈ N, etc.
even if we can assign numbers to the sets A1, A2, . . . , and functions f1, f2 . . .. Lebesgue integration
theory rectifies this disadvantage to a large extent.
In Riemann integration theory, we proceed by considering a partition of the domain of a function,
where as in Lebesgue integration theory, we proceed by considering a partition of the range of the
function - this is observed as another difference. Moreover, while Riemann’s theory is restricted to
the Euclidean space, the ideas involved in Lebesgue’s theory are applicable to more general spaces,
yielding an abstract measure theory. This abstract measure theory intersects with many branches
of mathematics and is very useful. There is even a philosophy that measures are easier to deal with
than sets.
2. Small subsets of Rd
It is possible to think about many mathematical notions expressing in some sense the idea that
a subset Y ⊂ Rd is a small set (or a big set) with respect to Rd. We will discuss this a little as a
warm-up. We will also use this opportunity to introduce Lebesgue outer measure.
Suppose you have a certain notion of smallness or bigness for a subset of Rd. Then there are
some natural questions. Two sample questions are:
MEASURE THEORY 3
1. If Y ⊂ Rd is big, is Rd \ Y small?
2. If Y1, Y2, . . . ⊂ Rd are small, is∪∞
n=1 Yn small?
For instance, consider the following two elementary notions. Saying that Y ⊂ Rd is unbounded
is one way of saying Y is big, and saying that Y ⊂ Rd is a finite set is one way of saying Y is small.
Note that the complement of an unbounded set can also be unbounded and a countable union of
finite sets need not be finite. So here we have negative answers to the above two questions.
Exercise-1: Find an uncountable collection Yα : α ∈ I of subsets of R such that Yα’s are pairwise
disjoint, and each Yα is bounded neither above nor below.
To discuss some other notions of smallness, we introduce a few definitions.
Definitions: (i) We say Y ⊂ Rd is a discrete subset of Rd if for each y ∈ Y , there is an open set
U ⊂ Rd such that U ∩ Y = y. For example, 1/n : n ∈ N is a discrete subset of R and Zd is a
discrete subset of Rd.
(ii) A subset Y ⊂ Rd is nowhere dense in Rd if int[Y ] = ∅, or equivalently if for any nonempty open
set U ⊂ Rd, there is a nonempty open set V ⊂ U such that V ∩Y = ∅. If f : R → R is a continuous
map, then its graph G(f) := (x, f(x)) : x ∈ R is nowhere dense in R2 (∵ G(f) is closed and does
not contain any open disc). Also, A ∈ Rd×d : det(A) = 0 is nowhere dense in Rd×d.
(iii) A subset Y ⊂ Rd is of first category in Rd if Y can be written as a countable union of nowhere
dense subsets of Rd; otherwise, Y is said to be of second category in Rd. For example, Y = Q× R
is of first category in R2 since Y can be written as the countable union Y =∪
r∈Q Yr, where
Yr := r × R is nowhere dense in R2.
(iv) (The following definition can be extended by considering ordinal numbers, but we consider only
non-negative integers). For Y ⊂ Rd and integer n ≥ 0, define the nth derived set of Y inductively as
Y (0) = Y , Y (n+1) = limit points of Y (n) in Rd. We say Y ⊂ Rd has derived length n if Y (n) = ∅
and Y (n+1) = ∅; and we say Y has infinite derived length if Y (n) = ∅ for every integer n ≥ 0. For
example, Q has infinite derived length (since Q = R), and (1/m, 1/n) : m,n ∈ N has derived
length 2.
(v) We say A ⊂ Rd is a d-box if A =∏d
j=1 Ij , where Ij ’s are bounded intervals. The d-dimensional
volume of a d-box A is V old(A) =∏d
j=1 |Ij |. For example, V ol3([1, 4)× [0, 1/2]× (−1, 3]) = 6.
Exercise-2: If U ⊂ Rd is a nonempty open set, then U can be written as a countable union of
pairwise disjoint d-boxes (these d-boxes may not be open). [Hint : Let B be the collection of all
open d-boxes whose vertices have rational coordinates. For each x ∈ U , there is B ∈ B with
4 T.K.SUBRAHMONIAN MOOTHATHU
x ∈ B ⊂ U . Hence U can be written as a countable union U =∪
nBn with Bn ∈ B. Now note that
Bn+1 \∪n
j=1Bj can be written as a finite union of pairwise disjoint d-boxes.]
(vi) The d-dimensional Jordan outer content µ∗J,d[Y ] of a bounded subset Y ⊂ Rd is defined as
µ∗J,d[Y ] = inf∑k
n=1 V old(An) : k ∈ N, and An’s are d-boxes with Y ⊂∪k
n=1An.
(vii) The d-dimensional Lebesgue outer measure µ∗L,d[Y ] of an arbitrary set Y ⊂ Rd is defined as
µ∗L,d[Y ] = inf∑∞
n=1 V old(An) : An’s are d-boxes with Y ⊂∪∞
n=1An.
[101]We have that µ∗L,d[Y ] ≤ µ∗J,d[Y ] for any bounded set Y ⊂ Rd, and µ∗L,d[A] = µ∗J,d[A] = V old(A)
for any d-box A ⊂ Rd.
Proof. Any finite union∪k
n=1An of d-boxes can be extended to an infinite union∪∞
n=1An of d-boxes
without changing the total volume by taking An’s to be singletons for n > k. This observation
yields that µ∗L,d[Y ] ≤ µ∗J,d[Y ]. It is easy to see µ∗J,d[A] = V old(A) if A is a d-box. It remains
to show µ∗L,d[A] ≥ V old(A) when A is a d-box. First suppose A is closed. Then A is compact
by Heine-Borel. Let ε > 0 and let A1, A2, . . . ⊂ Rd be d-boxes such that A ⊂∪∞
n=1An and∑∞n=1 V old(An) < µ∗L,d[A] + ε. For each n ∈ N, let Bn be an open d-box with An ⊂ Bn and
V old(Bn) < V old(An)+ε/2n. Then Bn : n ∈ N is an open cover for the compact set A. Extracting
a finite subcover, we have V old(A) ≤∑k
n=1 V old(Bn) ≤∑∞
n=1(V old(An) + ε/2n) < µ∗L,d[A] + 2ε.
Thus µ∗L,d[A] = V old(A) for closed d-boxes. Now if B is an arbitrary d-box and ε > 0, then there
is a closed d-box A ⊂ B with V old(B)− ε < V old(A) = µ∗L,d[A] ≤ µ∗L,d[B].
Other basic properties of Lebesgue outer measure and Jordan outer content are given below.
[102] (i) µ∗L,d[∅] = 0.
(ii) [Monotonicity] µ∗L,d[X] ≤ µ∗L,d[Y ] if X ⊂ Y ⊂ Rd.
(iii) [Translation-invariance] µ∗L,d[Y + x] = µ∗L,d[Y ] for every Y ⊂ Rd and every x ∈ Rd.
(iv) [Countable subadditivity] If Y1, Y2, . . . ⊂ Rd and Y =∪∞
n=1 Yn, then µ∗L,d[Y ] ≤
∑∞n=1 µ
∗L,d[Yn].
(v) µ∗L,d[Y ] = 0 for every countable set Y ⊂ Rd.
(vi) For any Y ⊂ Rd, we have µ∗L,d[Y ]
= inf∑∞
n=1 V old(An) : An’s are closed d-boxes with Y ⊂∪∞
n=1An
= inf∑∞
n=1 V old(An) : An’s are open d-boxes with Y ⊂∪∞
n=1An.
(vii) For any Y ⊂ Rd, we have µ∗L,d[Y ] = infµ∗L,d[U ] : Y ⊂ U and U is open in Rd.
(viii) µ∗L,d[Rd] = ∞.
(ix) If X,Y ⊂ Rd are such that dist(X,Y ) := inf∥x− y∥ : x ∈ X, y ∈ Y > 0, then µ∗L,d[X ∪ Y ] =
µ∗L,d[X] + µ∗L,d[Y ].
MEASURE THEORY 5
Proof. (i), (ii) and (iii) are clear. To prove (iv), without loss of generality we may assume∑∞n=1 µ
∗L,d[Yn] < ∞. Given ε > 0, there exist d-boxes A(n, k) such that Yn ⊂
∪∞k=1A(n, k) and∑∞
k=1 V old(A(n, k)) < µ∗L,d[Yn] + ε/2n. Then Y ⊂∪∞
n=1
∪∞k=1A(n, k) and we have the estimate∑∞
n=1
∑∞k=1 V old(A(n, k)) ≤
∑∞n=1
(µ∗L,d[Yn] + ε/2n
)= (
∑∞n=1 µ
∗L,d[Yn]) + ε.
Now (v) follows from (iv) since singletons have Lebesgue outer measure zero (or we can see it
directly by noting that singletons are d-boxes with zero volume). The first part of (vi) is clear
since any d-box and its closure have equal volume. To get the second part, note that if A1, A2, . . .
are d-boxes and ε > 0, there exist open d-boxes B1, B2, . . . such that An ⊂ Bn and V old(Bn) <
V old(An)+ε/2n. We may deduce (vii) using part (vi) and Exercise-2. And (viii) follows from [101]
and [102](ii).
Now we prove (ix). From countable subadditivity, we have µ∗L,d[X ∪ Y ] ≤ µ∗L,d[X] + µ∗L,d[Y ]. To
prove the other inequality, we may assume µ∗L,d[X ∪ Y ] < ∞. Let δ = dist(X,Y ). Given ε > 0,
find d-boxes A1, A2, . . . such that X ∪ Y ⊂∪∞
n=1An and∑∞
n=1 V old(An) < µ∗L,d[X ∪ Y ] + ε. By
partitioning the d-boxes into smaller d-boxes and throwing away the unnecessary ones, we may
assume that diam[An] < δ and (X ∪ Y ) ∩ An = ∅ for every n ∈ N. Let Γ = n ∈ N : X ∩ An = ∅
and Γ′ = n ∈ N : Y ∩ An = ∅. Then N = Γ ∪ Γ′ is a disjoint union, X ⊂∪
n∈ΓAn, and
Y ⊂∪
n∈Γ′ An. Hence µ∗L,d[X] +µ∗L,d[Y ] ≤
∑n∈Γ V old(An)+
∑n∈Γ′ V old(An) =
∑∞n=1 V old(An) <
µ∗L,d[X ∪ Y ] + ε.
[103] (i) µ∗J,d[∅] = 0.
(ii) [Monotonicity] µ∗J,d[X] ≤ µ∗J,d[Y ] if X ⊂ Y are bounded subsets of Rd.
(iii) [Translation-invariance] µ∗L,d[Y +x] = µ∗L,d[Y ] for every bounded set Y ⊂ Rd and every x ∈ Rd.
(iv) [Finite subadditivity] If X,Y ⊂ Rd are bounded subsets, then µ∗J,d[X ∪ Y ] ≤ µ∗J,d[X] + µ∗J,d[Y ].
(v) µ∗J,d[Y ] = 0 for every finite set Y ⊂ Rd.
(vi) For any bounded set Y ⊂ Rd, we have µ∗J,d[Y ]
= inf∑k
n=1 V old(An) : k ∈ N, and An’s are closed d-boxes with Y ⊂∪k
n=1An
= inf∑k
n=1 V old(An) : k ∈ N, and An’s are open d-boxes with Y ⊂∪k
n=1An
= inf∑k
n=1 V old(An) : k ∈ N, and An’s are pairwise disjoint d-boxes with Y ⊂∪k
n=1An.
(vii) If X,Y ⊂ Rd are bounded sets with dist(X,Y ) := inf∥x − y∥ : x ∈ X, y ∈ Y > 0, then
µ∗J,d[X ∪ Y ] = µ∗J,d[X] + µ∗J,d[Y ].
(viii) For any bounded set Y ⊂ Rd, we have µ∗J,d[Y ] = µ∗J,d[Y ].
Proof. Statements (i) to (vii) can be proved as in [102]. To prove (viii), use the first expression for
µ∗J,d[Y ] in (vi) and note that a finite union of closed sets is closed.
Example: Let Y = Qd ∩ [0, 1]d. Note that µ∗L,d[Y ] = 0 = 1 = µL,d[Y ]. But we have µ∗J,d[Y ] =
µ∗J,d[Y ] = 1 by [103](viii). So the Jordan outer content of a bounded countable set need not be
6 T.K.SUBRAHMONIAN MOOTHATHU
zero. This example also shows that µ∗L,d[Y ] < µ∗J,d[Y ] is possible for a bounded set, and that the
Jordan outer content does not satisfy countable subadditivity for bounded sets (since the Jordan
outer content of a singleton is zero). If X = [0, 1]d \ Y , then µ∗J,d[X] = 1 since X = [0, 1]d and
hence µ∗J,d[X] + µ∗J,d[Y ] = 2 = 1 = µ∗J,d[X ∪ Y ]. This says that the assumption dist(X,Y ) > 0 is
necessary in [103](viii). .
Some ways of saying that Y ⊂ Rd is a small set:
(i) Y is a countable set.
(ii) Y is a discrete subset of Rd.
(iii) Y is contained in a vector subspace of Rd of dimension ≤ d− 1.
(iv) Y is nowhere dense in Rd.
(v) Y is of first category in Rd.
(vi) Y has finite derived length.
(vii) Y is a bounded set with µ∗J,d[Y ] = 0.
(viii) µ∗L,d[Y ] = 0.
It is good to investigate various possible implications between pairs of notions given above. Some
are discussed below.
Exercise-3: If Y is a discrete subset of Rd, then Y is countable. [Hint : Let B = B(x, 1/n) : x ∈
Qd, n ∈ N. Then, B is countable and for each y ∈ Y , there is B ∈ B such that B ∩ Y = y.]
Exercise-4: If Y is contained in a vector subspace W of Rd with dim(W ) ≤ d − 1, then Y is a
nowhere dense subset of Rd. [Hint : W is closed in Rd (∵ fix a basis for W and argue with the
coefficients of each basis vector separately) and W does not contain any open ball of Rd.]
Exercise-5: For any Y ⊂ Rd, the set Y \ Y (1) is discrete and hence countable. In particular, every
uncountable subset of Rd has a limit point in Rd. [Hint : Let y ∈ Y \Y (1). If B(y, 1/n)∩Y contains
a point other than y for every n ∈ N, then y ∈ Y (1), a contradiction.]
Exercise-6: If Y ⊂ Rd has finite derived length, then Y is countable. [Hint : Suppose Y (n) = ∅
and Y (n+1) = ∅. Then Y (n) = Y (n) \ Y (n+1) must be countable by Exercise-5. Go backwards
inductively.]
Exercise-7: Every nonempty open set U ⊂ Rd contains a bounded countable set Y with µ∗J,d[Y ] > 0.
Exercise-8: If Y ⊂ Rd is a bounded subset with finite derived length, then µ∗J,d[Y ] = 0. [Hint :
Suppose that Y (n) = ∅ and Y (n+1) = ∅, and let ε > 0. Find a compact set K ⊂ Rd with Y ⊂ K.
Note that Y (n) must be finite using the compactness of K. Find finitely many open d-boxes of
A1, . . . , Ak of total volume < ε/2 covering Y (n). Then, observe that Y (n−1) \∪k
j=1Aj is finite by
MEASURE THEORY 7
the compactness of K. Next, find finitely many open d-boxes of total volume < ε/22 covering
Y (n−1) \∪k
j=1Aj . Proceed backwards like this.] [Converse is not true: consider [0, 1]× 0 ⊂ R2.]
Exercise-9: µ∗L,d[Rd−1 × 0] = 0. [Hint : Let B1, B2, . . . ⊂ Rd−1 be countably many (d − 1)-
boxes covering Rd−1 such that V old−1(Bn) = 1 for every n ∈ N. Given ε > 0, let An = Bn ×
[−ε/2n+1, ε/2n+1]. Then Rd−1 × 0 ⊂∪∞
n=1An and∑∞
n=1 V old(An) =∑∞
n=1 ε/2n = ε.] [With
more effort we can prove the following: if Y is contained in a vector subspace W of Rd with
dim(W ) ≤ d− 1, then µ∗L,d[Y ] = 0.]
Exercise-10: If Y1, Y2, . . .Rd are such that µ∗L,d[Yn] = 0 for every n ∈ N, then µ∗L,d[∪∞
n=1 Yn] = 0.
Now we discuss some interesting examples.
[104] Let K ⊂ [0, 1] be the middle-third Cantor set. Then, K is an uncountable, nowhere dense
compact set with µ∗J,1[K] = µ∗L,1[K] = 0. Moreover, K has no isolated points.
Proof. We recall the construction of K. Let K0 = [0, 1], K1 = [0, 1/3] ∪ [2/3, 1], K2 = [0, 1/9] ∪
[2/9, 1/3]∪ [2/3, 7/9]∪ [8/9, 1], and so on. That is, Kn is the disjoint union of 2n closed subintervals
of [0, 1], each having length 1/3n, and Kn+1 is obtained from Kn by removing the middle-third
open intervals from each of these 2n closed intervals. The middle-third Cantor set K is defined as
K =∩∞
n=0Kn. Being the intersection of compact sets, K is compact. Since the maximal length
of an interval contained in Kn is (1/3)n, K does not contain any open interval, and hence K is
nowhere dense. Also, since K ⊂ Kn, the above description yields µ∗J,1[K] ≤ (2/3)n. So µ∗J,1[K] = 0
and hence µ∗L,1[K] = 0 also.
It may be verified that K = ∑∞
n=1 xn/3n : xn ∈ 0, 2. That is, K is precisely the set of those
x ∈ [0, 1] whose ternary expansion (i.e., base 3 expansion) x = 0.x1x2 · · · contains only 0’s and 2’s.
Hence K is bijective with 0, 2N which is uncountable.
We show K has no isolated point. Let x ∈ K and let U be a neighborhood of x. Choose n large
enough so that one of the 2n closed intervals constituting Kn, say Jn, satisfies x ∈ Jn ⊂ U . Let
y ∈ Jn \ x be an end point of Jn. This end point is never removed in the later construction, so
y ∈ Km for every m ≥ n. Thus y ∈ K ∩ (U \ x).
Remark: It may be noted that for x ∈ K, the base 3 expansion x = 0.x1x2 · · · is eventually constant
iff x is an end point of a removed open interval. This helps to see that K contains points other
than the end points of the (countably many) removed open intervals.
The following theorem is relevant while considering big and small sets in a topological sense.
[105] [Baire Category Theorem] Let (X, ρ) be a complete metric space and let Un ⊂ X be open
and dense in X for n ∈ N. Then,∩∞
n=1 Un is also dense in X. In particular,∩∞
n=1 Un = ∅.
8 T.K.SUBRAHMONIAN MOOTHATHU
Proof. Let V ⊂ X be a nonempty open set. It suffices to show V ∩ (∩∞
n=1 Un) = ∅. Since U1 is open
and dense, U1 ∩ V is a nonempty open set. Let B1 be an open ball in X such that B1 ⊂ U1 ∩ V
and diam[B1] < 1. Since U2 is open and dense, B1 ∩ U2 = ∅. Let B2 ⊂ X be an open ball with
B2 ⊂ B1∩U2 and diam[B2] < 1/2. In general, let Bn+1 ⊂ X be an open ball with Bn+1 ⊂ Bn∩Un+1
and diam[Bn+1] < 1/(n+1). If xn is the center of the ball Bn, then we note that for every n,m ≥ k
we have xn, xm ∈ Bk and hence ρ(xn, xm) ≤ diam[Bk] < 1/k. So (xn) is a Cauchy sequence. Since
(X, ρ) is complete, there is x ∈ X such that (xn) → x. Now, for any n, we have xm ∈ Bn for m ≥ n
and hence x ∈ Bn. Thus x ∈∩∞
n=1Bn ⊂ V ∩ (∩∞
n=1 Un).
Remarks: (i) By considering the complements of Un’s in the above, we get the following conclusion:
if (X, ρ) is a complete metric space, then X cannot be written as a countable union of nowhere
dense (closed) subsets of X. That is, X is of second category in itself. (ii) Since Rd is a complete
metric space with respect to the Euclidean metric, Rd cannot be written as a countable union of
nowhere dense (closed) subsets of Rd. (iii) From a topological point of view, a first category subset is
considered as a small set and a dense Gδ subset is considered as a big set because of Baire Category
Theorem. However, a set that is topologically big (small) need not be measure theoretically big
(small); we will see this in [106] below. (iv) The uncountability of the middle-third Cantor set can
be proved with the help of Baire Category Theorem also, see the exercise below.
Exercise-11: Let (X, ρ) be a complete metric space without isolated points. Then X must be
uncountable. [Hint : If X is countable, write X = x1, x2, . . . and let Un = X \ xn. Then Un
is open, and Un is dense since xn is not open (∵ xn is not an isolated point). But∩
n Un = ∅, a
contradiction.]
Exercise-12: It is not possible to write Qd as a countable intersection of open subsets of Rd. [Hint : If
possible, let Qd =∩∞
n=1 Un, where Un is open in Rd. Write Qd = x1, x2, . . . and let Vn = Rd\xn.
Then, (∩∞
n=1 Un) ∩ (∩∞
n=1 Vn) = ∅, contradicting Baire Category Theorem.]
We observe in the following the distinction between topological bigness (smallness) and measure
theoretical bigness (smallness).
[106] (i) For every ε > 0, there is a dense open set U ⊂ Rd such that µ∗L,d[U ] < ε.
(ii) There is a dense Gδ subset Y ⊂ Rd with µ∗L,d[Y ] = 0.
(iii) There is an Fσ set X ⊂ Rd of first category with µ∗L,d[X] = ∞ and µ∗L,d[Rd \X] = 0.
(iv) For every closed d-box A and every ε > 0, there is a nowhere dense compact set K ⊂ Rd such
that K ⊂ A and µ∗L,d[K] > V old(A)− ε.
Proof. (i) Write Qd = x1, x2, . . .. For each n ∈ N, let An be an open d-box with xn ∈ An and
V old(An) < ε/2n. Put U =∪∞
n=1An.
MEASURE THEORY 9
(ii) Let Un ⊂ Rd be a dense open subset with µ∗L,d[Un] < 1/n and put Y =∩∞
n=1 Un.
(iii) Let Y be as in (ii) and take X = Rd \ Y .
(iv) Let U be as in (i) and let K = A \ U .
The next result shows that the Lebesgue outer measure does not satisfy finite additivity (and
hence it does not satisfy countable additivity), even though it satisfies countable subadditivity.
[107] Let X = Qd ∩ [0, 1]d. Then, there is a subset Y ⊂ [0, 1]d satisfying the following:
(i) The translations Y + x are pairwise disjoint for x ∈ X.
(ii) There exist finitely many distinct elements x1, . . . , xn ∈ X such that µ∗L,d[∪n
i=1(Y + xi)] =∑ni=1 µ
∗L,d[Y + xi].
Proof. Define an equivalence relation on [0, 1]d by the condition that a ∼ b iff a− b ∈ Qd. By the
axiom of choice, we can form a set Y ⊂ [0, 1]d whose intersection with each equivalence class is a
singleton.
(i) We verify that (Y + r) ∩ (Y + s) = ∅ for any two distinct r, s ∈ X. If (Y + r) ∩ (Y + s) = ∅ for
r, s ∈ X, then there are a, b ∈ Y such that a + r = b + s. Now we have a − b = s − r ∈ Qd, and
hence a ∼ b. So we must have a = b by the definition of Y , and then necessarily r = s.
(ii) If z ∈ Rd, there is r ∈ Qd such that z− r ∈ [0, 1]d. Then there is y ∈ Y such that y ∼ z− r and
so there is r′ ∈ Qd such that y + r′ = z − r or z = y + r+ r′. This shows that Rd =∪
r∈Qd(Y + r).
By [102](iii) and [102](viii), we conclude that µ∗L,d[Y ] > 0. Let δ = µ∗L,d[Y ] and n ∈ N be such that
nδ > 2d. Choose distinct elements x1, . . . , xn ∈ X. Then∑n
i=1 µ∗L,d[Y + xi] = nδ > 2d again by
translation invariance. On the other hand, Y+X ⊂ [0, 2]d and therefore µ∗L,d[∪n
i=1(Y+xi)] ≤ 2d.
Remark: The construction above is due to Vitali, and hence the set Y is called a Vitali set.
3. About functions behaving nicely outside a small set
There are a few classical results in Analysis with conclusion of the following form: “... the
function has nice behavior outside a small set”. We will consider some such results here.
We know that a function that is the pointwise limit of a sequence of continuous functions may
not be continuous. For instance, f : [0, 1] → R given by f(1) = 1 and f(x) = 0 for x < 1 is
the pointwise limit of (fn), where fn : [0, 1] → R is fn(x) = xn. However, the set of points of
discontinuity of the limit function must be a small set often, see [108] below.
Definition: Let X,Y be metric spaces and let f : X → Y be a function. Then the oscillation ω(f, x)
of f at a point x ∈ X is defined as ω(f, x) = limδ→0+ diam[f(B(x, δ))]. Clearly, f is continuous at
x iff ω(f, x) = 0.
10 T.K.SUBRAHMONIAN MOOTHATHU
Exercise-13: Let X,Y be metric spaces and let f : X → Y be a function. Then the set x ∈
X : f is continuous at x is a Gδ subset of X. [Hint : The given set is equal to∩∞
n=1 Un, where
Un = x ∈ X : ω(f, x) < 1/n, and Un is open.]
Exercise-14: There does not exist any function f : Rd → R such that x ∈ Rd : f is continuous at x =
Qd. [Hint : Use Exercise-12 and Exercise-13.] [Remark: However, for f : R → R defined as f(0) = 1,
f(x) = 0 for x ∈ R \ Q, and f(p/q) = 1/q for p ∈ Z \ 0, q ∈ N, and p, q are relatively prime,
satisfies x ∈ R : f is continuous at x = R \Q.]
[108] Let (X, ρ1) be a complete metric space, (Y, ρ2) be an arbitrary metric space, and let (fn) be
a sequence of continuous functions from X to Y , converging pointwise to a function f : X → Y .
Then the set x ∈ X : f is continuous at x is a dense Gδ subset of X.
Proof. Let ε > 0 and Dε = x ∈ X : ω(f, x) ≥ ε. We know from the hint of Exercise-13 that Dε
is a closed set. We claim that Dε is nowhere dense in X. Let U ⊂ X be a nonempty open set. We
have to find a nonempty open set V ⊂ U such that Dε ∩ V = ∅.
For n ∈ N, let Kn = x ∈ X : ρ2(fn(x), fj(x)) ≤ ε/4 for every j ≥ n. Then Kn is a closed set
and X =∪∞
n=1Kn. The continuity of the distance function ρ2 implies that ρ2(fn(x), f(x)) ≤ ε/4
for every x ∈ Kn. Let U1 ⊂ X be a nonempty open set with U1 ⊂ U . Since (U1, ρ1) is a complete
metric space, there is n ∈ N such that U2 := int[Kn ∩ U1] = ∅. Let V ⊂ U2 be a nonempty
open set with diam[fn(V )] ≤ ε/4. For any x, y ∈ V , we have ρ2(f(x), f(y)) ≤ ρ2(f(x), fn(x)) +
ρ2(fn(x), fn(y))+ρ2(fn(y), f(y)) ≤ ε/4+ε/4+ε/4 = 3ε/4. Hence diam[f(V )] ≤ 3ε/4 and therefore
ω(f, x) ≤ 3ε/4 for every x ∈ V . This shows Dε ∩ V = ∅, proving our claim.
The claim implies that D :=∪∞
n=1D1/n is an Fσ set of first category in X. This completes the
proof since x ∈ X : f is continuous at x = X \D, and X is a complete metric space.
We know that the derivative of a differentiable real function need not be continuous. However,
we can say the following.
[109] Let f : R → R be differentiable. Then there exists a sequence (gn) of continuous functions
from R to R converging to f ′ pointwise. Consequently, x ∈ R : f ′ is continuous at x is a dense
Gδ subset of R.
Proof. Since f is differentiable, f is continuous. Define gn(x) = [f(x+ 1/n)− f(x)]/(1/n) and use
[108].
Now we will show that a monotone real function (increasing or decreasing) is continuous and
differentiable at most of the points.
MEASURE THEORY 11
[110] Let −∞ ≤ a < b ≤ ∞ and let f : (a, b) → R be a monotone function. Then, Y = x ∈ (a, b) :
f is discontinuous at x is a countable set (possibly empty).
Proof. Suppose f is increasing. If x ∈ Y , then necessarily f(x−) < f(x+), and we may choose a
rational number between f(x−) and f(x+). This gives a one-one map from Y to Q.
To talk about the points of differentiability of a monotone real function, we need a technical
lemma due to Vitali.
Definition: A collection Γ of nondegenerate intervals is a Vitali cover for a set X ⊂ R if for each
ε > 0, the subcollection I ∈ Γ : 0 < |I| < ε is also a cover for X.
[111] [Vitali’s covering lemma] Let X ⊂ R be such that µ∗L,1[X] < ∞ and let Γ be a collection of
intervals forming a Vitali cover for X. Then,
(i) There are countably many pairwise disjoint intervals I1, I2, . . . ∈ Γ such that µ∗L,1[X \∪
n In] = 0.
(ii) For every ε > 0, there exist finitely many pairwise disjoint intervals I1, . . . , Ik ∈ Γ with the
property that µ∗L,1[X \∪k
n=1 In] < ε.
Proof. Write µ∗ = µ∗L,1 for simplicity.
(i) With out loss of generality assume that every I ∈ Γ is a (nondegenerate) closed interval. By
[102](vi), choose an open set U ⊂ R such that X ⊂ U and µ∗[U ] < ∞. Every x ∈ X has a
neighborhood contained in U . Hence Γ′ = I ∈ Γ : I ⊂ U is also a vitali cover for X. We will
choose the intervals In inductively. Let δ0 = sup|J | : J ∈ Γ′ (note that δ0 ≤ µ∗[U ] <∞) and let
I1 ∈ Γ′ be any interval with |I1| > δ0/2. Suppose that we have chosen pairwise disjoint intervals
I1, . . . , In ∈ Γ′. If X ⊂∪n
i=1 Ii, then we are done. Else, any x ∈ X \∪n
i=1 Ii is at a positive
distance from the closed set∪n
i=1 Ii. Let δn = sup|J | : J ∈ Γ′ and Ii ∩ J = ∅ for 1 ≤ i ≤ n.
Then 0 < δn ≤ µ∗[U ] <∞. Let In+1 ∈ Γ′ be an interval with |In+1| > δn/2. We will show that the
sequence (In) does the job.
Observation: For every J ∈ Γ′, there is n ∈ N such that In ∩ J = ∅ (∵∑
|In| ≤ µ∗[U ] <∞ so that
(|In|) → 0, and hence there is n ∈ N such that |In| < |J |/2).
Let Y = X \∪∞
n=1 In and ε > 0. We claim that µ∗[Y ] < ε. Let cn be the midpoint of In and let
Yn ⊂ R be the closed interval with midpoint cn and |Yn| = 6|In| (this Yn may not be in Γ′). Let
k ∈ N be so that∑∞
n=k+1 |In| < ε/6. If x ∈ Y , then in particular x does not belong to the closed
set∪k
n=1 In. Choose J ∈ Γ′ with x ∈ J and In ∩ J = ∅ for 1 ≤ n ≤ k. By our observation above,
Im ∩ J = ∅ for some m ≥ k + 1. Let m be the smallest such number. Then |J | ≤ δm−1 < 2|Im|
and hence |x− cm| ≤ |J |+ |Im| ≤ 3|Im|. Therefore, x ∈ Ym. We have shown that Y ⊂∪∞
n=k+1 Yn.
Since∑∞
n=k+1 |Yn| ≤ 6∑∞
n=k+1 |In| < ε, we have proved that µ∗[Y ] < ε.
12 T.K.SUBRAHMONIAN MOOTHATHU
Now, note that the argument given above actually shows that for every ε > 0, there is k ∈ N
such that µ∗[X \∪k
n=1 In] < ε. Hence we have established (ii) also.
When a mathematical problem is difficult, it is a good idea to divide the problem into many
subcases and to treat each case separately. If f : (a, b) → R is a function, then the four Dini
derivatives of f at a point x ∈ (a, b) are defined as follows.
D+f(x) = lim suph→0+
f(x+ h)− f(x)
h[upper right derivative]
D+f(x) = lim infh→0+
f(x+ h)− f(x)
h[lower right derivative]
D−f(x) = lim suph→0−
f(x+ h)− f(x)
h[upper left derivative]
D−f(x) = lim infh→0−
f(x+ h)− f(x)
h[lower left derivative].
Remark: Here, lim suph→0+
f(x+ h)− f(x)
h:= lim
y→0+
[sup
0<h<y
f(x+ h)− f(x)
h
], and similarly the others.
Remark: Note that f is differentiable at x iff all the four Dini derivatives are equal and real (i.e.,
different from ±∞). SinceD+f(x) ≤ D+f(x) andD−f(x) ≤ D−f(x) by definition, we also see that
f is differentiable at x iff the four Dini derivatives are real numbers satisfying D+f(x) ≤ D−f(x)
and D−f(x) ≤ D+f(x).
Example: Let f : (−1, 1) → R be f(0) = 0 and f(x) = x sin(1/x) for x = 0. Then, D+f(0) = 1 =
D−f(0) and D+f(0) = −1 = D−f(0) so that f is not differentiable at 0.
[112] [Lebesgue’s differentiation theorem] Let −∞ ≤ a < b ≤ ∞, let f : (a, b) → R be a monotone
function and let Y = x ∈ (a, b) : f is not differentiable at x. Then µ∗L,1[Y ] = 0.
Proof. Since (a, b) can be written as a countable union of bounded open intervals, we may as
well assume (a, b) itself is bounded in view of Exercise-10. Assume f is increasing and write
µ∗ = µ∗L,1. By the remark above, Y = Y1 ∪ Y2, where Y1 = x ∈ (a, b) : D−f(x) < D+f(x) and
Y2 = x ∈ (a, b) : D+f(x) < D−f(x). We will only show that µ∗[Y1] = 0; the case of Y2 is similar.
Let Γ = (r, s) ∈ Q2 : r < s, let X(r, s) = x ∈ (a, b) : D−f(x) < r < s < D+f(x) and note
that Y1 =∪
(r,s)∈ΓX(r, s). Hence it suffices to show there µ∗[X(r, s)] = 0 for every (r, s) ∈ Γ, by
Exercise-10. Fix (r, s) ∈ Γ, write X = X(r, s) and let ε > 0 be arbitrary. Choose an open set
U ⊂ (a, b) such that X ⊂ U and µ∗[U ] < µ∗[X] + ε by [102](vi).
Since D−f < r on X, for each x ∈ X and δ > 0 we can find a nondegenerate closed interval
I(x, δ) = [x − α, x] ⊂ U such that 0 < α < δ and f(x) − f(x − α) < rα. Then Γ = I(x, δ) : x ∈
X, δ > 0 is a Vitali cover for X. By Vitali’s lemma, we can find finitely many pairwise disjoint
intervals I1, . . . , Ik ∈ Γ such that µ∗[X \∪k
n=1 |In|] < ε.
MEASURE THEORY 13
Let V =∪k
n=1 int[In]. Then, V is open, V ⊂ U , and µ∗[X] − ε < µ∗[V ] ≤ µ∗[U ] < µ∗[X] + ε.
Let X ′ = V ∩X. Since D+f > s on X, and hence on X ′, for each y ∈ X ′ and δ > 0 we can find a
nondegenerate closed interval J(y, δ) = [y, y + β] ⊂ V (hence J(y, δ) ⊂ In for some n ∈ 1, . . . , k)
such that 0 < β < δ, and f(y+β)−f(y) > sβ. Then Γ = J(y, δ) : y ∈ X ′, δ > 0 is a Vitali cover
for X ′. Again by Vitali’s lemma, we can find finitely many pairwise disjoint intervals J1, . . . , Jm ∈ Γ
such that µ∗[X ′ \∪m
j=1 |Jj |] < ε. Then∑m
j=1 |Jj | ≥ µ∗[X ′]− ε ≥ µ∗[X]− 2ε.
Write In = [xn−αn, xn] and Jj = [yj , yj+βj ]. For each n ∈ 1, . . . , k, letDn = j ∈ 1, . . . ,m :
Jj ⊂ In. Then 1, . . . ,m is the disjoint union of Dn’s.
Note that∑
j∈Dn(f(yj + βj) − f(yj)) ≤ f(xn) − f(xn − αn) for each n ∈ 1, . . . , k since f
is increasing. Summing over n, we get∑m
j=1(f(yj + βj) − f(yj)) ≤∑k
n=1(f(xn) − f(xn − αn)),
and hence∑m
j=1 sβj <∑k
n=1 rαn, or s(∑m
j=1 |Jj |) < r(∑k
n=1 |In|). From the earlier estimates we
conclude that s(µ∗[X] − 2ε) < r(µ∗[X] + ε). Since ε > 0 was arbitrary and r < s, we must have
µ∗[X] = 0.
The conclusion of [112] can be extended to more general class of real functions.
Definition: If f : [a, b] → R is a function and P = a0 = a ≤ a1 ≤ · · · ≤ an−1 ≤ an = b is a
partition of [a, b], let V ba (f, P ) =
∑ni=1 |f(ai)− f(ai−1)|. Define the total variation of f as V b
a (f) =
supV ba (f, P ) : P is a partition of [a, b]. We say f is of bounded variation if V b
a (f) <∞. It is easy
to see that if f is of bounded variation, then f is bounded (∵ if x ∈ [a, b], take P = a ≤ x ≤ b to
see that |f(x)− f(a)| ≤ V ba (f)).
Remark: Imagine a particle moving along the graph of f from the point (a, f(a)) to the point
(b, f(b)) (the particle has to make a jump wherever there is a discontinuity in the graph). Loosely
speaking, f is of bounded variation iff the total distance travelled by the particle along the graph of
f is finite (∵ the total variation of f is just the vertical variation; but since the horizontal distance
b− a is finite, the above interpretation is justified).
Examples: (i) If f : [a, b] → R is a monotone function, then V ba (f, P ) = |f(b) − f(a)| for any
partition P of [a, b] and hence V ba (f) = |f(b) − f(a)| < ∞. So f is of bounded variation. (ii)
Suppose f : [a, b] → R is Lipschitz continuous (this happens if f is C1) with Lipschitz constant
λ > 0. Then, it may be seen that V ba (f) ≤ λ(b− a) <∞ and hence f is of bounded variation.
Non-example: A (uniformly) continuous function f : [a, b] → R need not be of bounded variation.
Let f : [0, 1] → R be the (uniformly) continuous function defined as f(0) = 0 and f(x) = x sin(1/x)
if x ∈ (0, 1]. Let ak = 2/kπ ∈ [0, 1] for k ∈ N. Observe that |f(a2k)−f(a2k−1)| = |0−a2k−1| = a2k−1.
Let m ∈ N and P = 0 ≤ a2m ≤ a2m−1 ≤ · · · ≤ a1 ≤ 1. Then V 10 (f, P ) ≥
∑mk=1 |f(a2k) −
14 T.K.SUBRAHMONIAN MOOTHATHU
f(a2k−1)| =∑m
k=1 a2k−1 = (2/π)∑m
k=1(2k − 1)−1 → ∞ as m→ ∞. Hence V 10 (f) = ∞, and thus f
is not of bounded variation. This example also shows that bounded ; bounded variation.
Remark: If f, g : [a, b] → R are of bounded variation, r, s ∈ R, and h : [a, b] → R is de-
fined as h = rf(x) + sg(x), then V ba (h) ≤ |r|V b
a (f) + |s|V ba (g) < ∞. Hence f : [a, b] → R :
f is of bounded variation is a real vector space (in fact, it is a normed space with the norm
∥f∥ = |f(a)|+ V ba (f)).
Exercise-15: If f, g : [a, b] → R are of bounded variation, then fg is of bounded variation. [Hint :
Let M > 0 be such that |f |, |g| ≤ M . Now, subtracting and adding the term f(ai)g(ai−1),
note that |(fg)(ai) − (fg)(ai−1)| ≤ |f(ai)||g(ai) − g(ai−1)| + |f(ai) − f(ai−1)||g(ai−1)| and hence
V ba (fg) ≤M(V b
a (f) + V ba (g)).]
Exercise-16: If f : [a, b] → R is a function and c ∈ [a, b], then V ba (f) = V c
a (f)+Vbc (f). [Hint : If P1 is
a partition of [a, c] and P2 is a partition of [c, b], then V ca (f, P1)+V
bc (f, P2) = V b
a (f, P1∪P2) ≤ V ba (f).
Conversely, if P is a partition of [a, b], first refine it by inserting c and then divide into partitions
P1 of [a, c] and P2 of [c, b]. Check that V ba (f, P ) ≤ V c
a (f, P1) + V bc (f, P2) ≤ V c
a (f) + V bc (f).]
[113] A function f : [a, b] → R is of bounded variation iff there exist monotone functions g, h :
[a, b] → R such that f(x) = g(x) − h(x) for every x ∈ [a, b]. Consequently, for any function
f : [a, b] → R of bounded variation, we have µ∗L,1[x ∈ [a, b] : f is not differentiable at x] = 0.
Proof. Suppose f = g−h, where g, h are monotone. Since g, h are of bounded variation, f is also of
bounded variation since the collection of functions of bounded variation on [a, b] is a vector space.
Conversely assume that f is of bounded variation and define g : [a, b] → R as g(x) = V xa (f). Then
g is monotone increasing. Let h = g − f , and consider points x < y in [a, b]. By Exercise-16, we
have g(y)− g(x) = V yx (f) ≥ |f(y)− f(x)| ≥ f(y)− f(x), and therefore h(y) ≥ h(x). Thus h is also
monotone increasing. Clearly, f = g − h.
The result below and related discussions are left as a reading assignment (see page 108 of H.L.
Royden, Real Analysis, 3rd Edn.).
[114] Let [a, b] be a compact interval. Then for a function f : [a, b] → R, we have the following
implications: f is Lipschitz continuous ⇒ f is absolutely continuous ⇒ f is of bounded variation.
Consequently, if f is either Lipschitz continuous or absolutely continuous, then µ∗L,1[Y ] = 0, where
Y = x ∈ [a, b] : f is not differentiable at x.
Remark: However, there is a limit to these type of results; there are continuous functions f :
[a, b] → R which are not differentiable at any point (see Theorem 7.18 of W. Rudin, Principles of
Mathematical Analysis).
MEASURE THEORY 15
Now we mention a characterization of Riemann integrable functions in terms of small sets. For
simplicity, we restrict ourselves to dimension one, even though the corresponding result is true
in higher dimensions as well. If f : [a, b] → R is a bounded function and if P = a0 = a ≤
a1 ≤ · · · ≤ an−1 ≤ an = b is a partition of [a, b], let Mi = supf(x) : ai−1 ≤ x ≤ ai and
mi = inff(x) : ai−1 ≤ x ≤ ai. The upper and lower Riemann sums with respect to the partition
P are defined as U(f, P ) =∑n
i=1Mi(ai − ai−1) and L(f, P ) =∑n
i=1mi(ai − ai−1). A bounded
function f : [a, b] → R is said to be Riemann integrable if for every ε > 0 there is a partition
P of [a, b] such that U(f, P ) − L(f, P ) < ε. The following characterization says that a Riemann
integrable function is not very different from a continuous function.
[115] Let f : [a, b] → R be a bounded function and let Y = x ∈ [a, b] : f is not continuous at x.
Then, f is Riemann integrable iff µ∗L,1[Y ] = 0.
Proof. Let ω(f, x) be the oscillation of f at x defined earlier.
⇒: Since Y =∪∞
k=1 Yk, where Yk = x ∈ [a, b] : ω(f, x) ≥ 1/k, it suffices to show µ∗L,1[Yk] = 0 for
every k ∈ N. Fix k ∈ N and let ε > 0. Let P = a0 = a ≤ a1 ≤ · · · ≤ an−1 ≤ an = b be a partition
of [a, b] with U(f, P ) − L(f, P ) < ε/2k. Let Ai = (ai−1, ai) and Γ = 1 ≤ i ≤ n : Yk ∩ Ai = ∅.
Note that Mi − mi ≥ 1/k for i ∈ Γ. Write Yk = Y ′k ∪ Y ′′
k , where Y ′k = Yk ∩ (
∪i∈ΓAi) and
Y ′′k = Yk∩a0, a1, . . . , an. We have ε/2k > U(f, P )−L(f, P ) ≥
∑i∈Γ(Mi−mi)|Ai| ≥ 1/k
∑i∈Γ |Ai|
and hence∑
i∈Γ |Ai| < ε/2. And since Y ′′k is a finite set, there are finitely many intervals B1, . . . , Bm
such that Y ′′k ⊂
∪mj=1Bj and
∑mj=1 |Bj | < ε/2. Thus Yk ⊂ [
∪i∈ΓAi] ∪ [
∪mj=1Bj ] and
∑i∈Γ |Ai| +∑m
j=1 |Bj | < ε. Since ε > 0 was arbitrary, µ∗L,1[Yk] = 0.
⇐: Let ε > 0 be given. We have to find a partition P of [a, b] such that U(f, P )−L(f, P ) < ε. Let
λ = sup|f(x)| : x ∈ [a, b] and let ε′ = ε/[2λ + 2(b − a)]. For each x ∈ [a, b] \ Y , choose an open
interval A(x) ⊂ R containing x such that |f(x)−f(z)| < ε′ for every z ∈ [a, b]∩A(x) by continuity.
Also choose countably many open intervals Bm such that Y ⊂∪∞
m=1Bm and∑∞
m=1 |Bm| < ε′.
Then A(x) : x ∈ [a, b] \ Y ∪ Bm : m ∈ N is an open cover for the compact set [a, b]. Extract a
finite subcover A(xj) : 1 ≤ j ≤ p∪Bm : 1 ≤ m ≤ q. The end points inside [a, b] of these finitely
many intervals determine a partition P = a0 = a ≤ a1 ≤ · · · ≤ an−1 ≤ an = b of [a, b]. Observe
that for each i ∈ 1, . . . , n, we have [ai−1, ai] ⊂ A(xj) for some j ∈ 1, . . . , p, or [ai−1, ai] ⊂ Bm for
some m ∈ 1, . . . , q. Let Γ = 1 ≤ i ≤ n : [ai−1, ai] ⊂ A(xj) for some j and Γ′ = 1, . . . , n \ Γ.
Note thatMi−mi ≤ 2ε′ if i ∈ Γ. Hence U(f, P )−L(f, P ) ≤∑
i∈Γ(Mi−mi)(ai−ai−1) +∑
i∈Γ′(Mi−
mi)(ai−ai−1) ≤ 2ε′∑
i∈Γ(ai−ai−1)+2λ∑
i∈Γ′(ai−ai−1) ≤ 2ε′∑n
i=1(ai−ai−1)+2λ∑q
m=1 |Bm| <
2ε′(b− a) + 2λε′ = ε.
16 T.K.SUBRAHMONIAN MOOTHATHU
A corollary of [115] is that any bounded function f : [a, b] → R with at most countably many
points of discontinuity (in particular, any continuous function) is Riemann integrable. The higher
dimensional generalization of [115] can be stated as follows.
[115′] Let A ⊂ Rd be a d-box, let f : A → R be a bounded function and let Y be the set
x ∈ A : f is not continuous at x. Then, f is Riemann integrable iff µ∗L,d[Y ] = 0.
Example: Let f : [0, 1] → R be f(0) = 0 and f(x) = sin (1/x) for x = 0. Even though the graph
of f has infinitely many ups and downs (in fact, f is not of bounded variation), f is Riemann
integrable since f is bounded and is discontinuous only at one point, namely 0.
Exercise-17: Let f : [a, b] → R be a bounded function. If f is either monotone or of bounded
variation, then f is Riemann integrable. [Hint : Use [110], [113], and [115].]
Exercise-18: If f, g; [a, b] → R are Riemann integrable, then maxf, g, minf, g, and af + bg
(where a, b ∈ R) are also Riemann integrable. [Hint : The set of discontinuities of maxf, g is
contained in x : f is not continuous at x ∪ x : g is not continuous at x.]
Definition: Let X be a set and A ⊂ X. The indicator function1 1A : X → R of the subset A is
defined as 1A(x) =
1, if x ∈ A,
0, if x ∈ X \A.
Example: We discuss an example that illustrates the main drawback of Riemann integration theory.
Write [0, 1] ∩ Q = r1, r2, . . ., let fn : [0, 1] → R be the indicator function of r1, . . . , rn, and let
f : [0, 1] → R be the indicator function of [0, 1] ∩ Q. We have 0 ≤ f1 ≤ f2 ≤ · · · ≤ f ≤ 1 and the
sequence (fn) converges to f pointwise. Each fn is Riemann integrable since fn is discontinuous
only at finitely many points. But f is discontinuous at every point of [0, 1], and the Lebesgue outer
measure of [0, 1] is positive. Hence f is not Riemann integrable by [115]. Thus even the pointwise
limit of a uniformly bounded, monotone sequence of Riemann integrable functions need not be
Riemann integrable.
Remark: Let f : [0, 1] → R be the indicator function of [0, 1] ∩Q. Since f is not continuous at any
point, it is not possible to realize f as the pointwise limit of a sequence of continuous functions
from [0, 1] to R, in view of [108].
Remark: Let (fn) be a sequence of continuous functions from [a, b] to R converging pointwise to
a function f : [a, b] → R, and let Y = x ∈ [a, b] : f is not continuous at x. From [108] we know
that Y is an Fσ set of first category in [a, b]. But Y can have positive outer Lebesgue measure by
1The indicator function of a set A is also called the characteristic function of A and is also denoted as χA.
MEASURE THEORY 17
[106]. Hence f may not be Riemann integrable. Thus even the pointwise limit of a sequence of
continuous functions may not be Riemann integrable (of course, we did not give an example).
Remark: Lebesgue integration theory is developed not just for the sake of making the indicator
function of [0, 1] ∩Q integrable. The limit theorems in Lebesgue’s theory allow us to integrate the
pointwise limit of a sequence of integrable functions, and to interchange limit and integration, under
very mild hypothesis. Moreover, the powerful tools in Lebesgue’s theory make many proofs simpler
(eg: the proof of the change of variable theorem in d-dimension), and provide us with new ways of
dealing with functions (eg: Lp spaces). Also, as we will see later, in Lebesgue’s theory we have a
more satisfactory version of the Fundamental Theorem of Calculus (describing differentiation and
integration as inverse operations of each other).
4. σ-algebras and measurable spaces
A d-box in Rd has a well-defined d-dimensional volume. We may ask whether it is possible to
define the notion of a d-dimensional volue for all subsets of Rd. Of course, we would like to have
consistency conditions such as monotonicity and countable additivity.
Question: Can we have a function µ : P(Rd) → [0,∞] such that
(i) µ[A] = V old(A) if A ⊂ Rd is a d-box,
(ii) [Monotonicity] µ[A] ≤ µ[B] for subsets A,B of Rd with A ⊂ B,
(iii) [Countable additivity] µ[∪∞
n=1An] =∑∞
n=1 µ[An] if An’s are pairwise disjoint subsets of Rd?
Remark: From [107] we know that the Lebesgue outer measure µ∗L,d does not satisfy countable
additivity. The key observation of Lebesgue’s theory is that µ∗L,d will satisfy all the three conditions
stated above if we restrict µ∗L,d to a slightly smaller collection A ⊂ P(Rd) by discarding some
pathological subsets of Rd. In order to describe the structure of this smaller collection A, it is
convenient to proceed in an abstract manner, which we do below.
Definition: Let X be a nonempty set. A collection A ⊂ P(X) of subsets of X is said to be a
σ-algebra on X if the following hold:
(i) ∅, X ∈ A.
(ii) A ∈ A ⇒ X \A ∈ A.
(iii) A1, A2, . . . ∈ A ⇒∪∞
n=1An ∈ A.
If A is a σ-algebra on X, then (X,A) is called a measurable space.
Example: ∅, X and P(X) are trivial examples of σ-algebras on any nonempty set X. The
following are some σ-algebras on Rd (verify):
A1 = A ⊂ Rd : A or Rd \A is countable,
18 T.K.SUBRAHMONIAN MOOTHATHU
A2 = A ⊂ Rd : A or Rd \A is of first category in Rd,
A3 = A ⊂ Rd : µ∗L,d[A] = 0 or µ∗L,d[Rd \A] = 0.
A4 = A ⊂ Rd : [0, 1]d ⊂ A or [0, 1]d ⊂ Rd \A.
Exercise-19: If A is a σ-algebra on a set X show that (i) A \ B,A∆B ∈ A if A,B ∈ A, (ii)∩∞n=1An ∈ A if A1, A2, . . . ∈ A.
Exercise-20: Let A = A ⊂ Rd : A is a countable (possibly finite or empty) union of d-boxes. Is
A a σ-algebra on Rd? [Hint : Let d = 1. Consider Q and R \Q, or the middle-third Cantor set and
its complement.]
Exercise-21: Are the following σ-algebras on Rd: A1 = A ⊂ Rd : A or Rd \A is open in Rd and
A2 = A ⊂ Rd : A or Rd \A is dense in Rd?
Exercise-22: If Aα’s are σ-algebras on a set X, then∩
αAα := A ⊂ X : A ∈ Aα for every α is
also a σ-algebra on X.
Definition: Let X be a nonempty set and C ⊂ P(X) be a collection of subsets of X. A σ-algebra A
on X is said to be generated by C if A is the smallest σ-algebra on X containing C. Here, A exists
and is unique since A is precisely the intersection of all σ-algebras on X containing C (note that
there is at least one σ-algebra on X containing C, namely P(X)).
Definition: Let X be a metric space. Then the σ-algebra on X generated by the collection of all
open subsets of X is called the Borel σ-algebra on X, and is denoted as B(X) (or just B, if X is
clear from the context). The subsets of X belonging to B(X) are called Borel subsets of X. For
example, open subsets, closed subsets, Gδ subsets and Fσ subsets of X are Borel subsets of X.
[116] [Characterizations of the Borel σ-algebra on Rd] Consider the following collections of subsets
of Rd:
C1 = A ⊂ Rd : A is closed,
C2 = A ⊂ Rd : A is compact,
C3 = A ⊂ Rd : A is closed d-box,
C4 = A ⊂ Rd : A is an open d-box,
C5 = A ⊂ Rd : A is a d-box,
C6 = A ⊂ Rd : A is an open ball,
C7 = f−1(W ) : f : Rd → R is continuous and W ⊂ R is open.
If Ai is the σ-algebra on Rd generated by Ci for 1 ≤ i ≤ 7, then Ai = B(Rd) for 1 ≤ i ≤ 7.
Proof. Clearly A3 ⊂ A2 ⊂ A1 = B(Rd). Since (a, b) =∪∞
n=n0[a+1/n, b− 1/n] (where n0 is chosen
so that a+1/n0 ≤ b− 1/n0), it follows that any open d-box is a countable union of closed d-boxes,
MEASURE THEORY 19
and therefore A4 ⊂ A3. Since [a, b] =∩∞
n=1(a − 1/n, b + 1/n), [a, b) =∩∞
n=1(a − 1/n, b), and
(a, b] =∩∞
n=1(a, b+1/n), we deduce that any d-box is a countable intersection of open d-boxes, and
hence A4 = A5. Since any open set in Rd can be written as a countable union of open d-boxes as
well as a countable union of open balls, we have A4 = A6 = B(Rd). Thus Ai = B(Rd) for 1 ≤ i ≤ 6.
By the definition of continuity, we have A7 ⊂ B(Rd). If U ⊂ Rd is an open set different from
Rd, let A = Rd \ U and define f : Rd → R as f(x) = dist(x,A) := inf∥x− a∥ : a ∈ A. Then f is
continuous, and A = f−1(0) because A is closed. Now, U = f−1(R \ 0) and R \ 0 is open in R.
Hence B(Rd) ⊂ A7, completing the proof.
Topological remarks: (i) If X is a separable metric space, then any base or subbase for the topology
of X will generate the Borel σ-algebra B(X). (ii) In the above characterization we used implicitly
the fact that Rd is second countable and locally compact. If a metric space X fails to be second
countable or locally compact, then the σ-algebra generated by all compact subsets of X will only
be a proper sub-collection of B(X). For example, try to figure out what happens for the spaces
(R, discrete metric) (which is not second countable), and (Q, Euclidean metric) (which is not
locally compact).
Exercise-23: Show that B(R) is generated by each of the following collections: (a,∞) : a ∈ R,
[a,∞) : a ∈ R, (−∞, b) : b ∈ R, (−∞, b] : b ∈ R, (a, b) : a < b and a, b ∈ Q.
Next our aim is to determine the cardinality of B(Rd). We need some set-theoretic preparation.
Definition: An order ≤ on a set X is a partial order if (i) x ≤ x for every x ∈ X, (ii) x ≤ y and
y ≤ x ⇒ x = y for every x, y ∈ X, (iii) x ≤ y and y ≤ z ⇒ x ≤ z for every x, y, z ∈ X. We say
(X,≤) is a totally ordered set if ≤ is a partial order and any two elements of X are comparable.
We say (X,≤) is a well-ordered set if (X,≤) is totally ordered and any nonempty subset Y ⊂ X
has a least element in Y .
Examples: (i) Let X be the collection of all nonempty subsets of N. Define an order ≤ on X as
A ≤ B iff the minimum of A is less than or equal to the minimum of B. Then this is not a partial
order since the second condition fails. (ii) If X is any nonempty set, then P(X) with inclusion as
order is partially ordered, but in general not totally ordered. (iii) R with the usual order is totally
ordered, but not well-ordered since the subset (0, 1) does not contain a least element. (iv) N with
the usual order is well-ordered.
Well-ordering principle (equivalent to the axiom of choice): Any nonempty set admits a well-
ordering.
Now we describe the construction of some ordinal numbers. Start with an uncountable set X
such that card(X) = card(R), and let ≤ be a well-ordering on X. Let θ denote the least element of
20 T.K.SUBRAHMONIAN MOOTHATHU
X. By adding one extra element to X if necessary, we may also assume that (X,≤) has a largest
element, say θ′. For each β ∈ X, let Lβ = α ∈ X : α < β be the left section of β in X. Let
Y = β ∈ X : Lβ is uncountable. Then Y = ∅ since θ′ ∈ Y . So Y has a least element, say Ω.
Then LΩ is uncountable, but Lβ is countable for every β < Ω. Here, Ω is called the first uncountable
ordinal, and each β ∈ LΩ is called a countable ordinal number since each β ∈ LΩ represents the
type of a countable well-ordered set through Lβ.
Fact: If A ⊂ LΩ is a nonempty countable set, then A has a least upper bound in LΩ. [Proof : If
B =∪
β∈A Lβ, then B is countable and hence LΩ \B = ∅. The least element of LΩ \B is the least
upper bound of A.]
If α ∈ LΩ, then the least element of the nonempty set β ∈ LΩ : α < β will be denoted as α+1.
Note that there are no elements between α and α + 1 in LΩ. On the other hand, given β ∈ LΩ,
there may or may not exist α ∈ LΩ such that α+ 1 = β. For example, if β ∈ LΩ is the least upper
bound of the countable set θ, θ+1, θ+2, . . . (where recall that θ is the least element of LΩ), then
there is no α ∈ X with α + 1 = β. We say β ∈ LΩ is a limit ordinal if there is no α ∈ LΩ with
α+ 1 = β.
Exercise-24: (i) If card(X) ≤ card(R), then card(XN) ≤ card(R). (ii) If card(J) ≤ card(R) and
card(Xβ) ≤ card(R) for each β ∈ J , then, card(∪
β∈J Xβ) ≤ card(R). [Hint : (i) AssumeX = (0, 1).
Define a one-one map f : (0, 1)N → (0, 1) as follows. If x = (xn) ∈ (0, 1)N and if xn = 0. xn,1xn,2 · · · ,
then f(x) = 0. x1,1x1,2x2,1x1,3x2,2x3,1 · · · . (ii) Let g : R → J and hβ : R → Xβ be surjections. Then
f : R2 →∪
β∈J Xβ defined as f(x, y) = hg(y)(x) is a surjection, and card(R2) = card(R).]
[117] card(B(Rd)) = card(R).
Proof. We will use transfinite induction (i.e., induction with respect to ordinal numbers) by using
LΩ described above. Recall that we denoted the least element of LΩ by the symbol θ. To start the
induction process, let Aθ = U ⊂ Rd : U is open. Let β ∈ LΩ and assume that we have defined Aα
for every α ∈ Lβ. If β is a limit ordinal, define Aβ =∪
α<β Aα. If β = α+ 1 for some α ∈ LΩ, let
A′α = A ⊂ Rd : Rd \A ∈ Aα, and Aβ = A ⊂ Rd : A is a countable union of members from Aα∪
A′α. This defines Aβ for every β ∈ LΩ. Finally, put A =
∪β<ΩAβ. From our construction, it is
clear that A ⊂ B(Rd).
We verify that A is a σ-algebra on Rd. It suffices to check only the third property. So consider
A1, A2, . . . ∈ A. Then there are β1, β2, . . . ∈ LΩ such that An ∈ Aβn for every n ∈ N. By the
Fact mentioned above, the countable set βn : n ∈ N has a least upper bound, say δ in LΩ. Then
An ∈ Aδ for every n ∈ N and hence∪∞
n=1An ∈ Aδ+1 ⊂ A. Thus A ⊂ B is a σ-algebra on Rd
containing all open subsets of Rd. Hence A = B(Rd).
MEASURE THEORY 21
Now, it suffices to show that card(A) = card(R). Since there is an open ball of radius 1 centered
at each point of Rd, we have card(A) ≥ card(Aθ) ≥ card(R). So it suffices to establish that
card(A) ≤ card(R). Since card(LΩ) = card(R) and A =∪
β<ΩAβ, it is enough to show that
card(Aβ) ≤ card(R) for each β ∈ LΩ in view of Exercise-24.
Let D be the collection of all open balls in Rd with rational radius and center in Qd. Then
D is countable, and any open set U ⊂ Rd can be written as a countable union of members of
D. Hence card(Aθ) ≤ card(DN) = card(R). Let β ∈ LΩ and suppose we have proved that
card(Aα) ≤ card(R) for every α < β. If β is a limit ordinal, then Aβ =∪
α<β Aα is a countable
union and hence card(Aβ) ≤ card(R). If there is α with α+1 = β, then any A ∈ Aβ can be written
as A =∪∞
n=1An with An ∈ Aα ∪ A′α. This gives a one-one map from Aβ into (Aα ∪ A′
α)N. Hence
card(Aβ) ≤ card((Aα ∪ A′α)
N) ≤ card(R), again by Exercise-24. This completes the proof.
[118] [Corollary] For any uncountable set Y ⊂ Rd, there is A ⊂ Y such that A is not a Borel subset
of Rd.
Proof. We have card(B(Rd)) = card(R) = card(Y ) < card(P(Y )).
Definition: Let N (Rd) = A ⊂ Rd : µ∗L,d[A] = 0. The members of N (Rd) are called Lebesgue null
sets. The σ-algebra L(Rd) on Rd generated by B(Rd) ∪ N (Rd) is called the Lebesgue σ-algebra on
Rd, and members of L(Rd) are called Lebesgue measurable subsets of Rd.
[119] card(N (Rd)) = card(L(Rd)) = card(P(Rd)) > card(R). Hence, N (Rd) * B(Rd) $ L(Rd).
Proof. Let K be the middle-third Cantor set. Then, for any subset A ⊂ K, we have µ∗L,1[A] ≤
µ∗L,1[K] = 0. So µ∗L,d[A] = 0 also. This shows that P(K) ⊂ N (Rd) ⊂ L(Rd). And card(P(K)) =
card(P(R)) since K is an uncountable subset of R.
[120] [Translation invariance] (i) A+ x ∈ B(Rd) for every A ∈ B(Rd) and x ∈ Rd.
(ii) A+ x ∈ N (Rd) for every A ∈ N (Rd) and x ∈ Rd.
(iii) A+ x ∈ L(Rd) for every A ∈ L(Rd) and x ∈ Rd.
Proof. First let us mention a general principle that will be used at many places. To establish that
the members of a certain σ-algebra D on a set X satisfies a certain property P , it suffices to do
the following: show that the collection A ⊂ X : A satisfies property P is a σ-algebra, and then
find a suitable collection C ⊂ P(X) generating D and show that every member of C satisfies the
property P .
Let A = A ⊂ Rd : A + x ∈ B(Rd) for every x ∈ Rd. It is easy to check that A is a σ-algebra
containing all d-boxes. And recall that the collection of all d-boxes generates B(Rd). This proves
22 T.K.SUBRAHMONIAN MOOTHATHU
(i). Next, statement (ii) is a consequence of the translation invariance property of the Lebesgue
outer measure, and (iii) follows from (i) and (ii) by applying the principle mentioned above.
We will give other characterizations of the Lebesgue measurable sets shortly, and we will also
show that L(Rd) = P(Rd).
5. Measures and outer measures
Definition: Let X be a nonempty set. A function µ∗ : P(X) → [0,∞] is called an outer measure if
the following are satisfied:
(i) µ∗[∅] = 0,
(ii) [Monotonicity] A ⊂ B ⊂ X ⇒ µ∗[A] ≤ µ∗[B],
(iii) [Countable subadditivity] µ∗[∪∞
n=1An] ≤∑∞
n=1 µ∗[An] for every A1, A2, . . . ⊂ X.
Example: µ∗L,d is an outer measure on Rd. But the outer Jordan content µ∗J,d is not an outer
measure on [0, 1]d since it does not satisfy countable subadditivity.
Remark: Note that we do not need the concept of a σ-algebra to define an outer measure.
Definition: Let (X,A) be a measurable space. A function µ : A → [0,∞] is called a measure on
(X,A) if the following are satisfied:
(i) µ[∅] = 0,
(ii) [Countable additivity] µ[∪∞
n=1An] =∑∞
n=1 µ[An] if A1, A2, . . . ∈ A are pairwise disjoint.
If µ is a measure on (X,A), the triple (X,A, µ) is called a measure space.
Remark: Note that countable additivity implies finite additivity (by taking An = ∅ for all large n),
and finite additivity implies monotonicity for a measure.
Definition: Let (X,A, µ) be a measure space. A set A ∈ A is said to be a µ-null set if µ[A] = 0.
We say (X,A, µ) is complete if any subset of any µ-null set also belongs to A. Sometimes we will
just say that µ is complete to mean (X,A, µ) is complete.
Examples: (i) Let (X,A) be a measurable space and let µ be the counting measure on (X,A)
defined as µ[A] = ∞ if A is infinite and µ[A] = card(A) if A is finite. Then (X,A, µ) is a complete
measure space. (ii) Let (X,A) be a measurable space, let x ∈ X, and define a measure µ on (X,A)
as µ[A] = 1 if x ∈ A, and µ[A] = 0 if x /∈ A. Here µ is called the dirac measure concentrated at x.
If (X,A) = (Rd,B(Rd)), then µ is not complete. (iii) Let µ1, µ2 be measures on a measurable space
(X,A) and let c1, c2 ∈ [0,∞). Then the linear combination µ := c1µ1 + c2µ2 is also a measure on
(X,A). If at least one of µ1, µ2 is complete and the corresponding ci = 0, then µ is also complete.
MEASURE THEORY 23
Exercise-25: [Completion of a measure space] Let (X,A, µ) be a measure space, let Nµ = B ⊂ X :
there is C ∈ A such that B ⊂ C and µ[C] = 0, and let A be the σ-algebra generated by A∪Nµ.
Let A1 = A ∪B : A ∈ A and B ∈ Nµ, A2 = A∆B : A ∈ A and B ∈ Nµ,
A3 = A \B : A ∈ A and B ∈ Nµ, and A4 = (A ∪B1) \B2 : A ∈ A and B1, B2 ∈ Nµ.
Show that Ai = A for 1 ≤ i ≤ 4 and µ has a natural extension µ to A. Here, (X, A, µ) is called the
completion of (X,A, µ). [Hint : Let A ∈ A, B ∈ Nµ, and let C ∈ A be so that B ⊂ C and µ[C] = 0.
If B′ = B \A, then B′ ∈ Nµ and A ∪B = A∆B′. If A′ = A \ C and B′′ = [A ∩ (C \B)] ∪ [B \A],
then A′ ∈ A, B′′ ∈ Nµ and A∆B = A′ ∪B′′. So A1 = A2. Similarly, A2 = A3 = A4. Next, use the
equalities of Ai’s to show Ai is a σ-algebra.]
Remark: We know that it is often advantageous to work with a complete metric space rather than
arbitrary metric spaces (to ensure, for example, that the limits of sequences do not go out of the
space). Similarly, we will see in our course that it is advantageous to work with a complete measure
space rather than an arbitrary measure space.
Now we would like to establish an important result which says that given an outer measure µ∗ on
a set X, there is a σ-algebra A on X such that the restriction of µ∗ to A is a complete measure. The
key idea is to take A as the collection of those subsets A of X for which the ‘boundary’ between A
and X \A is not very bad so that if an arbitrary subset E is divided into two pieces by intersecting
with A and X \ A, then the outer measures of the two pieces add up to give the outer measure of
E. Before proving the result, it is convenient to introduce another structure on a set X.
Definition: Let X be a nonempty set. A collection A ⊂ P(X) is an algebra on X if (i) ∅, X ∈ A,
(ii) A ∈ A ⇒ X \A ∈ A, (iii) A,B ∈ A ⇒ A ∪B ∈ A.
Remark: Let A be a collection of subsets of a set X satisfying the first two conditions in the
definition above. Then note that A satisfies condition (iii) iff A is closed under finite intersections.
Therefore, if A is an algebra, then sets of the form A0 \A1,∩n
j=0Aj , A0 \ [∪n
j=1Aj ], etc. belong to
A for every A0, A1, . . . , An ∈ A.
Example: Let X ⊂ Rd be a d-box and let A = A ⊂ X : A is a finite union of d-boxes. Then A
is an algebra on X (but not a σ-algebra on X).
Exercise-26: Let A be an algebra on a set X. If A is closed under countable union of disjoint
members, then A is a σ-algebra on X. [Hint : If A1, A2, . . . ∈ A, let B1 = A1 and Bn+1 =
An+1 \ [∪n
i=1Ai]. Then Bn’s are disjoint, and Bn ∈ A. So∪∞
n=1An =∪∞
n=1Bn ∈ A.]
We write Ac = X \A below.
24 T.K.SUBRAHMONIAN MOOTHATHU
[121] [Caratheodory’s restriction theorem] Let X be a nonempty set and µ∗ : P(X) → [0,∞] be an
outer measure. Let A = A ⊂ X : µ∗[E] = µ∗[A∩E]+µ∗[Ac∩E] for every E ⊂ X and µ = µ∗|A.
Then A is a σ-algebra on X and µ is a complete measure on (X,A).
Proof. Note that A can also be represented as A = A ⊂ X : µ∗[E] ≥ µ∗[A ∩ E] + µ∗[Ac ∩
E] for every E ⊂ X since µ∗ is subadditive.
Step-1 : First our aim is to show A is an algebra on X. Clearly, ∅, X ∈ A, and A is closed under
complementation. Now, let A,B ∈ A and note that (A ∩ B)c is the disjoint union of the three
sets Ac ∩ B,A ∩ Bc, Ac ∩ Bc. Hence for any E ⊂ X, we have µ∗[E] ≥ µ∗[B ∩ E] + µ∗[Bc ∩ E] =
µ∗[A∩B∩E]+µ∗[Ac∩B∩E]+µ∗[A∩Bc∩E]+µ∗[Ac∩Bc∩E] ≥ µ∗[(A∩B)∩E]+µ∗[(A∩B)c∩E]
by the subadditivity of µ∗. Hence A ∩B ∈ A for every A,B ∈ A. Thus A is an algebra on X.
Step-2 : We claim that if A,B ∈ A are disjoint, and E ⊂ X, then µ∗[(A ∪ B) ∩ E] = µ∗[A ∩ E] +
µ∗[B∩E]. This claim is true because µ∗[(A∪B)∩E] = µ∗[A∩[(A∪B)∩E]]+µ∗[Ac∩[(A∪B)∩E] =
µ∗[A ∩ E] + µ∗[B ∩ E]. Inductively, we have the following: if A1, . . . , Ak ∈ A are pairwise disjoint
and E ⊂ X, then µ∗[(∪k
n=1An) ∩ E] =∑k
n=1 µ∗[An ∩ E].
Step-3 : To show A is a σ-algebra on X, it is enough to check that A is closed under countable union
of pairwise disjoint members, in view of Exercise-26. Let A1, A2, . . . ∈ A be pairwise disjoint sets,
let A =∪∞
n=1An and Bk =∪k
n=1An. Note that Bk ∈ A since A is an algebra. For any E ⊂ X, we
have µ∗[E] = µ∗[Bk∩E]+µ∗[Bck∩E] ≥
∑kn=1 µ
∗[An∩E]+µ∗[Ac∩E] by Step-2 and by the inclusion
Ac ⊂ Bck. Letting k → ∞, we get µ∗[E] ≥
∑∞n=1 µ
∗[An ∩E] + µ∗[Ac ∩E] ≥ µ∗[A∩E] + µ∗[Ac ∩E]
by the countable subadditivity of µ∗. Thus A ∈ A and so A is a σ-algebra on X.
Step-4 : Let A1, A2, . . . ∈ A be pairwise disjoint and A =∪∞
n=1An. Then, taking E = A in the
argument of Step-3, we have µ∗[A] ≥∑∞
n=1 µ∗[An], and therefore µ∗[A] =
∑∞n=1 µ
∗[An] since µ∗ is
countably subadditive. This shows that µ∗ restricted to A is a measure.
Step-5 : Finally we show that the measure µ∗|A is complete. Let E ⊂ X, let A ∈ A be such that
µ[A] = 0 and let B ⊂ A. Then µ∗[B ∩ E] + µ∗[Bc ∩ E] ≤ µ∗[A ∩ E] + µ∗[E] = 0 + µ∗[E] = µ∗[E]
and hence B ∈ A.
6. Lebesgue measure on Rd
Since µ∗L,d is an outer measure on Rd, we can use [121] get a σ-algebra A on Rd such that the
restriction of µ∗L,d to A is a complete measure. We show below that A = L(Rd).
[122] [Characterization of Lebesgue measurable sets] Let µ∗ = µ∗L,d. Then, for a subset A ⊂ Rd,
the following are equivalent.
(i) A ∈ L(Rd), where L(Rd) is the σ-algebra generated by B(Rd) ∪N (Rd).
(ii) A satisfies Caratheodory’s condition, i.e., µ∗[E] = µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ Rd.
MEASURE THEORY 25
(iii) For every ε > 0, there is an open set U ⊂ Rd such that A ⊂ U and µ∗[U \A] < ε.
(iv) There is a Gδ set G ⊂ Rd such that A ⊂ G and µ∗[G \A] = 0.
(v) For every ε > 0, there is a closed set K ⊂ Rd such that K ⊂ A and µ∗[A \K] < ε.
(vi) There is an Fσ set F ⊂ Rd such that F ⊂ A and µ∗[A \ F ] = 0.
(vii) ∀ ε > 0, ∃ open U ⊂ Rd and closed K ⊂ Rd such that K ⊂ A ⊂ U and µ∗[U \K] < ε.
(viii) There exist a Gδ set G ⊂ Rd and an Fσ set F ⊂ Rd such that F ⊂ A ⊂ G and µ∗[G \ F ] = 0.
Proof. (i) ⇒ (ii): Let A = A ⊂ Rd : µ∗[E] = µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ Rd. Since µ∗
is an outer measure, A is a σ-algebra on Rd by [121]. Let C = A ⊂ Rd : A is a d-box ∪ N (Rd).
It is clear that C generates L(Rd). Hence it suffices to show C ⊂ A. If A ∈ N (Rd), then µ∗(E) ≥
µ∗(Ac ∩ E) = µ∗(A ∩ E) + µ∗(Ac ∩ E) for every E ⊂ Rd and hence A ∈ A. Thus N (Rd) ⊂ A. So
it suffices to show that every d-box A belongs to A.
Consider a d-box A and an arbitrary set E ⊂ Rd. We have to show µ∗[E] ≥ µ∗[A∩E]+µ∗[Ac∩E]
(∵ the other inequality is true by subadditivity). If µ∗[E] = ∞, we are done. So assume µ∗[E] <∞
and let ε > 0. Then there are d-boxes Bn such that E ⊂∪∞
n=1Bn and∑∞
n=1 µ∗[Bn] < µ∗[E]+ ε by
the definition. If n is such that Bn intersects both A and Ac, then we may partition Bn into finitely
many d-boxes so that each d-box of the partition is contained in either A or Ac (this is possible since
A is a d-box). After this sort of refinement, we may assume that each Bn is contained in either A or
Ac. Let Γ = n ∈ N : Bn ⊂ A and Γ′ = n ∈ N : Bn ⊂ Ac. Then A∩E ⊂∪
n∈ΓBn and Ac∩E ⊂∪n∈Γ′ Bn. Hence µ
∗[E]+ε >∑∞
n=1 µ∗[Bn] =
∑n∈Γ µ
∗[Bn]+∑
n∈Γ′ µ∗[Bn] ≥ µ∗[A∩E]+µ∗[Ac∩E].
Since ε > 0 is arbitrary, µ∗[E] ≥ µ∗[A ∩ E] + µ∗[Ac ∩ E] and thus A ∈ A.
(ii) ⇒ (iii): Let ε > 0. Let A = B ⊂ Rd : µ∗[E] = µ∗[B ∩E] + µ∗[Bc ∩E] for every E ⊂ Rd. We
know that A is a σ-algebra by [121]. From the argument above, we also know that all d-boxes belong
to A. Write Rd =∪∞
n=1Bn, where each Bn is a d-box, and let An = A ∩ Bn. Then, An ∈ A for
every n ∈ N (since A,Bn ∈ A and since A is a σ-algebra). Note that A =∪∞
n=1An and µ∗[An] <∞
for each n ∈ N. Choose an open set Un ⊂ Rd such that An ⊂ Un and µ∗[Un] < µ∗[An] + ε/2n by
[102](vii). Since An ∈ A, we have µ∗[Un] = µ∗[An ∩ Un] + µ∗[Acn ∩ Un] = µ∗[An] + µ∗[Un \ An].
Hence µ∗[Un \ An] < ε/2n. Put U =∪∞
n=1 Un. Then, U is open, A ⊂ U , and µ∗[U \ A] ≤∑∞n=1 µ
∗[Un \An] <∑∞
n=1 ε/2n = ε since U \A ⊂
∪∞n=1(Un \An) and µ
∗ is countably subadditive.
(iii) ⇒ (iv): Let Un ⊂ Rd be an open set such that A ⊂ Un and µ∗[Un\A] < 1/n. Let G =∩∞
n=1 Un.
Then A ⊂ G, and µ∗[G \A] ≤ µ∗[Un \A] < 1/n for every n ∈ N so that µ∗[G \A] = 0.
(iv) ⇒ (i): We have A = G \ (G \A), G ∈ B(Rd) and G \A ∈ N (Rd). Hence A ∈ L(Rd).
Thus (i)-(iv) are equivalent. The implications (i)+(iii) ⇒ (v) ⇒ (vi) ⇒ (i) can be proved
similarly. Hence statements (i)-(vi) are equivalent. It is easy to see that (iii)+(v) ⇔ (vii), and
(iv)+(vi) ⇔ (viii). This completes the proof.
26 T.K.SUBRAHMONIAN MOOTHATHU
Remark: For sets A ⊂ B and ε > 0, the statements µ∗[B \ A] ≤ ε and µ∗[B] ≤ µ∗[A] + ε are not
equivalent (problems: it can happen that µ∗[A] = ∞; even otherwise µ∗ is not finitely additive).
Definition: Let µL,d be the restriction of µ∗L,d to L(Rd). Then µL,d is a complete measure on the
measurable space (Rd,L(Rd)) by [121] and [122]. We call µL,d the d-dimensional Lebesgue measure.
Remark: (Rd,L(Rd), µL,d) is the completion of the measure space (Rd,B(Rd), µL,d). This can be
seen as follows. Let µ = µL,d and Nµ = A ⊂ Rd : A ⊂ B for some B ∈ B(Rd) with µ[B] = 0.
Then Nµ ⊂ N (Rd). To get the other inclusion, note that if A ∈ N (Rd), then by [122](iv) there is
a Gδ subset (hence a Borel subset) G ⊂ Rd such that A ⊂ G and µ[G] = 0. Thus Nµ = N (Rd).
Now apply Exercise-25. In particular, if A ∈ L(Rd), then there exist Bi ∈ B(Rd) and Yi ∈ N (Rd)
such that A = B1 ∪ Y1 = B2 \ Y2 = B3∆Y3 = (B4 ∪ Y4) \ Y5. But we have better representations
from [122]. If A ∈ L(Rd), then there exist Gδ set G ⊂ Rd, Fσ set F ⊂ Rd and Y1, Y2 ∈ N (Rd) such
that A = G \ Y1 = F ∪ Y2.
[123] [Uniqueness of Lebesgue measure] Let µ = µL,d and let µ′ be a measure on (Rd,L(Rd)) such
that µ′[A] = µ[A] for every d-box A. Then µ′ = µ.
Proof. If U ⊂ Rd is open, then U can be written as a countable union of pairwise disjoint d-boxes and
hence µ′[U ] = µ[U ]. If A ∈ L(Rd), then using [122] we get µ[A] = infµ[U ] : A ⊂ U and U open =
infµ′[U ] : A ⊂ U and U open ≥ µ′[A]. If A ∈ L(Rd) is a bounded set contained in a d-box B,
then µ[A] ≥ µ′[A] and µ[B \ A] ≥ µ′[B \ A] so that by additivity we obtain µ[A] = µ′[A]. For the
general case, note that any A ∈ L(Rd) can be written as a countable disjoint union A =∪∞
n=1An
where An ∈ L(Rd) are bounded.
Exercise-27: Let (X,A, µ) be a measure space and let An ∈ A for n ∈ N.
(i) If A1 ⊂ A2 ⊂ · · · , then µ[∪∞
n=1An] = limn→∞ µ[An].
(ii) If A1 ⊃ A2 ⊃ · · · , and if µ[An] <∞ for all large n ∈ N, then µ[∩∞
n=1An] = limn→∞ µ[An].
(iii) Give an example to say that the finiteness assumption ‘µ[An] <∞’ is necessary in (ii).
[Hint : (i) Let B1 = A1 and Bn+1 = An+1 \ An. Then Bn’s are pairwise disjoint and∪∞
n=1Bn =∪∞n=1An. Now use countable additivity. (ii) Assume µ[A1] < ∞ and apply the first part to the
sets Bn := A1 \An. (iii) Consider the Lebesgue measure on R and let An = (n,∞).]
[124] [Properties of Lebesgue measure] Let µ = µL,d. Then we have the following:
(i) µ[A] = V old(A) for every d-box A ⊂ Rd.
(ii) µ[K] <∞ for every compact set K ⊂ Rd.
(iii) [Translation invariance] µ[A+ x] = µ[A] for every A ∈ L(Rd) and x ∈ Rd.
(iv) ∀ A ∈ L(Rd) and ε > 0, ∃ open U ⊂ Rd and closed K ⊂ Rd with K ⊂ A ⊂ U and µ[U \K] < ε.
MEASURE THEORY 27
(v) ∀ A ∈ L(Rd), ∃ a Gδ set G ⊂ Rd and an Fσ set F ⊂ Rd such that F ⊂ A ⊂ G and µ[G\F ] = 0.
(vi) [Outer regularity] µ[A] = infµ[U ] : U ⊂ Rd is open and A ⊂ U for every A ∈ L(Rd).
(vii) [Inner regularity] µ[A] = supµ[K] : K ⊂ Rd is compact and K ⊂ A for every A ∈ L(Rd).
Proof. (i) is clear; (ii) follows from (i) since every compact subset is contained in some d-box; and
(iii) is a consequence of the translation invariance of µ∗L,d and [120]. Statements (iv) and (v) are
just taken from [121]; and (vi) is a consequence. So we just have to prove (vii).
Given A ∈ L(Rd), let An = A ∩ B(0, n). Let ε > 0 and let Kn ⊂ Rd be a closed set so that
Kn ⊂ An and µ[An \Kn] < ε/2n+1. Replacing Kn with∪n
i=1Ki if necessary, we may assume that
K1 ⊂ K2 ⊂ · · · . Note that Kn is compact since Kn is a closed subset of the compact set B(0, n).
If F =∪∞
n=1Kn (where F may not be compact), then we see that µ[A \ F ] ≤∑∞
n=1 ε/2n+1 = ε/2.
Note that µ[A] = µ[F ] + µ[A \ F ]. If µ[A] = ∞, then we must have µ[F ] = ∞, and hence
limn→∞ µ[Kn] = µ[F ] = ∞ = µ[A] by Exercise-27(i). If µ[A] < ∞, then µ[F ] + ε/2 ≥ µ[A], and
therefore µ[Kn] + ε > µ[A] for all large n ∈ N, again by Exercise-27(i).
Remark: Let X be a metric space. A measure µ defined on (X,B(X)) is said to be regular if
for every A ∈ B(X) we have µ[A] = infµ[U ] : U ⊂ X is open and A ⊂ U = supµ[K] : K ⊂
X is compact and K ⊂ A. The last two statements of the above result say that the Lebesgue
measure is regular. More generally, the following fact is true. Let X be a locally compact, second
countable metric space and let µ be a measure on B(X) having the property that µ[K] < ∞ for
every compact K ⊂ X. Then µ is regular.
Remark: Since B(Rd) ⊂ L(Rd), now we have a well-defined notion of d-dimensional volume for
each Borel subset of Rd. However, there are subsets of Rd for which the concept of a d-dimensional
volume is not well-defined, as the following result shows.
[125] [Existence of non-measurable set] There is a subset Y ⊂ Rd which is not Lebesgue measurable;
that is, L(Rd) = P(Rd).
Proof. We know that µ∗L,d is not finitely additive on P(Rd) by [107]. But the restriction of µ∗L,d to
L(Rd) is a measure and hence µ∗L,d is countably additive on L(Rd). Therefore, L(Rd) = P(Rd). In
fact, we may check that the Vitali set Y constructed in [107] does not belong to L(Rd).
There are other methods to produce sets which are not Lebesgue measurable.
Definition: A subset B ⊂ Rd is called a Bernstien set if for every uncountable closed set F ⊂ Rd,
we have B ∩ F = ∅ and Bc ∩ F = ∅.
[126] (i) There exists a Bernstien set B ⊂ Rd.
(ii) If B ⊂ Rd is a Bernstien set, then B /∈ L(Rd).
28 T.K.SUBRAHMONIAN MOOTHATHU
(iii) If A ∈ L(Rd) is with µL,d[A] > 0, then there is Y ⊂ A such that Y /∈ L(Rd).
Proof. Write µ = µL,d.
(i) Let F be the collection of all uncountable closed subsets of Rd. We have card(F) = card(R).
So we may index the members of F using the well-ordered set LΩ (where Ω is the first uncountable
ordinal) as F = Fβ : β ∈ LΩ. We will choose two distinct points xβ, yβ ∈ Fβ for each β ∈ LΩ
by transfinite induction. Let θ denote the least element of LΩ and let xθ, yθ ∈ Fθ be two distinct
points. Let β ∈ LΩ and assume that we have chosen points xα, yα ∈ Fα for every α < β. Then
xα : α < β ∪ yα : α < β is countable since Lβ is countable. Therefore, we may choose two
distinct points xβ, yβ from the uncountable set Fβ \ [xα : α < β ∪ yα : α < β]. This completes
the induction step. Finally put B = xβ : β ∈ LΩ. Then xβ ∈ B ∩ Fβ and yβ ∈ Bc ∩ Fβ for every
β ∈ LΩ so that B is a Bernstien set.
(ii) Let if possible B ∈ L(Rd). Then µ[B] > 0 or µ[Bc] > 0. Since both B and Bc are Bernstien
sets, we may assume µ[B] > 0. Then by [124], there is a closed set F ⊂ Rd such that F ⊂ B and
µ[F ] > 0. Now, F must be uncountable since µ[F ] > 0. This is a contradiction to the definition of
Bernstien set since Bc ∩ F = ∅.
(iii) If possible, let P(A) ⊂ L(Rd). Then, in particular, A ∩ B,A ∩ Bc ∈ L(Rd). So we must have
µ[A∩B] > 0 or µ[A∩Bc] > 0. This leads to a contradiction as in the proof of part (ii) since neither
A ∩B nor A ∩Bc contains an uncountable closed subset of Rd.
Remark: Thus for A ⊂ Rd, we observe that (i) P(A) ⊂ B(Rd) iff A is countable, and (ii) P(A) ⊂
L(Rd) iff A ∈ N (Rd). Below we discuss some more properties of the Lebesgue measure.
Exercise-28: Let A ⊂ L(Rd) be a collection of pairwise disjoint sets such that µL,d[A] > 0 for every
A ∈ A. Then A is countable. [Hint : For k, n ∈ N, let Ak,n = A ∈ A : µL,d[A ∩ B(0, n)] > 1/k.
Then Ak,n is finite and A =∪∞
k=1
∪∞n=1Ak,n.]
Exercise-29: Let µ = µL,d, and A ∈ L(Rd) be with µ[A] > 0. Then for every 0 < λ < 1, there
is an open d-box B ⊂ Rd such that µ[A ∩ B] > λµ[B]. [Hint : Replacing A with A ∩ B(0, r) for
some r > 0, assume µ[A] < ∞. Choose open d-boxes B1, B2, . . . such that A ⊂∪∞
n=1Bn and
λ∑∞
n=1 µ[Bn] < µ[A]. Since A =∪∞
n=1(A ∩ Bn), we get λ∑∞
n=1 µ[Bn] <∑∞
n=1 µ[A ∩ Bn], so
λµ[Bn] < µ[A ∩Bn] for some n ∈ N.]
[127] [Steinhaus’ theorem] If A ∈ L(Rd) is with µL,d[A] > 0, then A − A contains an open neigh-
borhood of 0 ∈ Rd, where A−A := x− y : x, y ∈ A.
Proof. Write µ = µL,d. Let 1/2 < λ < 1 and let B =∏d
j=1(aj , bj) be an open d-box such that
µ[A∩B] > λµ[B], chosen by Exercise-29. Replacing A with A∩B, we may assume that A ⊂ B and
µ[A] > λµ[B]. Let r > 0 be so that for C :=∏d
j=1(aj − r, bj + r) we have µ[C] < 2λµ[B]. We claim
MEASURE THEORY 29
that U :=∏d
j=1(−r, r) ⊂ A − A. To prove the claim, we need to show that A ∩ (A + z) = ∅ for
every z ∈ U . Note that for each z ∈ U , both A and A+ z are contained in C and µ[A]+µ[A+ z] =
2µ[A] > 2λµ[B] > µ[C].
Another proof is as follows. Since µ is inner regular, there is compact K ⊂ A with µ[K] > 0. By
the outer regularity of µ (applied to K), find an open neighborhood U of K with 2µ[K] > µ[U ].
Since the compact set K is contained in the open set U , there is δ > 0 such that B(x, δ) ⊂ U for
every x ∈ K. That is, K +B(0, δ) ⊂ U . If z ∈ B(0, δ), then both K and K + z are contained in U
and µ[K] + µ[K + z] = 2µ[K] > µ[U ]; hence K ∩ (K + z) = ∅. So B(0, δ) ⊂ K −K ⊂ A−A.
Exercise-30: Let µ = µL,d, and A ∈ L(Rd) be with µ[A] > 0. Then for every 0 < β < µ[A], there is
B ∈ L(Rd) such that B ⊂ A and µ[B] = β. [Hint : Let ε > 0 and write Rd =∪∞
n=1Cn, a pairwise
disjoint union of d-boxes with µ[Cn] < ε. Then there is k ∈ N such that β < µ[∪k
n=1(A∩Cn)] < β+ε.
Using this idea, inductively choose Lebesgue measurable sets B0 = A ⊃ B1 ⊃ B2 ⊃ · · · such that
β < µ[Bm] < β + 1/m and put B =∩∞
m=0Bm.]
Definition: Let µ = µL,d. If A ∈ L(Rd) and x ∈ Rd, define the Lebesgue density of A at x as
dens(A, x) = limr→0µ[A ∩B(x, r)]
µ[B(x, r)], if the limit exists.
[128] [Lebesgue density theorem] If A ∈ L(Rd), then there is Y ∈ N (Rd) such that dens(A, x) = 1
for every x ∈ A \ Y .
Proof. Write µ = µL,d. We may assume A is bounded (∵ consider A ∩ B(0, n) for each n ∈ N
separately). First suppose d = 1. Let (bn) be a sequence in (0, 1) converging to 1.
Let Yn = x ∈ A : for every δ > 0, there is r ∈ (0, δ) such that µ[A ∩ B(x, r)] < bnµ[B(x, r)]
and Y =∪∞
n=1 Yn. Clearly, dens(A, x) = 1 for every x ∈ A \ Y . Thus it suffices to show µ[Yn] = 0
for every n ∈ N. Fix n ∈ N and consider Yn. Given ε > 0, choose an open set U such that Yn ⊂ U
and µ[U ] < µ∗[Yn] + ε (at present we do not know whether Yn is measurable). For each x ∈ Yn
and δ > 0, there is an open interval I(x, δ) ⊂ U of the form (x − r, x + r) such that r < δ and
µ[A ∩ I(x, δ)] < bn|I(x, δ)|. So Γ = I(x, δ) : x ∈ Yn and δ > 0 is a Vitali cover for Yn. Hence
by Vitali’s covering lemma ([111]), there exist finitely pairwise disjoint intervals I1, . . . , Ik ∈ Γ
such that µ∗[Yn \∪k
j=1 Ij ] < ε. Hence µ∗[Yn] − ε < µ∗[Yn ∩ (∪k
j=1 Ij)] ≤∑k
j=1 µ∗[Yn ∩ Ij ] ≤∑k
j=1 µ∗[A ∩ Ij ] < bn
∑kj=1 |Ij | < bnµ[U ] ≤ bn(µ
∗[Yn] + ε), and hence µ∗[Yn] < (1 + bn)ε/(1 − bn).
Since ε > 0 is arbitrary, µ∗[Yn] = 0, or µ[Yn] = 0.
About the general case: Vitali’s covering lemma (which we stated as result [111] for d = 1) can
be generalized to dimension d by replacing intervals with balls, and then the above proof can be
imitated to establish Lebesgue density theorem in dimension d.
30 T.K.SUBRAHMONIAN MOOTHATHU
Remark: The above result says that if A ∈ L(Rd), then for almost every x ∈ Rd, either A or Ac
has full density at x.
Reading assignment: [See Theorem 7.7 of W. Rudin, Real and Complex Analysis, 3rd Edn.] Let
µ = µL,d and f ∈ L1(µ). Then µ-almost every x ∈ Rd is a Lebesgue point of of f in the sense that
limr→01
µ[B(x, r)]
∫B(x,r) |f(y)− f(x)|dµ(y) = 0.
7. Pre-measure → outer measure → measure
Definition: Let (X,A, µ) be a measure space. We say (i) µ is a probability measure if µ[X] = 1,
(ii) µ is a finite measure if µ[X] <∞, and (iii) µ is a σ-finite measure if there exist X1, X2, . . . ∈ A
such that X =∪∞
n=1Xn and µ[Xn] < ∞ for every n ∈ N. For example, µL,d restricted to
([0, 1]d,B([0, 1]d)) is a probability measure, µL,d restricted to ([0, 2]d,B([0, 2]d)) is a finite measure
(but not a probability measure), and µL,d on to (Rd,L(Rd)) is a σ-finite measure (but not a finite
measure). If X is an uncountable set, then the counting measure on (X,P(X)) is not σ-finite.
The following definition is motivated by considering the collection consisting of the empty set
and all bounded or unbounded intervals in R.
Definition: Let X be a nonempty set. A collection A ⊂ P(X) is called a semi-algebra on X if the
following are true:
(i) ∅ ∈ A.
(ii) A,B ∈ A ⇒ A ∩B ∈ A.
(iii) A ∈ A ⇒ X \A is a finite union of pairwise disjoint members of A.
Remark: Note that every algebra is a semi-algebra.
Definition: Let A be a semi-algebra on a nonempty set X. A function µ : A → [0,∞] is a
pre-measure if
(i) µ[∅] = 0,
(ii) [Countable additivity] IfA1, A2, . . . ∈ A are pairwise disjoint and∪∞
n=1An ∈ A, then µ[∪∞
n=1An] =∑∞n=1 µ[An].
Remark: The concepts of probability pre-measure, finite pre-measure, and σ-finite pre-measure can
be defined in the natural manner. In particular, if A is a (semi-)algebra on X, then a pre-measure
µ on A is said to be σ-finite if there are X1, X2, . . . ∈ A such that X =∪∞
n=1Xn and µ[Xn] < ∞
for every n ∈ N.
Exercise-31: Let A be a semi-algebra on a nonempty set X and let A′ be the collection of all
finite unions of pairwise disjoint members of A. Show that A′ is the algebra generated by A (i.e.,
the smallest algebra containing A). [Hint : If A1, . . . , An ∈ A are disjoint and B1, . . . , Bm ∈ A
MEASURE THEORY 31
are disjoint, then (∪n
i=1Ai) ∩ (∪m
j=1Bj) =∪n
i=1
∪mj=1(Ai ∩ Bj) ∈ A′. So A′ is closed under finite
intersections. If Aci =
∪nij=1Aij is a disjoint union, then Ac
i ∈ A′ and so (∪n
i=1Ai)c =
∩ni=1A
ci ∈ A′.]
We will show that a pre-measure on a semi-algebra can be extended first to a pre-measure on an
algebra, and next to an outer measure on P(X). Then we can obtain a measure by [121].
[129] [Outer measure from pre-measure] (i) Let A be a semi-algebra on a nonempty set X, let µ
be a pre-measure on A and let A′ be the algebra generated by A. Then A′ is an algebra on X and
µ extends uniquely to a pre-measure on A′.
(ii) Let A′ be an algebra on a nonempty set X, and let µ : A′ → [0,∞] be a pre-measure. Define
µ∗ : P(X) → [0,∞] as µ∗[Y ] = inf∑∞
n=1 µ[An] : An ∈ A′ and Y ⊂∪∞
n=1An. Then µ∗ is an outer
measure on X, and µ∗[A] = µ[A] for every A ∈ A′.
Proof. (i) If A ∈ A′ and if A =∪k
n=1An is any disjoint union with An ∈ A, define µ[A] =∑kn=1 µ[A]. If A =
∪mj=1Bj is another disjoint union with Bj ∈ A, then A =
∪kn=i
∪mj=1(An ∩ Bj)
and Bj =∪k
n=1(An∩Bj) are also a disjoint unions. Hence∑k
n=1 µ[An] = µ[A] =∑k
n=1
∑mj=1 µ[An∩
Bj ] =∑m
j=1[∑k
n=1 µ[An∩Bj ]] =∑m
j=1 µ[Bj ], and thus µ is well-defined on A′. Clearly, µ is finitely
additive on A′.
Now consider pairwise disjoint sets A1, A2, . . . ∈ A′ with A :=∪∞
n=1An ∈ A′. Write An =∪kni=1An,i, a disjoint union with An,i ∈ A. Also choose disjoint members B1, . . . , Bm ∈ A with
A =∪m
j=1Bj . Then Bj =∪∞
n=1
∪kni=1(An,i ∩Bj) is a disjoint union of members of A. Hence µ[A] =∑m
j=1 µ[Bj ] =∑m
j=1
∑∞n=1
∑kni=1 µ[An,i ∩Bj ] =
∑∞n=1(
∑mj=1
∑kni=1 µ[An,i ∩Bj ]) =
∑∞n=1 µ[An].
The extension is unique since a pre-measure on an algebra has to be finitely additive.
(ii) Clearly, µ∗ satisfies monotonicity and µ∗[∅] = 0. To establish countable subadditivity, imitate
the proof of [102](iv). Now consider A ∈ A′. By definition, µ∗[A] ≤ µ[A]. If An ∈ A′ are such that
A ⊂∪∞
n=1An, let B1 = A1 and Bn+1 = An+1\ [∪n
i=1Ai]; then A∩Bn ∈ A′ are pairwise disjoint and
A =∪∞
n=1(A ∩Bn) so that µ[A] =∑∞
n=1 µ[A ∩Bn] ≤∑∞
n=1 µ[An]. This implies µ[A] ≤ µ∗[A].
The following result is useful in establishing the uniqueness of a measure. The proof below is
essentially that of the π-λ theorem or monotone class theorem from the literature.
[130] [Dynkin’s theorem] Let A be a semi-algebra on a nonempty set X, and let µ be a σ-finite
pre-measure on A. Let A be the σ-algebra generated by A and let µ1, µ2 be two measures on (X, A)
such that µ1[A] = µ[A] = µ2[A] for every A ∈ A. Then µ1 = µ2.
Proof. If A′ is the algebra generated by A and if µ is the unique extended pre-measure on A′ given
by [129], then it is easily seen that µ1[A] = µ[A] = µ2[A] for every A ∈ A′. Therefore, we may as
well assume that A itself is an algebra.
32 T.K.SUBRAHMONIAN MOOTHATHU
Case-1: µ[X] < ∞. Let Aeq = A ∈ A : µ1[A] = µ2[A]. We have to show Aeq = A. Let Γ be the
collection of all C ⊂ P(X) such that
(i) A ⊂ C ⊂ A.
(ii) If B,C ∈ C and B ⊂ C, then C \B ∈ C.
(iii) If C1, C2, . . . ∈ C are pairwise disjoint, then∪∞
n=1Cn ∈ C.
Observe that that Aeq ∈ Γ (∵ for (ii), note that µ1[C \ B] = µ1[C] − µ1[B] = µ2[C] − µ2[B] =
µ2[C \B]; here we use the assumption µ[X] <∞). Let C0 be the intersection of all C ∈ Γ. Clearly
C0 ∈ Γ and thus A ⊂ C0 ⊂ Aeq ⊂ A. We will show below that C0 is a σ-algebra. If this is done,
then we will get C0 = A, and hence we will have Aeq = A. Since X ∈ A ⊂ C0, we see from (ii) that
C0 is closed under complementation. In view of (i) and (iii), to show that C0 is a σ-algebra, now it
suffices to show that C0 is closed under finite intersections. We do this in two steps.
Let C1 = C ∈ C0 : A∩C ∈ C0 for every A ∈ A. We wish to check that C1 ∈ Γ. Clearly C1 ⊂ A,
and we have A ⊂ C1 since A is an algebra. The conditions (ii) and (iii) are verified as follows.
If B,C ∈ C1 are with B ⊂ C and A ∈ A, then A ∩ (C \ B) = A ∩ (C ∩ Bc) = (A ∩ C) ∩ Bc =
(A ∩ C) ∩ (Ac ∪ Bc) = (A ∩ C) ∩ (A ∩ B)c = (A ∩ C) \ (A ∩ B) ∈ C0 since A ∩ C,A ∩ B ∈ C0and A ∩ B ⊂ A ∩ C. Hence C \ B ∈ C1. If C1, C2, . . . ∈ C1 are pairwise disjoint and A ∈ A, then
A ∩ (∪∞
n=1Cn) =∪∞
n=1(A ∩Cn) ∈ C0 since A ∩C1, A ∩C2, . . . are pairwise disjoint members of C0.
Hence∪∞
n=1Cn ∈ C1. Thus C1 ∈ Γ and therefore C1 = C0 by the minimality of C0.
Let C2 = C ∈ C0 : D ∩ C ∈ C0 for every D ∈ C0. We may verify in a similar manner that
C2 ∈ Γ (we have A ⊂ C2 by what is proved above). Then C2 = C0. So C0 is closed under finite
intersections.
Case-2: There are En ∈ A with µ[En] <∞ such that X =∪∞
n=1En. We may assume En ⊂ En+1.
Let An = A ∩ En : A ∈ A and An = A ∩ En : A ∈ An. Then An is an algebra on En, and
An is the σ-algebra on En generated by A. Applying the first part of the proof, we have that
µ1 = µ2 on each An. Now if A ∈ A, then A is the increasing union A =∪∞
n=1(A ∩ En) so that
µ1[A] = limn→∞ µ1[A ∩ En] = limn→∞ µ2[A ∩ En] = µ2[A].
Example: Let X = (0, 1] and A = ∅ ∪ (a, b] : 0 ≤ a < b ≤ 1. Then A is a semi-algebra on X
and B(X) is the σ-algebra generated by A. If µ1 is the counting measure on (X,B(X)) and µ2 is
the measure defined as µ2[A] = ∞ if A = ∅ and µ[∅] = 0, then µ1 = µ2. But the two measures
agree on A. This example shows that the assumption of σ-finiteness is necessary in [130].
[131] [Caratheodory’s extension theorem] Let A be a semi-algebra on a nonempty set X, let
µ : A → [0,∞] be a pre-measure, and let A be the σ-algebra on X generated by A. Then there is
a measure µ : A → [0,∞] such that µ|A = µ. If µ is σ-finite, then µ is unique.
MEASURE THEORY 33
Proof. The uniqueness part follows from [130]. We only have to establish the existence part. By
[129](i), we may assume that A is an algebra on X, and let µ∗ be the outer measure on X produced
from µ by [129](ii). Then by [121], there is a σ-algebra A∗ on X such that the restriction of
µ∗ to A∗ is a (complete) measure. For our purpose, it suffices to show that the algebra A is
contained in A∗ (then A ⊂ A∗ also). Let A ∈ A. To show A ∈ A∗, we need to establish that
µ∗[E] ≥ µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ X.
We may assume µ∗[E] <∞ (otherwise the inequality is trivial). Given ε > 0, choose A1, A2, . . . ∈
A such that E ⊂∪∞
n=1An and∑∞
n=1 µ[An] < µ∗[E] + ε. Note that A ∩ E ⊂∪∞
n=1(A ∩ An),
Ac∩E ⊂∪∞
n=1(Ac∩An), and A∩An, A
c∩An ∈ A since A is an algebra. µ∗[E]+ε >∑∞
n=1 µ[An] =∑∞n=1 µ[A∩An]+
∑∞n=1 µ[A
c∩An] ≥ µ∗[A∩E]+µ∗[Ac∩E]. As ε > 0 is arbitrary, we are done.
Example: We say A ⊂ Rd is a generalized d-box if A =∏d
j=1 Ij , where each Ij ⊂ R is a bounded
or unbounded interval. For example, (−∞, 3] × (1, 5] × (0,∞) is a generalized 3-box. Note that
every generalized d-box A has a well-defined d-dimensional volume, V old(A) =∏d
j=1 |Ij |. If A =
∅ ∪ A ⊂ Rd : A is a generalized d-box, then A is a semi-algebra on Rd. Define µ : A → [0,∞]
as µ[A] = V old(A). To check countable additivity, it is enough to work with d-boxes. Let A be a
d-box and suppose A =∪∞
n=1An is a pairwise disjoint union of d-boxes. Since µ[A] ≥∑k
n=1 µ[An]
for every k ∈ N by finite additivity, we get µ[A] ≥∑∞
n=1 µ[An]. For the other inequality, use the
proof of [101]. Thus µ is a pre-measure on A. Since the σ-algebra generated by A is B(Rd), by
[131] we obtain a measure µ on (Rd,B(Rd)) such that µ[A] = V old(A) for every generalized d-box
A. It follows by [130] that µ = µL,d. We could have constructed the Lebesgue measure on the Borel
σ-algebra in this way also (and then we could have done a completion process, as in Exercise-25).
To give the next example, we describe the space first. Let m ≥ 2 be an integer and let X =
1, . . . ,mN. Give the finite set 1, . . . ,m the discrete topology, and give X the product topology.
If x, y ∈ X, define ρ(x, y) = 0 if x = y, and ρ(x, y) = 2−n if n ∈ N is the smallest such that xn = yn.
It may be checked that ρ is a metric on the set X. If w = w1 · · ·wn ∈ 1, . . . ,mn, then w is
called a word of length n, and we write |w| = n. If w ∈ 1, . . . ,mn, let Cw = x = (xn)∞n=1 ∈ X :
x1 · · ·xn = w, and we call Cw a cylinder set.
Exercise-32: Let X = 1, . . . ,mN be the space given above. Then,
(i) The metric ρ induces the product topology on X.
(ii) Each cylinder set Cw is both open and closed in X.
(iii) If A ⊂ X is a finite union of cylinder sets, then A can also be written as A =∪k
j=1Cw(j),
where the cylinder sets Cw(j)’s are pairwise disjoint and |w(1)| = · · · = |w(k)|.
(iv) Cw : n ∈ N and w ∈ 1, . . . ,mn is a base for the topology on X.
(v) (X, ρ) is compact, totally disconnected, and is without isolated points.
34 T.K.SUBRAHMONIAN MOOTHATHU
Remark: With more topological arguments, it can be shown that 1, . . . ,mN is homeomorphic to
the middle-third Cantor set. Hence 1, . . . ,mN is called a Cantor space.
[132] [Uniform Bernoulli measure on the Cantor space] Let m ≥ 2 be an integer and let X =
1, . . . ,mN. Then there is a unique probability measure µ on B(X) such that µ[Cw] = m−|w| for
every cylinder set Cw ⊂ X.
Proof. Let A = ∅ ∪ Cw : n ∈ N and w ∈ 1, . . . ,mn. Check that A is a semi-algebra on
X. Note that the σ-algebra on X generated by A is B(X) because of Exercise-32(iv). Define
µ : A → [0,∞] as µ[∅] = 0 and µ[Cw] = m−|w|. Clearly µ is finitely additive on A. If A ∈ A is
A =∪∞
n=1An with An ∈ A, then there is k ∈ N such that A =∪k
n=1An since the members of
A are both compact and open. So countable additivity of µ is automatically satisfied. Thus µ is
a pre-measure, and by [131], µ extends to a measure µ on B(X). We have µ[X] =∑m
j=1 µ[Cj ] =∑mj=1 µ[Cj ] =
∑mj=1m
−1 = 1, and hence µ is unique by [130].
Similar argument gives many other probability measures on the Cantor space as stated below.
[132′] [Bernoulli measures on the Cantor space] Let m ≥ 2 be an integer, X = 1, . . . ,mN, and let
p1, . . . , pm be nonnegative reals such that∑m
j=1 pj = 1. Then there is a unique probability measure
µ on B(X) with the property that for every n ∈ N and w = w1 · · ·wn ∈ 1, . . . ,mn, we have
µ[Cw] = pw1 · · · pwn .
Definition: A measure µ on (R,B(R)) such that µ[I] < ∞ for any bounded interval (i.e., for any
1-box) is called a Lebesgue-Stieltjes measure. We characterize them below by Exercise-33 and [133].
Exercise-33: Suppose µ is a Lebesgue-Stieltjes measure on (R,B(R)), and define a function fµ :
R → R as fµ(x) = µ[(0, x]] if x > 0, fµ(0) = 0, and fµ(x) = −µ[(x, 0]] if x < 0. Then, fµ
is increasing, continuous from the right, and fµ(b) − fµ(a) = µ[(a, b]]. [Hint : for example, if
x > 0 and if (xn) is a decreasing real sequence converging to x, then (0, x] =∩∞
n=1(0, xn] so that
fµ(x) = µ[(0, x]] = limn→∞ µ[(0, xn]] = limn→∞ fµ(xn) by Exercise-27].
[133] [Lebesgue-Stieltjes measures] Let f : R → R be increasing and right-continuous. Then there
is a unique σ-finite measure µ on (R,B(R)) such that µ[(a, b]] = f(b)− f(a) for reals a < b.
Proof. Let A = ∅∪(a, b] : −∞ ≤ a < b <∞∪(a,∞) : −∞ ≤ a <∞. It may be verified that
A is a semi-algebra on R and the σ-algebra generated by A is B(R). Let f(−∞) = limx−∞ f(x)
and f(∞) = limx∞ f(x). Define µ : A → [0,∞] as µ[∅] = 0, µ[(a, b]] = f(b) − f(a), and
µ[(a,∞)] = f(∞) − f(a). Since f is increasing, we observe the following for any A ∈ A: (i) If
A1, . . . , Ak ∈ A are such that A ⊂∪k
n=1An, then µ[A] ≤∑k
n=1 µ[An]. (ii) If A1, . . . , Ak ∈ A are
pairwise disjoint and∪k
n=1Ak ⊂ A, then∑k
n=1 µ[An] ≤ µ[A].
MEASURE THEORY 35
From (i) and (ii), µ is finitely additive on A. Our aim is to show µ is a pre-measure on A.
Let A1, A2, . . . ∈ A be pairwise disjoint members such that A :=∪∞
n=1An ∈ A. We have µ[A] ≥
µ[∪k
n=1An] =∑k
n=1 µ[An] for every k ∈ N by (ii), and therefore µ[A] ≥∑∞
n=1 µ[An]. Now we need
to prove the other inequality.
Case-1: A = (a, b] where a ∈ R. Then necessarily An has the form (an, bn] for every n ∈ N.
Consider ε > 0. Using the right-continuity of f , we may find x ∈ (a, b) and cn > bn such that
f(x) − f(a) < ε/2 and f(cn) − f(bn) < ε/2n+1. Since [x, b] ⊂∪∞
n=1(an, cn), by the compactness
of [x, b] there is k ∈ N such that [x, b] ⊂∪k
n=1(an, cn) and hence (x, b] ⊂∪k
n=1(an, cn]. Then by
(i), µ[(x, b]] ≤∑k
n=1 µ[(an, cn]] =∑k
n=1(f(cn) − f(an)) =∑k
n=1(f(cn) − f(bn)) +∑k
n=1(f(bn) −
f(an)) <∑k
n=1 ε/2n+1 +
∑kn=1(f(bn)− f(an)) ≤ ε/2 +
∑∞n=1 µ[An]. Therefore, µ[A] = µ[(a, b]] =
µ[(a, x]] + µ[(x, b]] = (f(x) − f(a)) + µ[(x, b]] < ε/2 + ε/2 +∑∞
n=1 µ[An]. Since ε > 0 is arbitrary,
we get µ[A] ≤∑∞
n=1 µ[An].
Case-2: A = (−∞, b]. Let (xk) be a decreasing sequence in A converging to−∞ and letXk = (xk, b].
We have f(−∞) = limk→∞ f(xk), and therefore µ[A] = f(b)− f(−∞) = limk→∞(f(b)− f(xk)) =
limk→∞ µ[Xk]. Since Xk =∪∞
n=1(Xk ∩ An) is a disjoint union of members of A, by Case-1 we get
µ[Xk] ≤∑∞
n=1 µ[Xk ∩An] ≤∑∞
n=1 µ[An] for each k ∈ N. Hence µ[A] ≤∑∞
n=1 µ[An].
Case-3: Let A = (a,∞). Let (yk) be an increasing sequence in A converging to ∞ and let Yk =
(a, yk]. As in Case-2, we have µ[A] = limk→∞ µ[Yk], and µ[Yk] ≤∑∞
n=1 µ[Yk ∩ An] ≤∑∞
n=1 µ[An]
for each k ∈ N. Hence µ[A] ≤∑∞
n=1 µ[An].
Thus µ is a pre-measure on A. Since R =∪
n∈Z(n, n+1] and µ[(n, n+1]] = f(n+1)−f(n) <∞, µ
is a σ-finite pre-measure. Hence by [131], µ extends uniquely to a σ-finite measure on (R,B(R)).
Remark: In the above, µ[x] = 0 iff f is continuous at x.
Examples: In [133], we have the following: (i) If f = IR, then µ = µL,1 on B(R). (ii) If y ∈ R
and f = 1[y,∞), then µ is the dirac probability measure concentrated at the point y. That is,
µ[A] = 1 if y ∈ A, and µ[A] = 0 if y /∈ A. (iii) If f is the integer-part function, that is, f(x) = n if
x ∈ [n, n+1) for n ∈ Z, then µ[A] = |A∩Z| for any A ∈ B(R). Here note that µ[(0, 1)] < µ[(0, 1]] <
µ[[0, 1]], and the completion of B(R) with respect to µ is P(R) and not L(R) (∵ if A ⊂ R, then
A = (A ∩ Z) ∪ (A \ Z) ∈ B(R) ∪Nµ).
Remark: If µ is a Lebesgue-Stieltjes measure, then clearly µ[K] < ∞ for every compact K ⊂ R,
and therefore µ is regular by an earlier remark.
[134] [Approximation theorem for sets of finite measure] Let (X,A, µ) be a σ-finite measure space
and let A′ be a semi-algebra generating A. Then for any A ∈ A with µ[A] < ∞ and ε > 0, there
exist A1, . . . , Ak ∈ A′ such that µ[A∆(∪k
n=1An)] < ε.
36 T.K.SUBRAHMONIAN MOOTHATHU
Proof. Define an outer measure µ∗ using the pre-measure µ|A′ as in [129], and then get a measure µ
on A by restricting µ∗ to A as in the the proof of [131]. By σ-finiteness, we must have µ = µ because
of Dynkin’s theorem. Thus µ∗ = µ on A. So there are A1, A2, . . . ∈ A′ such that A ⊂∪∞
n=1An and∑∞n=1 µ[An] < µ[A] + ε/2, and therefore µ[(
∪∞n=1An) \A] < ε/2. Then for all large k ∈ N, we have
µ[A∆(∪k
n=1An)] ≤ µ[(∪∞
n=1An) \A] +∑∞
n=k+1 µ[An] < ε/2 + ε/2 = ε.
Remark: Using the above result, we may deduce the following. (i) If µ is a Lebesgue-Stieltjes
measure and A ∈ B(R) is with µ[A] < ∞, then for every ε > 0 there are finitely many intervals
I1, . . . , Ik ∈ B(R) such that µ[A∆(∪k
n=1 In)] < ε. (ii) If A ∈ L(Rd) is with µL,d[A] <∞ and ε > 0,
then there are finitely many d-boxes A1, . . . , Ak ⊂ Rd such that µL,d[A∆(∪k
n=1An)] < ε. (iii) If
X is the Cantor space X = 1, . . . ,mN (m ≥ 2) and µ is a Bernoulli measure on (X,B(X)),
then for any A ∈ B(X) and ε > 0, there are finitely many cylinder sets A1, . . . , Ak ⊂ X such that
µ[A∆(∪k
n=1An)] < ε.
Example: We give an example to say σ-finiteness is necessary in [134]. Let A′ = ∅ ∪ (a, b] : 0 ≤
a < b ≤ 1. Then the semi-algebra A′ generates B((0, 1]). If µ is the counting measure on B((0, 1])
and if A ∈ B((0, 1]) is any nonempty finite set, then it is not possible to approximate A with a
finite union∪k
n=1An of members of A′ as in [134] since µ[An] = 0 or ∞ for An ∈ A′.
8. Measurable functions
Definition: Let (X,A), (Y, C) be measurable spaces. A function f : X → Y is said to be A-
C measurable (or just measurable) if f−1(C) ∈ A for every C ∈ C. Note the similarity of the
definition of a measurable function with the definition of a continuous function between topological
spaces.
Remark: Let (Xi,Ai) be measurable spaces for 1 ≤ i ≤ 3, let f : X1 → X2 be A1-A2 measurable
and g : X2 → X3 be A2-A3 measurable. Then g f : X1 → X3 is A1-A3 measurable.
Exercise-34: Let (X,A), (Y, C) be measurable spaces, f : X → Y be a function and let D ⊂ P(Y )
be a collection generating C. If f−1(D) ∈ A for every D ∈ D, then f is A-C measurable. [Hint :
Check that C ∈ C : f−1(C) ∈ A is a σ-algebra containing D.]
Exercise-35: If X,Y are metric spaces, then every continuous function f : X → Y is B(X)-B(Y )
measurable. [Hint : Use Exercise-34 with D = open subsets of Y .]
Exercise-36: Let (X,A), (Y, C) be measurable spaces, f : X → Y be a function, and let X =∪∞j=1Xj with Xj ∈ A. Let Aj = A ∩Xj : A ∈ A and fj = f |Xj . Then f is A-C measurable iff
fj is Aj-C measurable for every j ∈ N.
MEASURE THEORY 37
Example: [Fat Cantor sets] We imitate the construction of the middle third Cantor set. Fix m ≥ 3.
First remove an open interval of length 1/m from the middle of [0, 1], then remove an open interval
of length 1/m2 from the middle of each of the remaining two closed intervals, then remove an open
interval of length 1/m3 from the middle of each of the remaining four closed intervals, and so on.
Finally, we are left with a nowhere dense closed subset of [0, 1]; call it Fm. Note that F3 is the
middle-third Cantor set. The total length of all open intervals removed in order to construct Fm
is 1/m + 2/m2 + 4/m3 + 8/m4 + · · · = 1/(m − 2) and therefore µ[Fm] = (m − 3)/(m − 2), where
µ = µL,1. Hence µ[Fm] > 0 for m ≥ 4, and we call Fm a Fat Cantor set for m ≥ 4 (this is another
way to obtain nowhere dense compact subsets of R having positive Lebesgue measure). It may be
proved with some effort that for any m ≥ 4, there is a homeomorphism hm : R → R such that
hm(Fm) = F3 (first define hm : R \ Fm → R \ F3 in the natural way, check that this is uniformly
continuous, and then extend, etc.). This has some interesting consequences. (i) hm is not L(R)-
L(R) measurable for m ≥ 4. Reason: P(F3) ⊂ N (R) ⊂ L(R), but Fm has non-measurable subsets
by [126] since µ[Fm] > 0 for m ≥ 4. (ii) Homeomorphisms of R need not take µ-null sets to µ-null
sets; the homeomorphism h−14 takes F3 to F4. On the other hand, it may be checked that Lipschitz
maps take µ-null sets to µ-null sets. (iii) We get another argument to say B(R) = L(R) as follows.
If B(R) = L(R), then P(F3) ⊂ L(R) = B(R) and hence P(F4) ⊂ B(R) since homeomorphisms must
take Borel sets to Borel sets. But we know by [126] that P(F4) \ L(R) = ∅.
Discussion: When we deal with real-valued functions on a measurable space (X,A), we have a
choice between the following two notions: A-B(R) measurability and A-L(R) measurability. Now,
we want the collection of measurable functions to be as large as possible, and we also want the
collection to include all continuous functions from X to R when X is a metric space. So we
prefer the notion of A-B(R) measurability. Moreover, we may have to deal with extended real-
valued functions sometimes (i.e., functions taking values in [−∞,∞]); for instance, while taking
the supremum of a sequence of real-valued functions. Next, when X = Rd, again we have a choice:
whether to take A = B(Rd) or A = L(Rd). When we deal with Lebesgue measure, it is convenient
to use L(Rd) because of completeness. These considerations motivate the following definitions.
Definition: Note that we can topologize [−∞,∞] by having the usual topology on R and taking
(a,∞] and [−∞, a) as neighborhoods of ∞,−∞ respectively for a ∈ R. So the topological structure
on [−∞,∞] is similar to that on [0, 1]. The Borel σ-algebra B([−∞,∞]) is the σ-algebra on
[−∞,∞] generated by open subsets of [−∞,∞]. It may be seen that B([−∞,∞]) is generated by
B(R) ∪ −∞, ∞ also.
Definition: Let (X,A) be a measurable space. We say f : X → [−∞,∞] is measurable if f is A-
B([−∞,∞]) measurable. If f(X) ⊂ R, then measurability of f just means A-B(R) measurability.
38 T.K.SUBRAHMONIAN MOOTHATHU
Definition: Consider X ∈ L(Rd). Let L(X) := X∩A : A ∈ L(Rd), and note that it is a σ-algebra
on X. We say f : X → [−∞,∞] is measurable if f is L(X)-B([−∞,∞]) measurable. For example,
if Y ∈ L(X), then 1Y is a (Lebesgue-Borel) measurable function.
Remark: Thus when we say f : Rd → [−∞,∞] is measurable, we actually mean f is Lebesgue-Borel
measurable, which is an asymmetric notion. Two disadvantages of this asymmetry are the following:
(i) If f : R → R is a measurable bijection, then f−1 need not be measurable. (ii) If f, g : R → R are
measurable, then f g need not be measurable. For instance, let g : R → R be a homeomorphism
with g(F4) = F3 (recall fat Cantor sets), let Y ⊂ F3 be such that g−1(Y ) /∈ L(R) and let f : R → R
be f = 1Y . Then, f, g are Lebesgue-Borel measurable, but f g is not Lebesgue-Borel measurable
since (f g)−1(1) /∈ L(R). However, we will see in the sequel some of the advantages of using the
notion of Lebesgue-Borel measurability (especially while discussing ‘almost everywhere pointwise
convergence’ of a sequence of functions).
[135] Let (X,A) be a measurable space. Then for a function f : X → [−∞,∞], the following are
equivalent:
(i) f is measurable.
(ii) f−1(U) ∈ A for every open U ⊂ [−∞,∞].
(iii) f−1(−∞), f−1(∞) ∈ A and f−1(U) ∈ A for every open U ⊂ R.
(iv) f−1((a,∞]) = x ∈ X : f(x) > a ∈ A for every a ∈ R.
(v) f−1([a,∞]) = x ∈ X : f(x) ≥ a ∈ A for every a ∈ R.
(vi) f−1([−∞, a)) = x ∈ X : f(x) < a ∈ A for every a ∈ R.
(vii) f−1([−∞, a]) = x ∈ X : f(x) ≤ a ∈ A for every a ∈ R.
Proof. Observe that each statement from (ii)-(vii) is of the form “f−1(D) ∈ A for every D ∈ D”
for a generating collection D of B([−∞,∞]), and use Exercise-34.
[136] Let (X,A) be a measurable space, and let F = f : X → R : f is measurable. Then,
(i) F is a real vector space under pointwise operations.
(ii) If f, g ∈ F , then the pointwise product fg ∈ F ; and f/g ∈ F provided g(x) = 0 ∀ x ∈ X.
(iii) If f ∈ F and g : R → R is continuous, then g f ∈ F .
Proof. (i) Let f, g ∈ F and c, a ∈ R. Since x ∈ X : (−f)(x) > a = x ∈ X : f(x) < a, we
have that −f is measurable. If c = 0, then trivially cf is measurable. If c > 0, then x ∈ X :
(cf)(x) > a = x ∈ X : f(x) > a/c, and hence cf is measurable. If c < 0, then cf = −(−cf) is
measurable. Note that x ∈ X : (f + g)(x) > a =∪
r∈Q[Ar ∩Br], where Ar = x ∈ X : f(x) > r
and Br = x ∈ X : g(x) > a − r. Since Ar, Br ∈ A, we have x ∈ X : (f + g)(x) > a ∈ A, and
thus f + g is measurable.
MEASURE THEORY 39
(ii) We wish to check f2 ∈ F . Consider a ∈ R. If a < 0, then x ∈ X : (f(x))2 > a = X ∈ A. If
a ≥ 0, then x ∈ X : (f(x))2 > a = x ∈ X : f(x) >√a ∪ x ∈ X : f(x) < −
√a ∈ A. Thus f2
is measurable. Now, fg = [(f + g)2 − (f − g)2]/4 is measurable.
Next suppose g(x) = 0 for every x ∈ X. We wish to show 1/g ∈ F . Note that if a > 0, then
x ∈ X : (1/g)(x) > a = x ∈ X : g(x) < 1/a; if a < 0, then x ∈ X : (1/g)(x) > a = x ∈
X : g(x) > 1/a; and x ∈ X : (1/g)(x) > 0 = x ∈ X : g(x) > 0. So (1/g) ∈ F , and therefore,
f/g = f(1/g) ∈ F . Statement (iii) This is clear.
Remark: In the above result, we considered only real-valued functions. For extended real-valued
functions, there is a problem in defining pointwise addition and pointwise multiplication. If f(x) =
∞ and g(x) = −∞, then f + g and fg are not defined at x.
Definition: For a function f : X → [−∞,∞], define the positive part and negative part as f+ =
maxf, 0, f− = max−f, 0. Note that f+, f− ≥ 0; f = f+ − f− and |f | = f+ + f−.
Definition: If (fn) is a sequence of functions from X to [−∞,∞], define
(supn∈N fn)(x) = supfn(x) : n ∈ N,
(infn∈N fn)(x) = inffn(x) : n ∈ N,
lim supn→∞ fn = limk→∞(supn≥k fn) = infk→∞(supn≥k fn), and
lim infn→∞ fn = limk→∞(infn≥k fn) = supk→∞(infn≥k fn).
[137] Let (X,A) be a measurable space, and let F = f : X → [−∞,∞] : f is measurable.
(i) If (fn) is a sequence in F , then sup fn, inf fn, lim sup fn, lim inf fn ∈ F .
(ii) If (fn) is a sequence in F converging pointwise to a function f : X → [−∞,∞], then f ∈ F .
(iii) Let f, g ∈ F with f, g ≥ 0, and let c ∈ R. Then cf, f + g ∈ F .
(iv) If f, g ∈ F , then fg ∈ F .
(v) If f ∈ F , then f+, f−, |f | ∈ F .
Proof. (i) Let a ∈ R. Note that x ∈ X : (sup fn)(x) > a =∪∞
n=1x ∈ X : fn(x) > a and
x ∈ X : (inf fn)(x) < a =∪∞
n=1x ∈ X : fn(x) < a. Hence sup fn, inf fn ∈ F . Applying this
again, we get lim sup fn, lim inf fn ∈ F .
(ii) Note that f = lim sup fn = lim inf fn.
(iii) This is similar to the proof of [136].
(iv) Here we cannot imitate the proof of [136] since f+g, f−g may not be defined. Let fn(x) = f(x)
if |f(x)| ≤ n, fn(x) = n if f(x) ≥ n and fn(x) = −n if f(x) ≤ −n. Similarly define gm. Check
that the real-valued functions fn, gm are measurable. Hence fngm is measurable by [136]. Since
fgm = limn→∞ fngm, we get fgm is measurable by (ii). Then fg = limm→∞ fgm is also measurable.
(v) By part (i), f+ = maxf, 0 ∈ F and f− = max−f, 0 ∈ F . Then, |f | = f+ + f− ∈ F .
40 T.K.SUBRAHMONIAN MOOTHATHU
9. Approximating measurable functions with simple functions
In the previous section, we did not use the concept of a measure. Now we will consider properties
of measurable functions on measure spaces. In this and the coming sections, some results are for
extended real-valued functions; while some other results are only for real-valued functions - keep
an eye on this distinction.
An arbitrary measurable function can be quite complicated. But we can tackle measurable
functions by expressing them as limits of functions which are easier to handle.
Definition: Let (X,A) be a measurable space, and ϕ : X → R be a measurable function. We say ϕ
is a simple function if ϕ(X) is a finite set, i.e., if there are A1, . . . , An ∈ A and c1, . . . , cn ∈ R such
that ϕ =∑n
i=1 ci 1Ai (∵ if ϕ(X) = c1, . . . , cn, let Aj = ϕ−1(cj)).
Definition: Let (X,A, µ) be a measure space, and ϕ : X → R be a measurable function. We say ϕ is
a µ-bounded simple function if ϕ is simple and if there is A ∈ A with µ[A] <∞ such that ϕ(x) = 0
for every x ∈ X \ A (this is equivalent to saying that there are A1, . . . , An ∈ A with µ[Ai] < ∞,
and c1, . . . , cn ∈ R such that and ϕ =∑n
i=1 ci 1Ai). Eg: ϕ : R → R, ϕ = 21Q + 31[4,5].
Definition: A simple function ψ : Rd → R is called a step function if ψ has the form ψ =∑n
i=1 ci 1Ai ,
where each Ai ⊂ Rd is a d-box.
Remark: Simple functions are real-valued and they do not assume the values ±∞ by definition.
[138] [Approximating measurable functions with simple functions] Let (X,A) be a measurable
space and let f : X → [−∞,∞] be measurable. Then there is a sequence (ϕn) of simple functions
on X converging to f pointwise. Moreover, the following additional properties are true:
(i) If f ≥ 0, we may choose ϕn’s in such a way that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ f .
(ii) If f is bounded, we may choose ϕn’s so that (ϕn) → f uniformly.
(iii) If µ is a σ-finite measure on (X,A), we may choose ϕn’s as µ-bounded simple functions.
(iv) If the assumptions of (i) &(ii) hold, then ϕn’c can be chosen to satisfy the conclusions of both
(i) & (ii). Similarly for the combination (i) & (iii); but not for the combination (ii) & (iii).
Proof. The idea is to partition the range of f . Let I(n, k) = [k/2n, (k + 1)/2n) and Γn = k ∈ Z :
−n2n ≤ k < n2n. Note that Pn := I(n, k) : k ∈ Γn defines a partition of [−n, n) in such a way
that Pn+1 restricted to [−n, n) is a refinement of Pn. Let A(n, k) = f−1(I(n, k)) for n ∈ N and
k ∈ Γn. Also, let A(n,−∞) = f−1([−∞,−n)) and A(n,∞) = f−1([n,∞]). Define ϕn : X → R as
ϕn = −n1A(n,−∞) +∑
k∈Γnk2−n 1A(n,k) + n1A(n,∞). Check that (ϕn) → f pointwise.
Statements (i) and (ii) are clear from the construction.
MEASURE THEORY 41
(iii) Write X =∪∞
j=1Xj , where Xj ∈ A and µ[Xj ] < ∞. We may also assume Xj ’s are pairwise
disjoint. Let fj be the restriction of f to Xj . For each j, choose a sequence (ϕj,n)∞n=1 of simple
functions on Xj as described above, converging to fj pointwise. Define ϕn : X → R as ϕn(x) =
ϕj,n(x) if x ∈ Xj for 1 ≤ j ≤ n, and ϕn(x) = 0 if x ∈∪∞
j=n+1Xj . Since µ[Xj ] < ∞, each ϕj,n is
µ-bounded and hence ϕn is also µ-bounded. And evidently, (ϕn) → f pointwise.
(iv) The assertions about the combinations (i) & (ii), and (i) & (iii) can be seen from the above
constructions. If f : R → R is f ≡ 1, then the assumptions of (ii) & (iii) hold, but there does not
exist any sequence of µL,1-bounded simple functions converging to f uniformly on R.
Exercise-37: Let f : [−2, 2] → [−∞,∞] be defined as f(x) = |x + 1| if x ∈ [−2, 0], f(x) = 1/x if
x ∈ (0, 1), and f(x) = log(x− 1) if x ∈ [1, 2]. Check that f is measurable and draw the graphs of
the simple functions ϕn constructed in the proof of [138] for n = 1, 2, 3.
Definition: Let (X,A, µ) be a measure space. We say that some property holds µ-almost everywhere
(µ-a.e. for short) if there is A ∈ A with µ[A] = 0 such that the property holds on X \ A. For
instance, for functions fn, f : X → [−∞,∞], we say the sequence (fn) converges to f pointwise
µ-almost everywhere if there is A ∈ A with µ[A] = 0 such that (fn) → f pointwise on X \A.
One advantage of working with a complete measure space is the following.
[139] Let (X,A, µ) be a complete measure space and let f, g, fn : X → [−∞,∞].
(i) If f is measurable and f = g µ-a.e., then g is measurable.
(ii) If fn’s are measurable, and (fn) → f pointwise µ-a.e., then f is measurable.
Proof. (i) Let A ∈ A be so that µ[A] = 0 and f(x) = g(x) for every x ∈ Ac. Note that x ∈ X :
g(x) > a = [x ∈ X : f(x) > a ∩Ac] ∪ x ∈ A : g(x) > a. Now, x ∈ X : f(x) > a ∈ A since f
is measurable, and x ∈ A : g(x) > a ∈ A by completeness. Hence g is measurable.
(ii) Let A ∈ A be so that µ[A] = 0 and (fn) → f pointwise on Ac. Define gn : X → [−∞,∞] as
gn = fn on Ac, and gn = f on A. Then gn’s are measurable by part (i). Since (gn) → f pointwise
on the whole of X, f is measurable by [137].
Example: Let (X,A, µ) be a measure space which is not complete, and Y ⊂ A ∈ A be with µ[A] = 0
and Y /∈ A. Then 1A is measurable and 1A = 1Y µ-a.e., but 1Y is not measurable.
[140] If f : [a, b] → R is Riemann integrable, then there are step functions ϕn : [a, b] → R such that
ϕ1 ≤ ϕ2 ≤ · · · ≤ f , and (ϕn) → f pointwise µL,1-a.e. Hence f is measurable.
Proof. Assume [a, b] = [0, 1] for simplicity. Let I(n, k) = [(k − 1)/2n, k/2n) for 1 ≤ k < 2n,
I(n, 2n) = [1 − 1/2n, 1] and Pn = I(n, k) : 1 ≤ k ≤ 2n. If z(n, k) = inff(y) : y ∈ I(n, k), then
z(n, k) ∈ R since f is bounded. Define the step function ϕn : [0, 1] → R as ϕn =∑2n
k=1 z(n, k)1I(n,k).
42 T.K.SUBRAHMONIAN MOOTHATHU
Since Pn+1 refines Pn, we have ϕn ≤ ϕn+1 ≤ f . If Y = x ∈ [0, 1] : f is discontinuous at x, then
µL,1[Y ] = 0 by [115]. Given x ∈ [0, 1] \ Y and ε > 0, choose δ > 0 by continuity. If n, k are such
that 2−n < δ and x ∈ I(n, k), then we get |f(x) − ϕn(x)| = |f(x) − z(n, k)| ≤ ε. Thus (ϕn) → f
pointwise on [0, 1] \ Y , and so f is measurable by [139].
10. Limit theorems for measurable functions
If fn ≡ n and f ≡ ∞, then (fn) → f pointwise, but |f(x)− fn(x)| cannot be made small. Now
we introduce some other notions of convergence, where the definition requires that |f(x) − fn(x)|
is small in some sense. So we consider only real-valued functions.
Definition: Let (X,A, µ) be a measure space and let fn, f : X → R be measurable functions. (i) We
say (fn) → f in measure if for every α > 0, we have limn→∞ µ[x ∈ X : |f(x) − fn(x)| ≥ α = 0;
equivalently, if for every α > 0, there is n0 ∈ N such that µ[x ∈ X : |f(x) − fn(x)| ≥ α] ≤ α
for every n ≥ n0. (ii) We say (fn) → f uniformly with negligible error (or almost uniformly) if for
every α > 0, there is A ⊂ A with µ[A] < α such that (fn) → f uniformly on X \ A. For example,
if fn : [0, 1] → R is fn(x) = xn, then (fn) → 0 uniformly with negligible error w.r.to µL,1.
[141] Let (X,A, µ) be a measure space, and let f, fn : X → R be measurable functions such that
(fn) → f uniformly with negligible error. Then, (fn) → f pointwise µ-a.e. and (fn) → f in
measure.
Proof. Let Ak ∈ A be so that µ[Ak] < 1/k and (fn) → f uniformly on X \ Ak. Now see that
(fn) → f pointwise outside the µ-null set∩∞
k=1Ak.
Next, choose n(k, α) ∈ N so that x ∈ X : |f(x)− fn(x)| ≥ α ⊂ Ak for every n ≥ n(k, α) using
the uniform convergence on X \Ak. Then, µ[x ∈ X : |f(x)−fn(x)| ≥ α] ≤ µ[Ak] < 1/k for every
n ≥ n(k, α), and thus (fn) → f in measure.
Partial converse to [141] is given by the following two theorems.
[142] [Egorov’s theorem] Let (X,A, µ) be a measure space with µ[X] <∞, and let f, fn : X → R be
measurable functions such that (fn) → f pointwise µ-a.e. Then (fn) → f uniformly with negligible
error (and hence (fn) → f in measure).
Proof. Let α > 0. We need to find A ∈ A with µ[A] < α such that (fn) → f uniformly on X \ A.
Let B ∈ A be such that µ[B] = 0 and (fn) → f pointwise on X \ B. Let A(k,m) =∪∞
n=mx ∈
X : |f(x) − fn(x)| ≥ 1/k. Then, A(k,m) ∈ A, A(k,m) ⊃ A(k,m + 1) and∩∞
m=1A(k,m) ⊂ B
for each k ∈ N because of pointwise convergence on X \ B. Since µ[X] < ∞ and µ[B] = 0, we
get limm→∞ µ[A(k,m)] = µ[∩∞
m=1A(k,m)] = 0 for each k ∈ N. Hence there is a subsequence (mk)
such that µ[A(k,mk)] < α/2k. Let A =∪∞
k=1A(k,mk) ∈ A. Then µ[A] <∑∞
k=1 α/2k = α. It may
MEASURE THEORY 43
be seen that for every x ∈ X \ A and every n ≥ mk, we have |f(x) − fn(x)| < 1/k; from this it
follows that (fn) → f uniformly on X \A.
[143] [Riesz theorem] Let (X,A, µ) be a measure space and let fn, f : X → R be measurable
functions such that (fn) → f in measure. Then there is a subsequence (fnk) such that (fnk
) → f
uniformly with negligible error (and hence (fnk) → f pointwise µ-a.e.).
Proof. Choose an increasing sequence (nk) such that if Ak = x ∈ X : |f(x)− fnk(x)| ≥ 2−k, then
µ[Ak] < 2−k. Given α > 0, choose k0 ∈ N with∑∞
k=k02−k < α and put A =
∪∞k=k0
Ak. Then
µ[A] < α and moreover |f(x)−fnk(x)| < 2−k for every x ∈ X \A and k ≥ k0, from which it follows
that (fnk) → f uniformly on X \A.
Example: If fn : R → R is fn = 1[n,∞), then (fn) → 0 pointwise, but the convergence is not
uniform on R\A for any A ∈ L(R) with µL,1[A] <∞. So the assumption µ[X] <∞ is necessary in
Egorov’s theorem. This example also shows that pointwise convergence does not imply convergence
in measure.
Example: Let fn : [0, 1] → R be fn = 1(1/(n+1),1/n). Then (fn) → 0 pointwise on [0, 1], and
uniformly on [ε, 1] for any ε > 0. But if A ⊂ [0, 1] is a Lebesgue null set, then (1/(n + 1), 1/n) is
not contained in A for any n ∈ N and it follows that the convergence (fn) → 0 cannot be uniform
on [0, 1] \A. Thus the conclusion of Egorov’s theorem cannot be improved to uniform convergence
outside a null set.
Example: Let (Jn) be a sequence of subintervals of [0, 1] with the following properties: (i) |Jn| → 0
as n → ∞, (ii) for each x ∈ [0, 1], there are infinitely many n with x ∈ Jn, and (iii) for each
x ∈ [0, 1], there are infinitely many n with x ∈ [0, 1] \ Jn. For instance, consider the sequence
[0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3], [2/3, 1], [0, 1/4], . . .. Let fn : [0, 1] → R be fn = 1Jn . Then
(fn) → 0 in measure, but (fn(x)) does not converge for any x ∈ [0, 1]. Hence convergence in
measure does not even imply pointwise convergence µ-a.e. In particular, we cannot replace (fnk)
with (fn) in the conclusion of Riesz theorem.
Example: Let f : R2 → R be f(x, y) = x. Then f is measurable, and µL,2[(x, y) ∈ R2 : |f(x, y)| ≥
n] = ∞ for every n ∈ N. As a consequence, a sequence of simple functions can never converge
to f in measure (∵ simple functions are bounded). Compare with [138]. Also, construct a similar
example f : R → R.
Exercise-39: [Linearity of convergence] Let (X,A, µ) be a measure space and let f, g, fn, gn : X → R
be measurable functions such that (fn) → f and (gn) → g in measure. Then (afn) → af and
(fn + gn) → f + g in measure, where a ∈ R. Prove similar results for pointwise convergence µ-a.e.
44 T.K.SUBRAHMONIAN MOOTHATHU
and uniform convergence with negligible error. [Hint : x ∈ X : |(f + g)(x)− (fn + gn)(x)| ≥ α ⊂
x ∈ X : |f(x)− fn(x)| ≥ α/2 ∪ x ∈ X : |g(x)− gn(x)| ≥ α/2.]
Remark: Let X be a metric space and let A,B ⊂ X be nonempty closed subsets with A ∩ B = ∅.
Then note there is a continuous function f : X → [0, 1] such that f ≡ 0 on A, and f ≡ 1 on B. In
fact, we may take f(x) = dist(x,A)/[dist(x,A) + dist(x,B)].
Exercise-40: (i) If A ∈ L(Rd), then there are continuous functions fn : Rd → R such that (fn) → 1A
in measure. (ii) If A ∈ L(Rd) is such that µ[A] < ∞, then there are step functions ψn : Rd → R
such that (ψn) → 1A in measure. [Hint : (i) Let Fn ⊂ A ⊂ Un ⊂ Rd be such that Fn is closed, Un
is open and µL,d[Un \ Fn] < 1/n. Let fn : Rd → [0, 1] be a continuous function such that fn ≡ 1
on Fn and fn ≡ 0 on Rd \ Un. Then, x ∈ Rd : 1A(x) = fn(x) ⊂ Un \ Fn. (ii) Fix n ∈ N.
By the remark given immediately after [134], there are finitely many d-boxes A1, . . . , Ak such that
µL,d[A∆∪k
j=1Aj ] < 1/n. We may assume Aj ’s are pairwise disjoint. Then for the step function
ψn :=∑k
j=1 1Aj , we have that x ∈ Rd : 1A(x) = ψn(x) = A∆∪k
j=1Aj .]
Exercise-41: Let (X,A, µ) be a measure space and let f, fn, fn,m : X → R be measurable functions.
(i) If (fn) → f in measure, and (fn,m) → fn in measure for each n ∈ N, then there are increasing
sequences (nk), (mk) such that (fnk,mk) → f in measure. (ii) Assuming µ[X] < ∞ if necessary,
prove similar results for pointwise convergence µ-a.e., and uniform convergence with negligible
error. [Hint : (i) Fix k ∈ N. First choose nk such that µ[x ∈ X : |f(x) − fnk(x)| ≥ 1/k] < 1/k.
Then choose mk such that µ[x ∈ X : |fnk(x) − fnk,mk
(x)| ≥ 1/k] < 1/k. Hence µ[x ∈ X :
|f(x) − fnk,mk(x)| ≥ 2/k < 2/k. Now it is not difficult to see (fnk,mk
) → f in measure. (ii) Use
part (i) and results [141], [142], [143]. See also the proof below.]
[144] Let B ⊂ Rd be a d-box and f : B → R be a measurable function. Then,
(i) There are step functions ψn : B → R such that (ψn) → f pointwise µL,d-a.e. on B.
(ii) There are continuous functions fn : B → R such that (fn) → f pointwise µL,d-a.e. on B. If B
is closed, then we may choose fn’s so that fn|∂B ≡ 0 also (where ∂B is the boundary of B).
Proof. By [138], there is a sequence (ϕn) of simple functions converging to f pointwise on B. Since
µ[B] <∞, (ϕn) → f in measure by [142] and [141].
(i) By Exercise-39 and Exercise-40, we may find step functions ψn,m : B → R such that (ψn,m) → ϕn
in measure for each n ∈ N. Then by Exercise-41, there are (nk), (mk) so that (ψnk,mk) → f in
measure. By Riesz theorem, a subsequence of (ψnk,mk) converges to f pointwise µL,d-a.e. on B.
(ii) By Exercise-39 and Exercise-40, for each n ∈ N we may find continuous functions fn,m : B → R
such that (fn,m) → ϕn in measure. Then by Exercise-41, there are (nk), (mk) so that (fnk,mk) → f
in measure. By Riesz theorem, a subsequence of (fnk,mk) converges to f pointwise µL,d-a.e. on B.
MEASURE THEORY 45
Now suppose that B is closed and (fn) is a sequence of continuous functions converging to f
pointwise on B \ Y , where Y is a Lebesgue null set. Let (An) be a sequence of closed d-boxes such
that A1 ⊂ A2 ⊂ · · · ⊂ int[B] and∪∞
n=1An = int[B]. Let hn : B → [0, 1] be a continuous function
so that hn ≡ 1 on An and hn ≡ 0 on ∂B. Then fn(x)hn(x) = 0 for every x ∈ ∂B, and (fnhn) → f
pointwise outside the Lebesgue null set Y ∪ ∂B.
[145] [Approximating measurable functions with step functions and with continuous functions of
compact support] Let f : Rd → R be measurable. Then,
(i) There are step functions ψn : Rd → R such that (ψn) → f pointwise µL,d-a.e.
(ii) There are continuous functions fn : Rd → R such that (fn) → f pointwise µL,d-a.e. We may
also arrange fn’s to have compact support.
Proof. Write Rd =∪∞
j=1Bj , a disjoint union of d-boxes, and let gj = f |Bj . By [144], there are
step functions ψj,n : Bj → R so that (ψj,n) → gj pointwise µ-a.e. on Bj for each j ∈ N. Define
ψn : Rd → R as ψn(x) = ϕj,n(x) if x ∈ Bj and 1 ≤ j ≤ n; and ψn(x) = 0 otherwise.
(ii) Write Rd =∪∞
j=1Bj , a union of closed d-boxes with pairwise disjoint interiors. Let gj = f |Bj .
For each j, let fj,n : Bj → R be continuous functions such that (fj,n) → gj pointwise µL,d-a.e. on
Bj , and fj,n ≡ 0 on ∂Bj (by [144]). Define fn : Rd → R as fn(x) = fj,n(x) if x ∈ Bj and 1 ≤ j ≤ n;
and fn(x) = 0 otherwise.
[146] [Luzin’s theorem] Let f : Rd → R be a measurable function and ε > 0. Then,
(i) There is Y ∈ L(Rd) such that µL,d[Y ] < ε, and f |Rd\Y : Rd \Y → R is continuous. We may also
choose Y to be open.
(ii) There is a continuous function g : Rd → R such that µL,d[x ∈ Rd : f(x) = g(x)] < ε.
Proof. (i) Write µ = µL,d. Choose continuous functions fn : Rd → R so that (fn) → f pointwise
µ-a.e. By Egorov’s theorem, (fn) → f uniformly with negligible error on B(0, j) for each j ∈ N.
Let Yj ∈ L(Rd) be such that Yj ⊂ B(0, j), µ[Yj ] < ε/2j and (fn) → f uniformly on B(0, j) \ Yj . If
Y =∪∞
j=1 Yj , then µ[Y ] < ε. Observe that B(0, j) \ Y is open in Rd \ Y , and (fn) → f uniformly
on B(0, j) \Y . Hence f |B(0,j)\Y is continuous. Since B(0, j) \Y : j ∈ N is a relatively open cover
for Rd \ Y , we conclude that f |Rd\Y is continuous. Now, replacing Y with a slightly bigger open
superset (by the outer regularity of Lebesgue measure), we may assume Y is open also.
(ii) By part (i), there is a closed set A ⊂ Rd such that µ[Rd \A] < ε and f |A is continuous (∵ take
A = Rd \ Y ). By Tietze extension theorem, there is a continuous g : Rd → R with g|A = f .
Remarks: (i) For a set A, saying that f |A is continuous is strictly weaker than saying that f is
continuous at every point of A. For instance, if f = 1Q : R → R, then f |Q is continuous, but f
is not continuous at any point of Q (or of R). (ii) If F ⊂ R is a fat Cantor set, f : R → R is
46 T.K.SUBRAHMONIAN MOOTHATHU
f = 1F , and Y ⊂ R is any Lebesgue null set, then f restricted to R \ Y is not continuous at any
point of the nonempty set F \Y (∵ F \Y = ∅ since F has positive measure; and for any x ∈ F \Y ,
there is a sequence in R \ (F ∪ Y ) converging to x since F is nowhere dense and Y is a null set).
This shows that we cannot improve Luzin’s theorem by choosing Y to be a null set in [146]. (iii)
If f : Rd → [−∞,∞] is a measurable function with µL,d[x ∈ Rd : f(x) = ±∞] > 0, then clearly
the conclusion of Luzin’s theorem fails for f .
Exercise-42: [Converse of Luzin] Let f : Rd → R be a function having the property that for each
ε > 0, there is Y ∈ L(Rd) such that µL,d[Y ] < ε, and f |Rd\Y : Rd \ Y → R is continuous. Then f is
measurable. [Hint : Choose Yn for ε = 2−n, and let fn : Rd be fn = 0 on Yn and fn = f on Rd \ Yn.
Check that fn’s are measurable, and (fn) → f pointwise outside the null set∩∞
k=1
∪∞n=k Yn. Then
use [139].]
The following result can be thought of as a functional analogue of the fact that for any A ∈ L(Rd)
there is B ∈ B(Rd) such that A∆B is a Lebesgue null set.
[147] If f : Rd → R is (Lebesgue-Borel) measurable, then there is a B(Rd)-B(R) measurable
function g : Rd → R such that f = g µL,d-a.e.
Proof. By [145], there are step functions ψn : Rd → R and D ∈ L(Rd) with µL,d[Rd \ D] = 0
so that (ψn) → f pointwise on D. We may assume D is an Fσ set (hence Borel) by [124](v).
Note that any step function is Borel-Borel measurable. Define gn = ψn 1D. Then gn’s are B(Rd)-
B(R) measurable. If g = f 1D, then (gn) → g pointwise everywhere on Rd. If K ⊂ R is a
nonempty compact set, let Um = y ∈ R : dist(y,K) < 1/m (which is an open set) and verify
that g−1(K) =∩∞
m=1
∪∞n=m g
−1n (Um). Hence g−1(K) ∈ B(Rd). Since B(R) is generated by compact
subsets of R, we get that g is B(Rd)-B(R) measurable. And by definition, f = g µL,d-a.e.
11. Integration with respect to a measure
Let (X,A, µ) be a measure space. We describe below an integration theory for measurable
functions f : X → [−∞,∞]. Strictly speaking, the measurability of f is not needed to define the
integral, but measurability is needed to derive many properties of the integral. The integral of f
with respect to µ will be defined in three steps: (i) when f is simple and ≥ 0, (ii) when f ≥ 0, (iii)
for arbitrary f .
Definition: Let (X,A) be a measurable space and let ϕ : X → R be a simple function. Suppose
y1, . . . , ym ∈ R are the distinct elements of ϕ(X), and let Xj = x ∈ X : ϕ(x) = yj ∈ A for
1 ≤ j ≤ m. Then∑m
j=1 yj 1Xj is called the canonical representation of ϕ. If∑m
j=1 yj 1Xj is the
canonical representation of ϕ, then note that X =∪m
j=1Xj is a disjoint union.
MEASURE THEORY 47
Integration, step-1: Let (X,A, µ) be a measure space. If ϕ : X → [0,∞) is a simple function with
canonical representation ϕ =∑m
j=1 yj 1Xj , define the integral of ϕ with respect to µ as∫X ϕdµ =∑m
j=1 yjµ[Xj ].
Remark: The reason for the assumption ϕ ≥ 0 is that ∞−∞ is undefined. For instance consider
the simple function ϕ : R → R given by ϕ = 21(∞,0)+(−3)1[0,∞), and try to imitate the definition
given above to see what goes wrong.
[148] Let (X,A, µ) be a measure space, let ϕ ≥ 0, ψ ≥ 0 be simple functions on X, and let
c ∈ [0,∞). Then,
(i)∫X(cϕ)dµ = c
∫X ϕdµ.
(ii)∫X(ϕ+ ψ)dµ =
∫X ϕdµ+
∫X ψdµ.
(iii) If∑n
i=1 ai 1Ai is any representation (not necessarily canonical) of ϕ with ai ≥ 0, then∫X ϕdµ =∑n
i=1 aiµ[Ai].
(iv)∫X ϕdµ = 0 iff ϕ = 0 µ-a.e.
(v) If ϕ ≤ ψ µ-a.e., then∫X ϕdµ ≤
∫X ψdµ.
(vi) If ϕ = ψ µ-a.e., then∫X ϕdµ =
∫X ψdµ.
Proof. (i) The case c = 0 is clear. If c > 0 and if∑m
j=1 yj 1Xj is the canonical representation of ϕ,
then∑m
j=1 cyj 1Xj is the canonical representation of cϕ.
(ii) Let ϕ =∑m
j=1 aj 1Aj , ψ =∑n
k=1 bk 1Bkbe the canonical representations. Then, we see that
X =∪m
j=1
∪nk=1(Aj ∩ Bk) is a disjoint union and (ϕ + ψ)(x) = aj + bk if x ∈ Aj ∩ Bk. Let
c1, . . . , ct ∈ R be the distinct elements of (ϕ+ ψ)(X). Let Γi = (j, k) : aj + bk = ci and let Ci =∪(j,k)∈Γi
(Aj∩Bk) for 1 ≤ i ≤ t. Then∑t
i=1 ci 1Ci is the canonical representation of ϕ+ψ. Therefore,∫X(ϕ + ψ)dµ =
∑ti=1 ciµ[Ci] =
∑ti=1 ci(
∑(j,k)∈Γi
µ[Aj ∩ Bk]) =∑t
i=1(∑
(j,k)∈Γi((aj + bk)µ[Aj ∩
Bk]) =∑m
j=1
∑nk=1((aj+bk)µ[Aj∩Bk]) =
∑mj=1 aj(
∑nk=1 µ[Aj∩Bk])+
∑nk=1 bk(
∑mj=1 µ[Aj∩Bk]) =∑m
j=1 ajµ[Aj ] +∑n
k=1 bkµ[Bk] =∫X ϕdµ+
∫X ψdµ.
(iii) We have∫X ϕdµ =
∫X(
∑ni=1 ai 1Ai)dµ =
∑ni=1 ai
∫X 1Aidµ =
∑ni=1 aiµ[Ai] by (i) and (ii).
(iv) This is direct from the definition of∫X ϕdµ, considering the canonical representation of ϕ.
(v) In view of (iv), we may assume ϕ ≤ ψ everywhere. Also we may suppose∫X ψdµ < ∞. If
ϕ1 = ψ−ϕ, then note that ϕ1 ≥ 0 is a simple function. Hence by (ii),∫X ψdµ =
∫X ϕdµ+
∫X ϕ1dµ.
Since all the terms are ≥ 0 and <∞, we get∫X ψdµ ≥
∫X ϕdµ.
(vi) This follows from (v) since ϕ ≤ ψ and ψ ≤ ϕ µ-a.e.
Integration, step-2: Let (X,A, µ) be a measure space. If f : X → [0,∞] is a measurable function,
we define the integral of f with respect to µ as∫X fdµ = sup
∫X ϕdµ : ϕ is simple and 0 ≤ ϕ ≤ f.
If f ≥ 0 is a simple function, then this definition agrees with the previous one by [148](iii). Also
48 T.K.SUBRAHMONIAN MOOTHATHU
note that if f, g : X → [0,∞] are measurable functions with f ≤ g, then∫X fdµ ≤
∫X gdµ by
definition.
Exercise-43: [Markov’s inequality] Let (X,A, µ) be a measure space, f : X → [0,∞] be a measurable
function, and let At = x ∈ X : f(x) ≥ t. Then tµ[At] ≤∫X fdµ for 0 < t < ∞. [Hint :
0 ≤ t1At ≤ f and hence∫X t1Atdµ ≤
∫X fdµ.]
Exercise-44: Let (X,A, µ) be a measure space, f : X → [0,∞] be a measurable function, let
A = x ∈ X : f(x) = ∞ and B = x ∈ X : f(x) > 0. (i) If∫X fdµ < ∞, then µ[A] = 0.
(ii) If∫X fdµ = 0, then µ[B] = 0. [Hint : (i) If µ[A] > 0, choose t ∈ (0,∞) large enough
so that tµ[A] >∫X fdµ; this contradicts Markov’s inequality. (ii) Write B =
∪∞n=1Bn, where
Bn = x ∈ X : f(x) > 1/n and use Markov’s inequality.]
Integration, step-3: Let (X,A, µ) be a measure space. If f : X → [−∞,∞] is a measurable
function, and if min∫X f+dµ,
∫X f−dµ < ∞, we define the integral of f with respect to µ as∫
X fdµ =∫X f+dµ−
∫X f−dµ. If both
∫X f+dµ and
∫X f−dµ are equal to ∞, the integral
∫X fdµ
is undefined (since ∞−∞ is undefined).
Definition Let (X,A, µ) be a measure space. A measurable function f : X → [−∞,∞] is µ-
integrable if∫X fdµ is defined and |
∫X fdµ| <∞; equivalently if max
∫X f+dµ,
∫X f−dµ <∞.
Definition: Let (X,A, µ) be a measure space, let A ∈ A and let f : A→ [−∞,∞] be a measurable
function. We define the integral of f on A with respect to µ as∫A fdµ =
∫X f 1Adµ, if it exists.
We say f is µ-integrable on A if∫A fdµ is defined and |
∫A fdµ| <∞.
Remark: Let (X,A, µ) be a measure space and let f = f1 + if2 : X → C be a function. If both
f1, f2 are µ-integrable, we say f is µ-integrable and define∫X fdµ =
∫X f1dµ+ i
∫X f2dµ. Thus we
can carry over our integration theory to complex valued functions in a natural way, but we will not
discuss this here.
Definition: LetX ∈ L(Rd) and let f : X → [−∞,∞] be a measurable function. We say f is Lebesgue
integrable if f is µL,d-integrable on X. Note that the theory of Lebesgue integrable functions is a
special case of the general integration theory; however, the general integration theory is motivated
by Lebesgue integration theory.
12. Properties of the integral, and limit theorems
As seen in the theory of measurable functions, often a complicated function can be approximated
by a sequence of functions which are more elementary in nature and easier to handle. Therefore,
we are primarily interested in knowing whether the equality∫X fdµ = limn→∞
∫X fndµ holds if
(fn) → f in some sense. That is, we want to know whether ‘limit’ and ‘integration’ can be
MEASURE THEORY 49
interchanged (i.e., whether∫X(limn→∞ fn)dµ = limn→∞
∫X fndµ). We give some examples to say
that this need not happen always.
Example-1: Let X = [0, 2], µ = µL,1, and fn = n1(1/n, 2/n). Then (fn) → 0 pointwise on X, but
limn→∞∫X fndµ = limn→∞ 1 = 1 = 0 =
∫X 0dµ.
Example-2: Let µ = µL,1 and fn = (1/n)1(0,n). Then fn’s are uniformly bounded by the constant
function 1, and (fn) → 0 uniformly on R. But limn→∞∫R fndµ = limn→∞ 1 = 1 = 0 =
∫R 0dµ.
Example-3: Let µ = µL,1 and fn = (1/n)1(n,∞). Then f1 > f2 > · · · > 0 and (fn) → 0 uniformly
on R. But limn→∞∫R fndµ = limn→∞∞ = ∞ = 0 =
∫R 0dµ.
However, we can interchange limit and integration in certain situations.
[149] [Monotone (increasing) convergence theorem] Let (X,A, µ) be a measure space and let fn :
X → [0,∞] be measurable functions on X with 0 ≤ f1 ≤ f2 ≤ · · · . If (fn) converges pointwise to
a function f : X → [0,∞], then∫X fdµ = limn→∞
∫X fndµ.
Proof. By [137], f is measurable. Since fn ≤ f , we have∫X fndµ ≤
∫X fdµ and therefore
limn→∞∫X fndµ ≤
∫X fdµ. To prove the other inequality it suffices to show the following: if
ϕ is a simple function on X with 0 ≤ ϕ ≤ f , then limn→∞∫X fndµ ≥
∫X ϕdµ.
Suppose ϕ =∑m
j=1 yj 1Xj is the canonical representation. Let 0 < λ < 1 and Xj,n = x ∈ Xj :
fn(x) ≥ λyj. Then Xj,n ⊂ Xj,n+1 since fn ≤ fn+1. Consider x ∈ Xj . Since f(x) ≥ ϕ(x) = yj , we
have that fn(x) ≥ λyj for all large n ∈ N. This says that∪∞
n=1Xj,n = Xj for each j. As this is an
increasing union, we obtain (µ[Xj,n]) → µ[Xj ] as n→ ∞ for each j.
If ϕn = λ∑m
j=1 yj 1Xj,n , then ϕn is a simple function with 0 ≤ ϕn ≤ fn. Therefore∫X fndµ ≥∫
X ϕndµ = λ∑m
j=1 yjµ[Xj,n]. Taking the limit n→ ∞, we get limn→∞∫X fndµ ≥ λ
∑mj=1 yjµ[Xj ] =
λ∫X ϕdµ. Since λ ∈ (0, 1) can be arbitrarily close to 1, we are through.
Exercise-45: Let (X,A, µ) be a measure space and let f : X → [0,∞] be a measurable function.
If (ϕn) is any sequence of simple functions on X such that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ f and (ϕn) → f
pointwise, then∫X fdµ = limn→∞
∫X ϕndµ. [Hint : Direct application of [149].]
Remark: Also note that for any measurable f : X → [0,∞], a sequence (ϕn) as mentioned in
Exercise-45 exists by [138]; moreover observe from the proof of [138] that ϕn’s are constructed by
partitioning the range of f (where as in Riemann integration theory we partition the domain of f).
Before continuing with the limit theorems and their corollaries, we deduce below some basic
properties of integration.
[150] [Basic properties of integration] Let (X,A, µ) be a measure space and let f, g : X → [−∞,∞]
be µ-integrable functions. Then,
50 T.K.SUBRAHMONIAN MOOTHATHU
(i) [Linearity] af + bg is µ-integrable and∫X(af + bg)dµ = a
∫X fdµ+ b
∫X gdµ for every a, b ∈ R.
(ii)∫X fdµ = 0 iff f = 0 µ-a.e.
(iii) [Monotonicity] If f ≤ g µ-a.e., then∫X fdµ ≤
∫gdµ.
(iv) If f = g µ-a.e., then∫X fdµ =
∫X gdµ.
Proof. The proofs of (ii), (iii), and (iv) are left as exercises (use [148] and the definition of inte-
gration). We prove (i) below. After modifying on a set of measure zero, we may assume f, g are
real-valued (see Exercise-44). We will consider f + g and cf separately, c ∈ R.
First suppose f ≥ 0, g ≥ 0. Let (ϕn), (ψn) be increasing sequences of nonnegative simple
functions converging pointwise to f, g respectively. Then (ϕn + ψn) is an increasing sequence
of nonnegative simple functions converging pointwise to f + g. Hence by Exercise-45 we have∫X(f + g)dµ = limn→∞
∫X(ϕn + ψn)dµ = limn→∞[
∫X ϕndµ +
∫X ψndµ] = limn→∞
∫X ϕndµ +
limn→∞∫X ψndµ =
∫X fdµ+
∫X gdµ.
Now consider the general case, and let h = f +g. Note that h++h− = |h| = |f +g| ≤ |f |+ |g| =
f++f−+g++g−. Hence by the additivity proved above, we get∫X h+dµ+
∫X h−dµ <∞, and this
implies h is µ-integrable. One common mistake now is to think that (f+g)+ = f++g+ which is not
true. However, we have h+−h− = h = f+g = (f+−f−)+(g+−g−) or h++f−+g− = h−+f++g+.
Using additivity, we have∫X h+dµ+
∫X f−dµ+
∫X g−dµ =
∫X h−dµ+
∫X f+dµ+
∫X g+dµ. Since all
the functions involved are integrable, we may group them back to get∫X hdµ =
∫X fdµ+
∫X gdµ.
Next we would like to show∫X(cf)dµ = c
∫X fdµ. First suppose f ≥ 0 and c ∈ [0,∞). From
[148] and the definition of integration, we easily get∫X(cf)dµ = c
∫X fdµ. Since (−f)+ = 0 and
(−f)− = f , we have∫X(−f)dµ =
∫X 0dµ −
∫X fdµ = −
∫X fdµ. It follows that
∫X(−cf)dµ =
−c∫X fdµ also. Use these and the additivity proved above to establish the general case.
Remark: The proof of [150](i) also gives the identities∫X(f + g)dµ =
∫X fdµ +
∫X gdµ and∫
X(cf)dµ = c∫X fdµ for measurable f, g : X → [0,∞] and c ∈ [0,∞). Similar remark about
properties (ii), (iii), (iv) for nonnegative measurable functions.
Now we discuss some consequences of the monotone convergence theorem. The following result
says that Lebesgue integration theory is indeed a generalization of Riemann integration theory.
[151] Let f : [a, b] → R be Riemann integrable. Then, f is Lebesgue integrable and∫[a,b] fdµL,1 =∫ b
a f(x)dx.
Proof. Write µ = µL,1 and assume [a, b] = [0, 1]. By [140] we know that f is measurable. Since f
is Riemann integrable, f is bounded. Therefore, replacing f with f + c for some constant c > 0,
we may assume f ≥ 0. Let I(n, k) = [(k− 1)/2n, k/2n) for 1 ≤ k < 2n and I(n, 2n) = [1− 1/2n, 1].
Let y(n, k) = inff(x) : x ∈ I(n, k), and define the step (simple) function ϕn : [0, 1] → R as ϕn =
MEASURE THEORY 51∑2n
k=1 y(n, k)1I(n,k). From the proof of [140] we know that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ f and (ϕn) → f
pointwise µ-a.e. Hence by [149] and [150](iii), we conclude that∫[0,1] fdµ = limn→∞
∫[0,1] ϕndµ.
Now∫[0,1] ϕndµ =
∑2n
k=1 y(n, k)|I(n, k)| =∑2n
k=1 y(n, k)(k/2n − (k − 1)/2n), which is the lower
Riemann sum L(f, Pn) corresponding to the partition Pn = I(n, k) : 1 ≤ k ≤ 2n. Since the
maximal length of an interval in Pn goes to 0 as n → ∞, we have (L(f, Pn)) →∫ 10 f(x)dx by
Riemann’s theory.
Exercise-46: Let (X,A, µ) be a measure space and let f : X → [−∞,∞] be measurable. Then, (i)
f is µ-integrable iff |f | is µ-integrable, (ii) if f is µ-integrable, then |∫X fdµ| ≤
∫X |f |dµ. [Hint : (i)
If f is µ-integrable, then use |f | = f++f− and additivity from [150](i) to conclude |f | is integrable.
For the converse, note that maxf+, f− ≤ |f |. (ii) Use similar ideas.]
Remark: Thus in our integration theory, f is integrable iff |f | is integrable. Now, infinite summation
is a discrete analogue of integration, and for a real series we know that convergence does not imply
absolute convergence. There is an integration theory known as Henstock-Kurzweil theory (which
is in a sense a further generalization of Lebesgue integration theory) where the integrability of f
does not necessarily imply the integrability of |f |.
Exercise-47: Let (X,A, µ) be a measure space and let fn : X → [0,∞] be measurable for n ∈ N.
Then∫X(
∑∞n=1 fn)dµ =
∑∞n=1
∫X fndµ. [Hint : Apply [149] and additivity to gn = f1 + · · ·+ fn.]
Example: Let µ = µL,1. Let (In) be a sequence of pairwise disjoint intervals in R with |In| = n−3 and
let f =∑∞
n=1 n1In . Then by Exercise-47,∫R fdµ =
∑∞n=1 n
−2 <∞ and∫R f
2dµ =∑∞
n=1 n−1 = ∞.
This shows that the product of two µ-integrable functions need not be µ-integrable.
Exercise-48: [Another way of constructing measures] Let (X,A, µ) be a measure space, let f : X →
[0,∞] be a measurable function, and define β : A → [0,∞] as β[A] =∫A fdµ. Then (i) β is a
measure on (X,A), and (ii) µ[A] = 0 implies β[A] = 0 for every A ∈ A (we say β is absolutely
continuous with respect to µ) [Hint : (i) For countable additivity, use Exercise-47 with fn = f 1An .]
Remark: The converse of Exercise-48 is also true if µ, β are σ-finite measure. That is, β satisfying
property (ii) can be realized in the form β[A] =∫A fdµ for some measurable function f : X → [0,∞].
This non-trivial converse is known as Radon-Nikodym theorem.
Exercise-49: [Borel-Cantelli lemma] Let (X,A, µ) be a measure space, let A1, A2, . . . ∈ A be such
that∑∞
n=1 µ[An] < ∞ and let A =∩∞
k=1
∪∞n=k An = x ∈ X : x ∈ An for infinitely many n ∈ N.
Then µ[A] = 0. [Hint : If f =∑∞
n=1 1An , then∫X fdµ < ∞ by Exercise-47; hence µ[A] = µ[x ∈
X : f(x) = ∞] = 0. Or note that µ[A] ≤ µ[∪∞
n=k An] ≤∑∞
n=k µ[An] → 0 as k → ∞.]
52 T.K.SUBRAHMONIAN MOOTHATHU
Remark: Taking (An) to be the sequence [0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3], [2/3, 1], [0, 1/4], . . .,
we see that the hypothesis∑∞
n=1 µ[An] <∞ cannot be weakened to (µ[An]) → 0 in Borel-Cantelli.
Exercise-50: Let (X,A, µ) be a measure space and f : X → [−∞,∞] be µ-integrable. Then for
every ε > 0 there is δ > 0 such that∫A |f |dµ < ε for every A ∈ A with µ[A] < δ. [Hint : We may
assume f ≥ 0. Let An = x ∈ X : f(x) ≤ n and fn = f 1An . Then µ[X \∪∞
n=1An] = µ[x :
f(x) = ∞] = 0, and (fn) increasingly converges to f pointwise on∪∞
n=1An. By [149], there is n
with∫X(f − fn)dµ < ε/2. Choose any δ ∈ (0, ε/2n) and use the splitting
∫f =
∫(f − fn) +
∫fn.]
For a sequence (fn) of nonnegative measurable functions, even when we cannot say lim∫fn =∫
(lim fn), we can say which side is bigger.
[152] [Fatou’s lemma] Let (X,A, µ) be a measure space and let fn : X → [0,∞] be measurable
functions for n ∈ N. Then∫X(lim infn→∞ fn)dµ ≤ lim infn→∞
∫X fndµ.
Proof. Let g = lim infn→∞ fn and gk = infn≥k fn. Then g and gk’s are measurable, 0 ≤ g1 ≤ g2 ≤
· · · ≤ g and (gk) → g pointwise. Hence by [149], we have∫X gdµ = limk→∞
∫X gkdµ. If n ≥ k, then
gk ≤ fn and therefore∫X gkdµ ≤ infn≥k
∫X fndµ. Use this above.
Remark: One way to remember the direction of the inequality in Fatou’s lemma is to keep the
example fn = 1(n,∞) in mind.
In monotone convergence theorem and Fatou’s lemma, we considered only nonnegative functions.
But often we may have to deal with a sequence of functions taking both positive and negative values.
A basic convergence theorem for such cases is the following.
[153] [Lebesgue’s dominated convergence theorem] Let (X,A, µ) be a mesure space and let fn :
X → [−∞,∞] be µ-integrable functions for n ∈ N. Suppose (fn) converges pointwise µ-a.e. to a
measurable function f : X → [−∞,∞]. Also suppose there is a µ-integrable function g : X → [0,∞]
such that |fn| ≤ g µ-a.e. for every n ∈ N. Then,
(i) f is µ-integrable.
(ii)∫X fdµ = limn→∞
∫X fndµ.
(iii) limn→∞∫X |f − fn|dµ = 0.
Proof. After modifying on a set of measure zero, we may assume that the µ-integrable functions
fn and g are real valued (see Exercise-44(i)). By a similar modification, we may also assume that
(fn) → f pointwise everywhere and |fn| ≤ g everywhere.
(i) We have |f | ≤ g since (fn) → f pointwise and |fn| ≤ g. Hence∫X |f |dµ ≤
∫X gdµ < ∞ by
Fatou’s lemma, and thus f is µ-integrable by Exercise-46.
MEASURE THEORY 53
(ii) Note that g + fn ≥ 0 and g − fn ≥ 0, so we may apply Faou’s lemma to these functions.
We have∫X gdµ +
∫X fdµ =
∫X(g + f)dµ ≤ lim infn→∞
∫X(g + fn)dµ = lim infn→∞[
∫X gdµ +∫
X fndµ] =∫X gdµ+lim infn→∞
∫X fndµ. Hence
∫X fdµ ≤ lim infn→∞
∫X fndµ. Similarly,
∫X gdµ−∫
X fdµ =∫X(g − f)dµ ≤ lim infn→∞
∫X(g − fn)dµ = lim infn→∞[
∫X gdµ −
∫X fndµ] =
∫X gdµ −
lim supn→∞∫X fndµ, and hence
∫X fdµ ≥ lim supn→∞
∫X fndµ.
(iii) Let hn = |f − fn|. Then hn’s are µ-integrable, (hn) → 0 pointwise and |hn|’s are bounded
by the µ-integrable function |f | + |g|. So by applying part (ii) to hn’s we obtain∫X |f − fn|dµ →∫
X 0dµ = 0.
Remark: In [153], the measurability of f is part of the hypothesis. But it can also be obtained as
a conclusion by [139] provided the measure space (X,A, µ) is complete.
Exercise-51: Let (X,A, µ) be a measure space and f : X → [−∞,∞] be µ-integrable. (i) Let
A,An ∈ A be such that A1 ⊂ A2 ⊂ · · · and A =∪∞
n=1An. Then∫A fdµ = limn→∞
∫Anfdµ. (ii)
Let Bn ∈ A be pairwise disjoint and let B =∪∞
n=1Bn. Then∫B fdµ =
∑∞n=1
∫Bnfdµ. [Hint : For
(i), apply [153] with fn = f 1An and g = |f 1A|. For (ii), use (i) with An = B1 ∪ · · · ∪Bn.]
Remark: Let µ = µL,1 and f : R → R be µ-integrable. If R =∪∞
n=1(an, bn) is an increasing union,
then∫R fdµ = limn→∞
∫(an,∞) fdµ = limn→∞
∫(−∞,bn)
fdµ = limn→∞∫(an,bn)
fdµ by Exercise-51.
13. Lp spaces
Definition: [Lebesgue spaces Lp(µ)] Let (X,A, µ) be a complete measure space and 1 ≤ p < ∞.
We define Lp(µ) = f : X → R : f is measurable and∫X |f |pdµ < ∞, with the understanding
that we identify f and g if f = g µ-a.e. (so the members of Lp(µ) are equivalence classes, strictly
speaking). Note that L1(µ) is just the space of µ-integrable real-valued functions. Let L∞(µ) =
f : X → R : f is measurable and there is M ∈ [0,∞) such that |f | ≤M µ-a.e., again with the
understanding that we identify f and g if f = g µ-a.e.
Seminar topic: For 1 ≤ p ≤ ∞, the space Lp(µ) is a real vector space under pointwise operations.
If we define ∥f∥p = (∫X |f |pdµ)1/p on Lp(µ) for 1 ≤ p < ∞, and ∥f∥∞ = infM ∈ [0,∞) : |f | ≤
M µ-a.e. on L∞(µ), then it can be proved with some effort that ∥ ·∥p is a complete norm on Lp(µ)
for 1 ≤ p ≤ ∞ (Minkowski inequality, Riesz-Fischer theorem). Also if f ∈ Lp(µ), g ∈ Lq(µ) and
1/p+ 1/q = 1, then fg ∈ L1(µ) and ∥fg∥1 ≤ ∥f∥p · ∥g∥q (Holder inequality).
Example: In general, there is no inclusion relation between Lp(µ) and Lq(µ) for 1 ≤ p < q ≤ ∞.
Let µ = µL,1 on R and let α ∈ (1,∞). (i) If f : R → R is f =∑∞
n=1 n−(1/α) 1(n,n+1), then f /∈ Lp(µ)
for 1 ≤ p ≤ α and f ∈ Lp(µ) for α < p ≤ ∞. (ii) Let (In) be a sequence of pairwise disjoint intervals
54 T.K.SUBRAHMONIAN MOOTHATHU
with |In| = n−(1+α). If f : R → R is f =∑∞
n=1 n1In , then f ∈ Lp(µ) for 1 ≤ p < α but f /∈ Lp(µ)
for α ≤ p ≤ ∞.
Exercise-52: Let (X,A, µ) be a complete measure space with µ[X] < ∞. Then Lp(µ) ⊃ Lq(µ) for
1 ≤ p ≤ q ≤ ∞. [Hint : |f |p ≤ |f |q on A := x ∈ X : |f(x)| > 1 and |f |p ≤ 1 on X \A.]
Example: [Series summation is discrete analogue of integration] If µ is the counting measure on
(N,P(N)), then Lp(µ) = lp = x ∈ RN :∑∞
n=1 |xn|p <∞ and ∥x∥p = (∑∞
n=1 |xn|p)1/p for x ∈ Lp(µ)
and 1 ≤ p <∞. Similarly, L∞(µ) = l∞ = x ∈ RN : supn∈N |xn| <∞ and ∥x∥∞ = supn∈N |xn| for
x ∈ L∞(µ). Also, lp ⊂ lq if 1 ≤ p ≤ q ≤ ∞ (direction of inclusion opposite to that in Exercise-52).
We will discuss two properties of L1(µ) below.
[154] [Riesz-Fischer theorem] If (X,A, µ) is a complete measure space, then (L1(µ), ∥ · ∥1) is a
complete normed space (corresponding result is true for 1 < p ≤ ∞ also).
Proof. The verification that (L1(µ), ∥ · ∥1) is a normed space is left to the reader. We prove only
completeness. Consider a Cauchy sequence (fn) from (L1(µ), ∥ · ∥1). Since a Cauchy sequence
converges iff a subsequence converges, we may assume ∥fn+1 − fn∥1 ≤ 2−n for every n ∈ N, after
passing onto a subsequence. Let g : X → [0,∞] be defined as g(x) = |f1(x)|+∑∞
n=1 |fn+1(x)−fn(x)|
and let A = x ∈ X : g(x) = ∞. By Exercise-47, we have∫X |g|dµ < ∞, and hence µ[A] = 0 by
Exercise-44. Define f : X → R as f(x) = 0 if x ∈ A and f(x) = f1(x) +∑∞
n=1(fn+1(x)− fn(x)) if
x ∈ X \ A. Then f is measurable, and |f | ≤ g so that∫X |f |dµ ≤
∫X gdµ < ∞. Thus f ∈ L1(µ).
Observe that f1(x) +∑∞
k=1(fk+1(x) − fk(x)) = fn(x) +∑∞
k=n(fk+1(x) − fk(x)) for any n ∈ N.
Hence (fn) → f pointwise on X \ A, and |f − fn| ≤ g for every n ∈ N. Applying [153] to the
sequence (f − fn), we have∫X |f − fn|dµ→ 0, that is, ∥f − fn∥1 → 0 as n→ ∞.
Remark: If 1 ≤ p <∞ and if ϕ =∑n
j=1 aj 1Aj is the canonical representation of a simple function,
then ϕ ∈ Lp(µ) iff µ[Aj ] <∞ for aj = 0.
[155] [Approximation in L1] Let µ = µL,d. Then the following collections are dense in (L1(µ), ∥·∥1):
(i) F1 = ϕ ∈ L1(µ) : ϕ is a simple function.
(ii) F2 = ψ ∈ L1(µ) : ψ is a step function.
(iii) F3 = f ∈ L1(µ) : f is a continuous function with compact support.
Proof. (i) Consider g ∈ L1(µ). First suppose g ≥ 0. By [138](i), there are simple functions ϕn such
that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ g and (ϕn) → g pointwise. We have ϕn ∈ L1(µ) since g ∈ L1(µ). Hence
by [153], we have∫Rd |g − ϕn|dµ → 0, i.e., ∥g − ϕn∥1 → 0 as n → ∞. In the general case write
g = g+ − g−, and consider g+, g− separately to see F1 is dense in (L1(µ), ∥ · ∥1).
MEASURE THEORY 55
(ii) If ϕ ∈ F1 and ε > 0, then we can find ψ ∈ F2 with ∥ϕ−ψ∥1 < ε (∵ write ϕ =∑n
j=1 cj 1Aj with
µ[Aj ] <∞ and approximate each Aj with a finite union of d-boxes by [134]). Hence F2 is dense in
(L1(µ), ∥ · ∥1).
(iii) If A is a d-box and ε > 0, choose an open d-box B such that A ⊂ B and µ[B \ A] < ε. Let
f : Rd → [0, 1] be a continuous function such that f ≡ 1 on A and f ≡ 0 on Rd \B (see Exercise-40
also). Then f ∈ F3 and ∥f − 1A∥1 ≤∫B\A |f |dµ ≤
∫B\A 1dµ = µ[B \A] < ε. By linearity, members
of F3 can approximate any step function ψ ∈ F2. Hence F3 is also dense in (L1(µ), ∥ · ∥1).
Remark: Result [155] remains valid for 1 ≤ p <∞; however the case of L∞(µ) is a little different.
Simple functions are dense in L∞(µ) because of [138](ii), but step functions and continuous functions
are not dense in L∞(µ). If g : R → R is g = 1[0,∞) ∈ L∞(µ), then ∥g−ψ∥ ≥ 1 for any step function
ψ ∈ L∞(µ) and ∥g − f∥∞ ≥ 1/3 for any continuous function f ∈ L∞(µ).
Remark: Let µ = µL,d and F ′2 be the collection of all step functions in L1(µ) of the form
∑nj=1 aj 1Aj ,
where aj ∈ Q and Aj ’s are d-boxes with vertices in Qd. Then, by the help of [155](ii), we may see
that F ′2 is a countable dense subset of L1(µ) (more generally, of Lp(µ) for 1 ≤ p <∞).
Exercise-53: [Translation invariance of Lebesgue integral] Let µ = µL,d and f ∈ L1(µ). Then,∫Rd f(x+y)dµ(x) =
∫Rd f(x)dµ(x) for every y ∈ Rd. [Hint : If A ∈ L(Rd), then
∫Rd 1A(x+y)dµ(x) =∫
Rd 1A−y(x)dµ(x) = µ[A − y] = µ[A] =∫Rd 1A(x)dµ(x). By the linearity given by [150], we have∫
Rd ϕ(x+ y)dµ(x) =∫Rd ϕ(x)dµ(x) for every simple function ϕ. Now use [155].]
Exercise-54: [Decay of Fourier coefficients] Let µ = µL,1. Then we have limn→∞∫R f(x) sin(nx)dµ =
0 = limn→∞∫R f(x) cos(nx)dµ = 0 for every f ∈ L1(µ). [Hint : Let [a, b] ⊂ R and ε > 0. Let n0 ∈ N
be such that 2π/n0 < ε/2. If n ≥ n0, find integers k,m such that a′ = 2kπ/n ∈ (a − ε/2, a) and
b′ = 2mπ/n ∈ (b, b+ε/2). We have |∫[a,b] sin(nx)dµ| ≤ |
∫[a′,b′] sin(nx)dµ|+
∫[a′, b′]\[a,b] | sin(nx)|dµ ≤
0 + ε. This shows limn→∞∫R 1[a,b](x) sin(nx)dµ = 0. Now use linearity and [155](ii).]
Now we mention a few points about convergence in Lp.
Remark: Suppose (fn) → f in L∞(µ) and let nk be such that ∥f − fn∥∞ ≤ 1/k for every n ≥ nk.
Then there are µ-null sets Ak ⊂ X such that |f(x)−fn(x)| ≤ 1/k for every x ∈ X \Ak and n ≥ nk.
So (fn) → f uniformly outside the µ-null set∪∞
k=1Ak. In particular, we see that convergence in
L∞(µ) implies uniform convergence with negligible error, convergence in measure and pointwise
convergence µ-a.e. Behavior in Lp(µ) for 1 ≤ p <∞ is different, as shown below.
Example: Let µ = µL,1 and let (Jn) be the sequence of intervals [0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3],
[2/3, 1], [0, 1/4], . . .. If fn = 1Jn , then (fn) → 0 in Lp(µ) for 1 ≤ p < ∞. Here, this means only
that the area under the graph of fpn goes to 0, so we cannot expect even pointwise convergence. In
56 T.K.SUBRAHMONIAN MOOTHATHU
fact, we know already that (fn(x)) does not converge for any x ∈ [0, 1] as the sequence contains
infinitely many 0’s and infinitely many 1’s. Also (fn) does not converge in L∞(µ).
[156] [Lp convergence - necessity] Let 1 ≤ p <∞ and suppose (fn) → f in Lp(µ). Then,
(i) (fn) → f in measure.
(ii) There is a subsequence (fnk) such that (fnk
) → f uniformly with negligible error.
(iii) There is a subsequence (fnk) such that (fnk
) → f pointwise µ-a.e.
Proof. (i) Let α > 0. If An = x ∈ X : |f(x) − fn(x)| ≥ α, then µ[An] =∫An
1dµ ≤∫Anα−p|f −
fn|pdµ ≤ α−p∥f − fn∥pp → 0 as n→ ∞ for 1 ≤ p <∞. Now (ii) and (iii) follow from [143].
Examples: Let µ = µL,1 on R. (i) Let fn = (1/n)1(0,en). Then (fn) → 0 uniformly and hence in
L∞(µ). But (fn) does not converge in Lp(µ) for 1 ≤ p < ∞ since ∥fn∥pp = en/np → ∞ as n → ∞.
(ii) Fix α ∈ [1,∞) and let fn = (1/n)1(0,nα). Then (fn) → 0 uniformly. Also (fn) → 0 in Lp(µ)
iff p > α. (iii) Fix α ∈ [1,∞), and let fn = n1(n−α,2n−α). Then (fn) → 0 pointwise; and (fn)
converges in Lp(µ) iff p < α. (iv) If fn = 1(n,n+1), then (fn) → 0 pointwise, but not in Lp(µ) for
1 ≤ p ≤ ∞. (v) If fn = n1(1/n,2/n), then (fn) → 0 pointwise as well as in measure, but not in
Lp(µ) for 1 ≤ p ≤ ∞. (vi) Let fn(x) = xn 1(0,1)(x). Then |fn| ≤ 1; (fn) → 0 pointwise but not
uniformly; (fn) → 0 in Lp for 1 ≤ p <∞, but (fn) does not converge in L∞(µ).
[157] [Lp convergence - sufficiency] Let 1 ≤ p <∞ and f, fn ∈ Lp(µ) be such that (fn) → f in one
of the following sense - (i) uniformly with negligible error, (ii) in measure, or (iii) pointwise µ-a.e.
If there is g ∈ Lp(µ) such that |fn| ≤ g µ-a.e., then (fn) → f in Lp(µ).
Proof. Since (i) implies both (ii) and (iii), it suffices to consider only (ii) and (iii). First suppose
(fn) → f pointwise µ-a.e. If 1 ≤ p < ∞, then |f − fn|p ≤ 2pgp µ-a.e. and 2pgp ∈ L1(µ). Hence
by Lebesgue’s dominated convergence theorem, ∥f − fn∥pp =∫X |f − fn|pdµ → 0. Now suppose
(fn) → f in measure. If the convergence does not happen in Lp(µ), there exist δ > 0 and a
subsequence (fnk) such that ∥f − fnk
∥p ≥ δ. By passing onto a further subsequence, we may
assume (fnk) → f pointwise µ-a.e. by [143]. Then (fnk
) → f in Lp(µ) by what we proved above, a
contradiction.
14. Product spaces and product measures
Definition: Let (Xα,Aα) be measurable spaces for α ∈ J , and let X =∏
α∈J Xα. The product
σ-algebra A on X is defined as the smallest σ-algebra on X such that each projection πα : X → Xα
is A-Aα measurable. That is, A is the σ-algebra generated by π−1α (Aα) : α ∈ J and Aα ∈ Aα.
We are mainly interested in finite products, especially the product of two measurable spaces.
Let (Xi,Ai) be measurable spaces for i = 1, 2. Then by definition, the product σ-algebra A on
MEASURE THEORY 57
X = X1 ×X2 is generated by A1 ×X2 : A1 ∈ A1 ∪ X1 × A2 : A2 ∈ A2. It is then easy to see
that A is also generated by A1 ×A2 = A1 ×A2 : A1 ∈ A1 and A2 ∈ A2.
Remarks: (i) Let (X,A1), (Y,A2) be measurable spaces and A be the product σ-algebra on X×Y .
If A ∈ A, then its projections πX(A), πY (A) may not belong to A1,A2 respectively. This is
essentially because projections do not commute with complementation and countable intersection.
To understand it better, consider A′ = A ∈ A : πX(A) ∈ A1, make an attempt to prove A′ is a
σ-algebra, and find out what goes wrong. To see what can be said positively, see Exercise-55 below.
(ii) Lebesgue assumed wrongly that projections would commute with countable intersections, and
this led him to the wrong belief that projections of Borel sets are Borel. Luzin corrected this
mistake. It is a non-trivial fact that a projection of a Borel set need not be Borel.
Example: The projection of a Lebesgue measurable set need not be Lebesgue measurable also. If
A ∈ P(R) \ L(R), then µ∗L,2[A× 0] = 0 and hence A× 0 ∈ L(R2). And A× 0 projects to A.
Definition: For A ⊂ X × Y , let Ax = y ∈ Y : (x, y) ∈ A and Ay = x ∈ X : (x, y) ∈ A.
Exercise-55: Let (X,A1), (Y,A2) be measurable spaces, and let A be the product σ-algebra on
X × Y . If A ∈ A, then Ax ∈ A2 for every x ∈ X and Ay ∈ A1 for every y ∈ Y . [Hint : Let
A′ = A ∈ A : A has the given property. Check that A′ is a σ-algebra containing A1 ×A2.]
Exercise-56: Let (X,A1), (Y,A2) be measurable spaces, and let A be the product σ-algebra on
X × Y . If C1 ⊂ A1, C2 ⊂ A2 are generating collections, then C1 × C2 generates A. [Hint : Let
A′ be the σ-algebra generated by C1 × C2. Clearly, A′ ⊂ A. Now fix C1 ∈ C1 and note that
A′2 := A2 ∈ A2 : C1 × A2 ∈ A′ is a σ-algebra containing C2. Hence A′
2 = A2. Use this to show
similarly that if A′1 = A1 ∈ A1 : A1 × A2 ⊂ A′, then A′
1 = A1. Thus A1 ×A2 ⊂ A′ also.]
[158] [Product measure] Let (Xi,Ai, µi) be measure spaces for i = 1, 2. Let X = X1 × X2 and
A be the product σ-algebra on X. Then there is a measure µ on (X,A) such that µ[A1 × A2] =
µ1[A1]µ2[A2] for every A1 ×A2 ∈ A1 ×A2. This µ is unique if µ1 and µ2 are σ-finite measures.
Proof. Verify that A1 × A2 is a semi-algebra on X. Define µ : A1 × A2 → [0,∞] as µ = µ1 ×
µ2. Suppose A × B ∈ A1 × A2 is such that A × B =∪∞
n=1(An × Bn), a disjoint union with
An × Bn ∈ A1 × A2. Then 1A(x)1B(y) =∑∞
n=1 1An(x)1Bn(y) for (x, y) ∈ X1 × X2. Fixing
x ∈ X1 and integrating with respect to µ2, we obtain by Exercise-47 that 1A(x)∫X2
1B(y)dµ2(y) =∫X2
∑∞n=1 1An(x)1Bn(y)dµ2(y) =
∑∞n=1 1An(x)
∫X2
1Bn(y)dµ2(y) =∑∞
n=1 1An(x)µ2[Bn]. Thus
1A(x)µ2[B] =∑∞
n=1 1An(x)µ2[Bn]. Now integrating this similarly with respect to µ1, we have
µ1[A]µ2[B] =∑∞
n=1 µ1[An]µ2[Bn]. That is, µ[A×B] =∑∞
n=1 µ[An×Bn]. Thus µ is a pre-measure
on the semi-algebra A1 × A2. Hence µ extends to a measure on A by Caratheodory’s extension
theorem [131]. Now suppose µ1 and µ2 are σ-finite. So there are An ∈ A1, Bm ∈ A2 such that
58 T.K.SUBRAHMONIAN MOOTHATHU
X1 =∪∞
n=1An, X2 =∪∞
m=1Bm, µ1[An] < ∞ and µ2[Bm] < ∞. Then X =∪∞
n=1
∪∞m=1(An ×Bm)
and µ[An ×Bm] <∞. Thus µ is σ-finite, and hence the extension is unique by [131].
[159] (i) If B is the product σ-algebra on Rd+k generated by B(Rd)× B(Rk), then B = B(Rd+k).
(ii) L(Rd)× L(Rk) ⊂ L(Rd+k)
(iii) If A is the product σ-algebra on Rd+k generated by L(Rd) × L(Rk), then A = L(Rd+k), but
the completion of A w.r.to µL,d+k is L(Rd+k) and µL,d+k is the unique product measure.
Proof. (i) This follows from Exercise-56 (take C1 = d-boxes and C2 = k-boxes) by noting that
A ⊂ Rd+k is a (d+ k)-box iff A is the product of a d-box with a k-box.
(ii) Any Y ∈ L(Rd)×L(Rk) can be written as Y = (B1 ∪C1)× (B2 ∪C2), where B1 ∈ B(Rd), C1 ∈
N (Rd), B2 ∈ B(Rk), and C2 ∈ N (Rk). Then we see that Y can be written as Y = (B1 ×B2) ∪ Z,
where Z ∈ N (Rd+k). Also B1 ×B2 ∈ B(Rd+k) by part (i). Hence Y ∈ L(Rd+k).
(iii) We have A ⊂ L(Rd+k) by part (ii). If A ∈ P(Rd)\L(Rd), and y ∈ Rk, then µ∗L,d+k[A×y] = 0
and hence A × y ∈ L(Rd+k). But A × y /∈ A by Exercise-55 and thus A = L(Rd+k). Since
B(Rd)×B(Rk) ⊂ L(Rd)×L(Rk), we have B(Rd+k) ⊂ A by part (i). Thus B(Rd+k) ⊂ A ⊂ L(Rd+k).
Now take completions to see L(Rd+k) is the completion of A. Uniqueness of µL,d+k is clear since it
is determined by its value on (d+ k)-boxes.
Remark: If A ∈ B(Rd+k), then Ax ∈ B(Rk) and Ay ∈ B(Rd) for every (x, y) ∈ Rd+k by Exercise-
55 and [159](i). We cannot replace Borel σ-algebra with Lebesgue σ-algebra here because of the
problem explained in the proof of [159](iii). However, it can be proved (with some effort) that if
A ∈ L(Rd+k), then Ax ∈ L(Rk) for almost every x ∈ Rd and Ay ∈ L(Rd) for almost every y ∈ Rk.
[160] [Fubini-Tonelli Theorem] (Rough statement) Let (X,A1, µ1), (Y,A2, µ2) be σ-finite measure
spaces, let µ be the unique product measure onX×Y and let f : X×Y → [−∞,∞] be a measurable
function. If f is either µ-integrable or ≥ 0, then∫X×Y fdµ =
∫X(
∫Y fdµ2)dµ1 =
∫Y (
∫X fdµ1)dµ2.
Remark: The proof of [160] is left as a reading assignment. We just remark that the main step in
the proof is to establish the result when f = 1A for some measurable A ⊂ X × Y . Once this is
done, then the result extends to simple functions by linearity and to arbitrary measurable functions
by approximation. The necessity of the σ-finite assumption can be seen from the following.
Example: Let X = Y = [0, 1], A1 = A2 = B([0, 1]), let µ1 = µL,1 and let µ2 be the counting
measure. If A = (x, y) ∈ [0, 1]2 : x = y and f = 1A, then∫Y f(x, y)dµ2 = µ2[Ax] = µ2[x] = 1
and∫X f(x, y)dµ1 = µ1[A
y] = µ1[y] = 0. Hence∫X(
∫Y fdµ2)dµ1 = 1 = 0 =
∫Y (
∫X fdµ1)dµ2.
MEASURE THEORY 59
15. Fundamental Theorem of Calculus (FTC)
Let µ = µL,1 for this section. We will write∫ ba fdµ for
∫[a,b] fdµ. Let f : [a, b] → R be Lebesgue
integrable and let g : [a, b] → R be g(x) =∫ xa fdµ. Can we say that g is differentiable and g′ = f?
The answer is no; consider for example f = 1[0,1/2] on [0, 1]. An obstruction to getting a positive
answer is the fact that if g is differentiable, then g′ should have intermediate value property (by
Darboux theorem), but f may not have intermediate value property. However, if we ignore what
happens on a small set, we can say something positive.
Definition: (Recall) We say f : [a, b] → R is absolutely continuous if for every ε > 0 there is δ > 0
such that for any finite collection of points x1 ≤ y1 ≤ x2 ≤ y2 ≤ · · · ≤ xn ≤ yn in [a, b] with∑nj=1 |yj − xj | < δ, we have
∑nj=1 |f(yj)− f(xj)| < ε.
Exercise-57: Let µ = µL,1 and let f : [a, b] → R be µ-integrable. (i) If g : [a, b] → R is g(x) =∫ xa fdµ,
then g is absolutely continuous. (ii) If∫ xa fdµ = 0 for µ-a.e. x ∈ [a, b], then f = 0 µ-a.e. [Hint : (i)
Use Exercise-50. (ii) By the continuity of g,∫ xa fdµ = 0 for every x ∈ [a, b]. Hence
∫ yx fdµ = 0 for
a ≤ x ≤ y ≤ b. If U ⊂ [a, b] is open, then U is a disjoint union of countably many intervals, and
hence∫U fdµ = 0 by Exercise-51(ii). IfK ⊂ [a, b] is closed, then
∫K fdµ =
∫ ba fdµ−
∫[a,b]\K fdµ = 0.
Now An = x : f(x) > 1/n and Bn = x : f(x) < −(1/n) are µ-null by inner regularity.]
Exercise-58: Let g : [a, b] → R be increasing. Then∫ ba g
′dµ ≤ g(b) − g(a). [Hint : By [112], g is
differentiable µ-a.e. Also g′ ≥ 0 since g is increasing. Extend g to [a, b+ 1] by setting g(x) = g(b)
for x ≥ b. If hn : [a, b] → R is hn(x) = n(g(x + 1/n) − g(x)), then hn’s are measurable and
(hn) → g′ pointwise µ-a.e. in [a, b]; so g′ is measurable. And, hn ≥ 0. So by Fatou’s lemma,∫ ba g
′dµ ≤ lim infn→∞∫ ba hndµ. Now,
∫ ba hndµ = n
∫ ba (g(x + 1/n) − g(x))dµ = n[
∫ b+1/na+1/n g(x)dµ −∫ b
a g(x)dµ] = n[∫ b+1/nb gdµ−
∫ a+1/na gdµ] ≤ n[
∫ b+1/nb g(b)dµ−
∫ a+1/na g(a)dµ] = g(b)− g(a).]
[161] [FTC - first part] Let µ = µL,1. Let f : [a, b] → R be µ-integrable and g : [a, b] → R be
g(x) =∫ xa fdµ. Then g is absolutely continuous and g′(x) = f(x) for µ-a.e. x ∈ [a, b].
Proof. By Exercise-57, g is absolutely continuous. Now we may assume f ≥ 0 by considering f+
and f− separately. Then g is increasing, and hence differentiable µ-a.e. by [112]. From the hint of
Exercise-58, we also see that g′ is measurable. For later use observe that limδ→0(δ−1
∫ c+δc gdµ) =
g(c) for any c ∈ (a, b) since g is continuous and g(c) = δ−1∫ c+δc g(c)dµ.
Case-1: f is bounded, say |f | ≤ M . Let hn(x) = (g(x + 1/n) − g(x))/(1/n). Then (hn) → g′
pointwise µ-a.e. We have |hn| = |n∫ x+1/nx fdµ| ≤ |n
∫ x+1/nx Mdµ| = M and hence |g′| ≤ M µ-a.e.
also. The constant function M is µ-integrable over [a, b] since [a, b] has finite measure. Hence by
Lebesgue dominated convergence theorem, for any c ∈ (a, b) we have∫ ca g
′dµ =
limn→∞∫ ca hndµ = limn→∞[n
∫ c+1/nc g(x)dµ− n
∫ a+1/na g(x)dµ] = g(c)− g(a) = g(c) =
∫ ca fdµ.
60 T.K.SUBRAHMONIAN MOOTHATHU
So∫ ca (g
′ − f)dµ = 0 for all c ∈ (a, b) and hence g′ = f µ-a.e. by Exercise-57.
Case-2: f is unbounded. Let fn = minf, n and gn(x) =∫ xa fndµ. By case-1, g′n = fn µ-a.e. If
sn(x) =∫ xa (f − fn)dµ, then sn is monotone increasing (∵ f − fn ≥ 0), and hence s′n ≥ 0 µ-a.e.
Since g(x) =∫ xa fdµ = gn(x) + sn(x), we conclude that g′ = g′n + s′n ≥ fn µ-a.e., for each n ∈ N.
By tending n → ∞, we get g′ ≥ f µ-a.e. Now to show that g′ = f µ-a.e., it suffices to show∫ ba (g
′ − f)dµ = 0 because of Exercise-44(ii).
Since g′ ≥ f µ-a.e., we have∫ ba (g
′ − f)dµ ≥ 0. On the other hand, by Exercise-58 we get∫ ba g
′dµ ≤ g(b)− g(a) = g(b) =∫ ba fdµ, or
∫ ba (g
′ − f)dµ ≤ 0, so we are through.
Exercise-59: [Devil’s staircase aka Cantor function] There is an increasing (hence of bounded vari-
ation), continuous surjection f : [0, 1] → [0, 1] such that f ′ = 0 µ-a.e. Let K be the middle-third
Cantor set. First define f on [0, 1] \ K by setting f ≡ 1/2 on (1/3, 2/3), f ≡ 1/4 on (1/9, 2/9),
f ≡ 3/4 on (7/9, 8/9), and so on. Then f on [0, 1] \K is uniformly continuous and hence it has a
continuous extension to [0, 1]. Verify the stated properties. A similar, strictly increasing example
g : [0, 1] → [0, 2] with certain other interesting properties is obtained by taking g(x) = f(x) + x.
Learn more about this from books/other sources.
Remark: The Cantor function is not absolutely continuous because of the following.
Exercise-60: Let f : [a, b] → R be absolutely continuous. If f ′ = 0 µ-a.e., then f is constant. [Hint :
We have to use Vitali’s lemma for this. See p.109 of H.L. Royden, Real Analysis, 3rd Edn.]
[162] [FTC - second part] Let f : [a, b] → R be absolutely continuous. Then f ′ exists µ-a.e., f ′ is
µ-integrable, and∫ ba f
′dµ = f(b)− f(a).
Proof. We know from [114] that f is of bounded variation, and f ′ exists µ-a.e. We may write
f = g − h where g, h are monotone increasing, by (the proof of) [113]. Now f ′ is measurable
because g′, h′ are measurable (see the hint of Exercise-58). Also, |f ′| ≤ g′ + h′. Using Exercise-58
we obtain∫ ba |f ′|dµ ≤
∫ ba (g
′ + h′)dµ ≤ g(b) + h(b)− g(a)− h(a) <∞, and hence f ′ is µ-integrable.
If s(x) =∫ xa f
′dµ, then s is absolutely continuous by Exercise-57, and s′ = f ′ µ-a.e. by [161]. So
the function f −s is absolutely continuous and (f −s)′ = f ′−s′ = 0 µ-a.e. Therefore there is c ∈ R
such that f(x) = s(x)+ c for every x ∈ [a, b] by Exercise-60. Since s(a) = 0, we have c = f(a), and
thus f(x)− f(a) =∫ xa f
′dµ for every x ∈ [a, b].
Remarks: (i) The example of Cantor function shows that we cannot weaken the hypothesis of
absolute continuity in [162] by either the bounded variation or the monotone increasing property.
(ii) From [161] and [162] we can conclude the following: a function f : [a, b] → R is absolutely
continuous iff f ′ exists µ-a.e., f ′ is µ-integrable and f(x) = f(a) +∫ xa f
′dµ for every x ∈ [a, b].
MEASURE THEORY 61
16. Measures on metric spaces
The following definitions are motivated by considering the construction and properties of Lebesgue
(outer) measure on Rd.
Definition: Let X be a metric space and let B(X) denote the Borel σ-algebra on X, i.e., the
σ-algebra generated by open subsets of X (equivalently, by closed subsets of X).
(i) Let µ∗ be an outer measure on X. We say A ⊂ X is µ∗-measurable if A satisfies the Caratheodory
condition: µ∗[E] = µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ X. Recall from [121] that µ∗ restricted
to the σ-algebra of all µ∗-measurable subsets of X is a complete measure.
(ii) An outer measure µ∗ on X is a Borel outer measure if every Borel subset of X is µ∗-measurable.
Similarly, a measure µ on some σ-algebra A on X is called a Borel measure if B(X) ⊂ A.
(iii) A Borel outer measure µ∗ on X is Borel regular if for any A ⊂ X, there is B ∈ B(X) such that
A ⊂ B and µ∗[A] = µ∗[B].
(iv) A Borel measure µ on X is locally finite if X can be covered by open subsets of finite measure.
When X is locally compact, µ is locally finite ⇔ µ[K] <∞ for every compact K ⊂ X.
(v) A Borel measure µ on X is inner regular if µ[B] = supµ[K] : K ⊂ X compact and K ⊂ B
for every B ∈ B(X).
(vi) A Borel measure µ on X is outer regular if µ[B] = infµ[U ] : U ⊂ X open and B ⊂ U for
every B ∈ B(X). Note that inner regularity implies outer regularity when µ[X] < ∞ (apply inner
regularity to Bc), and outer regularity implies inner regularity when X is compact and µ[X] <∞.
Examples: (i) The Lebesgue outer measure on Rd is Borel regular, and the Lebesgue measure on
Rd is locally finite, inner regular, and outer regular.
(ii) Fix y ∈ R and let µy be the Dirac measure at y on (R,P(R)) defined as µy[A] = 1 if y ∈ A
and µy[A] = 0 if y /∈ A. As an outer measure µy is Borel regular. As a measure, µy is locally
finite, inner regular, and outer regular. If we modify the definition and put µ[A] = ∞ if y ∈ A and
µ[A] = 0 if y /∈ A, then µ is not locally finite, though µ remains inner regular and outer regular.
(iii) Write Q = y1, y2, . . . and let µ be the finite Borel measure on (R,P(R)) given by µ =∑∞j=1 2
−jµyj , where µyj is the Dirac measure at yj . For A ⊂ R, the cocountable (hence Borel) set
B := A ∪ (R \ Q) satisfies A ⊂ B and µ[A] = µ[B]. Thus µ is Borel regular. Given B ∈ B(R),
put Kj = B ∩ y1, . . . , yj and Uj = B ∪ y1, . . . , yjc. Then Kj is compact, Uj is open since it is
cofinite, and µ[B] = supµ[Kj ] : j ∈ N = infµ[Uj ] : j ∈ N. Thus µ is inner regular and outer
regular. If we modify µ and define µ =∑∞
j=1 µyj , then µ is neither locally finite nor outer regular
since µ[[0, 1]] = ∞ but µ[0] = 1. However, µ is inner regular: argue as in the case of µ.
62 T.K.SUBRAHMONIAN MOOTHATHU
(iv) LetX be a metric space and µ be the counting measure on (X,P(X)), i.e., µ[A] is the cardinality
of A if A ⊂ X is a finite set, and µ[A] = ∞ otherwise. As an outer measure µ is Borel regular
since any non-Borel subset of X must be infinite. As a measure µ is inner regular since any infinite
subset of X contains finite sets of arbitrarily large cardinality. Except when X is a discrete metric
space, µ is neither locally finite nor outer regular.
(v) Let µ be a Borel measure on (R,B(R)) defined by the condition that µ[B] = 0 or = ∞ according
to whether B ∈ B(R) is of first category or second category in R. Then µ is neither locally finite
nor outer regular since every nonempty open subset of R is of second category in R. Any compact
subset of R contained in R \Q is nowhere dense in R, and hence µ is not inner regular.
(vi) Fix a non-Borel set Y ⊂ R and define a measure µ on (R,P(R)) as µ[A] = ∞ if A∩ Y = ∅ and
µ[A] = 0 if A ∩ Y = ∅ for A ⊂ R. It may be seen that µ, considered as an outer measure, is not
Borel regular since for any Borel set B ⊂ R containing R \ Y we have µ[R \ Y ] = 0 = ∞ = µ[B].
By considering singletons in Y , we see that µ is not locally finite. However, µ is inner regular. Now
suppose Y = Z ∪ (Q+√2), where Z ⊂ R is a non-Borel subset disjoint with Q ∪ (Q+
√2). Then
Y is (non-Borel and) dense in R. Hence µ[U ] = ∞ for any nonempty open set U ⊂ R containing
Q. On the other hand, µ[Q] = 0, and thus µ is not outer regular.
Definition: Let X be a metric space. An outer measure µ∗ on X is called a metric outer measure
if µ∗[A ∪ B] = µ∗[A] + µ∗[B] whenever A,B ⊂ X are subsets with dist(A,B) > 0. For example,
the Lebesgue outer measure on Rd is a metric outer measure by [102](ix).
Exercise-61: Let µ∗ be a metric outer measure on a metric space X, and Aj ⊂ X for j ∈ N.
(i) If dist(Ai, Aj) > 0 for i = j, then µ∗[∪∞
j=1Aj ] =∑∞
j=1 µ∗[Aj ].
(ii) If dist(Aj , Aj+2k) > 0 for every j, k ∈ N, then limn→∞ µ∗[∪n
j=1Aj ] = µ∗[∪∞
j=1Aj ].
[Hint : Enough to show ‘≥’ in both. (i) µ∗[∪∞
j=1Aj ] ≥ µ∗[∪n
j=1Aj ] =∑n
j=1 µ∗[Aj ] for every n ∈ N.
(ii) Let B1 =∪
j∈2N−1Aj and B2 =∪
j∈2NAj . We may assume µ∗[B1] < ∞ and µ∗[B2] < ∞, for
otherwise the result can be deduced using part (i). Now∑
j∈2N−1 µ∗[Aj ] = µ∗[B1] <∞ by (i), and
similarly,∑
j∈2N µ∗[Aj ] = µ∗[B2] < ∞. Therefore, given ε > 0 we have
∑j>n µ
∗[Aj ] < ε for all
large n so that µ∗[∪n
j=1Aj ] + ε ≥ µ∗[∪n
j=1Aj ] + µ∗[∪
j>nAj ] ≥ µ∗[∪∞
j=1Aj ].]
[163] On a separable metric space X, any metric outer measure µ∗ is a Borel outer measure.
Proof. Observe that B(X) is generated by open sets, every open set is a countable union of open
balls by separability, and every open ball is a countable intersection of closed balls. Therefore it
is enough to show every closed ball A := x ∈ X : d(a, x) ≤ r is µ∗-measurable, where a ∈ X
and r > 0. For E ⊂ X, we have Ac ∩ E =∪∞
j=1Ej , where E1 = x ∈ E : d(a, x) > r + 1 and
Ej = x ∈ E : r + 1j < d(a, x) ≤ r + 1
j−1 for j ≥ 2. Letting Yn =∪n
j=1Ej , we see µ∗(Ac ∩ E) =
MEASURE THEORY 63
limn→∞ µ∗[Yn] by Exercise-61(ii). Moreover µ∗[E] ≥ µ∗[(A ∩ E) ∪ Yn] = µ∗[A ∩ E] + µ∗[Yn] since
dist(A ∩ E, Yn) > 0. Thus µ∗[E] ≥ µ∗[A ∩ E] + µ∗[Ac ∩ E]. And ‘≤’ is true always.
Exercise-62: Let µ∗ be a Borel outer measure on a metric space X, and let Y ⊂ X be an arbitrary
subset. Then we may define another outer measure µ∗Y on X as µ∗Y [A] = µ∗[A ∩ Y ]. Then,
(i) µ∗Y is a Borel outer measure (even if Y is not Borel).
(ii) If µ∗ is Borel regular and Y is Borel, then µ∗Y is Borel regular. [Hint : (ii) Given A ⊂ X,
choose B ∈ B(X) with A ∩ Y ⊂ B and µ∗[A ∩ Y ] = µ∗[B]. Then A ∩ Y ⊂ B ∩ Y ⊂ B so that
B1 := [B ∩ Y ] ∪ Y c is a Borel superset of A with µ∗Y [B1] = µ∗[B ∩ Y ] = µ∗[A ∩ Y ] = µ∗Y [A].]
[164] Let µ be a finite Borel measure on a metric space X. Then for any B ∈ B(X) and ε > 0,
there exist a closed set F ⊂ X and an open set U ⊂ X such that F ⊂ B ⊂ U and µ[U \ F ] < ε (in
particular, µ[U \B] < ε and µ[B \ F ] < ε.)
Proof. Let B0 be the collection of all B ∈ B(X) such that for each ε > 0 there exist a closed set
F ⊂ X and an open set U ⊂ X with F ⊂ B ⊂ U and µ[U \F ] < ε. To show B0 = B(X), it suffices
to prove B0 is a σ-algebra containing all closed subsets of X. Clearly ϕ,X ∈ B0. If B ∈ B0, then
Bc ∈ B0 since F c \ U c = U \ F . Suppose Bj ∈ B0 for j ∈ N and ε > 0. Choose Fj ⊂ Bj ⊂ Uj with
Fj closed, Uj open, and µ[Uj \ Fj ] < 2−j−1ε. Let U =∪∞
j=1 Uj , which is open. Choose k ∈ N large
with µ[∪∞
j=1 Fj \∪k
j=1 Fj ] < ε/2 and put F =∪k
j=1 Fj , which is closed. Then F ⊂∪∞
j=1Bj ⊂ U
and µ[U \ F ] ≤ µ[U \∪∞
j=1 Fj ] + µ[∪∞
j=1 Fj \ F ] < ε/2 + ε/2 = ε. Thus∪∞
j=1Bj ∈ B0. If B ⊂ X is
closed, then for Un := x ∈ X : dist(x,B) < 1/n we have µ[B] = limn→∞ µ[Un] since µ[X] < ∞.
Given ε > 0, taking F = B and U = Un for large n, we see B ∈ B0.
We have some corollaries:
[165] Let µ be a Borel measure on a metric space X.
(i) If B ∈ B(X) is with µ[B] <∞, then for any ε > 0 there is a closed set F ⊂ X such that F ⊂ B
and µ[B \ F ] < ε; and consequently, there is an Fσ set E ⊂ X such that E ⊂ B and µ[B \E] = 0.
(ii) [Ulam’s theorem] Suppose X is a complete separable metric space and µ[X] < ∞. Then for
every B ∈ B(X) and ε > 0, there is a compact set K ⊂ X such that K ⊂ B and µ[B \K] < ε.
Consequently, µ is inner regular and outer regular.
(iii) Suppose X is locally compact and second countable and µ is locally finite. Then µ is inner
regular and outer regular. Moreover, for every B ∈ B(X), there exist a σ-compact set E ⊂ X and
a Gδ set G ⊂ X such that E ⊂ B ⊂ G and µ[E] = µ[B] = µ[G].
Proof. (i) Apply [164] to µB as defined in Exercise-62.
64 T.K.SUBRAHMONIAN MOOTHATHU
(ii) Let ε > 0 be given. We claim there is compact K0 ⊂ X with µ[X \ K0] ≤ ε/2. Let xj :
j ∈ N ⊂ X be dense. For each k ∈ N, choose nk ∈ N large enough so that the closed set
Ak :=∪nk
j=1B(xj , 1/k) satisfies µ[X \ Ak] < ε/2k+1. Then K0 :=∩∞
k=1Ak is closed and totally
bounded, and hence compact. Moreover, µ[X \K0] ≤∑∞
k=1 ε/2k+1 = ε/2, proving our claim. Now
for B ∈ B(X), there is closed F ⊂ X with F ⊂ B and µ[B\F ] < ε/2 by (i). Then K := F ∩K0 ⊂ B
is compact and µ[B \K] < ε. Hence µ is inner regular, and also outer regular since µ[X] <∞.
(iii) Write X =∪∞
j=1Kj where Kj ’s are compact and Kj ⊂ int[Kj+1]. Note that µ[Kj ] < ∞ for
every j ∈ N. Consider B ∈ B(X) and ε > 0. By [164], for each j there is a closed set Fj ⊂ X with
Fj ⊂ B and µKj [B \ Fj ] < 1/j. Replacing Fj with Fj ∩Kj , we may assume Fj ⊂ Kj , and then Fj
is also compact. If µ[B] = ∞, then µ[Fj ] → ∞ = µ[B] as j → ∞. If µ[B] < ∞, then for a given
ε > 0 choose j0 large enough with µ[B \Kj0 ] < ε/2 and 1/j0 < ε/2. Then we see µ[B \ Fj0 ] < ε,
and therefore, µ[B] = limj→∞ µ[Fj ], thereby proving inner regularity. To prove outer regularity, it
is enough to consider the case µ[B] <∞. Let ε > 0 be given. Let Vj = int[Kj+1], and choose open
supersets Uj of B with µVj [Uj \B] < 2−jε. Replacing Uj with Uj ∩Vj , we may assume Uj ⊂ Vj , and
put U =∪∞
j=1 Uj . Then U is open, B ⊂ U , and µ[U \B] ≤∑∞
j=1 µ[Uj \B ∩ Vj ] <∑∞
j=1 2−jε = ε.
This establishes outer regularity.
Now we prove the last assertion. If µ[B] = ∞, take G = X, and E =∪∞
j=1 Fj , where Fj ⊂ X is
compact with Fj ⊂ B and µ[Fj ] ≥ j. If µ[B] <∞, let Fj ⊂ Bj ⊂ Uj be such that Fj is compact in
X, Uj is open in X, and µ[Uj \ Fj ] < 1/j. Put E =∪∞
j=1 Fj and G =∩∞
j=1 Uj .
Definition: Let X be a locally compact second countable metric space. Let C(X,R) = f : X →
R : f is continuous and Cc(X,R) = f ∈ C(X,R) : f has compact support.
Exercise-63: Let X be locally compact second countable metric space and let µ, β be locally finite
Borel measures on X.
(i) If µ[K] = β[K] for every compact K ⊂ X, or if µ[U ] = β[U ] for every open U ⊂ X, then µ = β.
(ii) If∫X fdµ =
∫X fdβ for every f ∈ Cc(X,R) with f ≥ 0, then µ = β.
[Hint : (i) is a corollary to [165](iii). To prove (ii), consider compact K ⊂ X, let Un = x ∈ X :
dist(x,K) < 1/n, and fn : X → [0, 1] be continuous with fn ≡ 1 on K and fn ≡ 0 on X \ Un.
Then there is k ∈ N with Un is compact and fn ∈ Cc(X,R) for n ≥ k. Also, (fn) → 1K pointwise.
Using 1Uk∈ L1(µ) ∩ L1(β) as the dominating function, or replacing fn with f1 · · · fn, assuming
0 ≤ fn+1 ≤ fn and using fk ∈ L1(µ)∩L1(β) as the dominating function, apply Lebesgue dominated
convergence theorem to (fn)n≥k. Then, µ[K] =∫X 1Kdµ = limn→∞
∫X fndµ = limn→∞
∫X fndβ =∫
X 1Kdβ = β[K]. Now use (i).]
MEASURE THEORY 65
Exercise-64: Let X be a locally compact second countable metric space and µ be a locally finite
Borel measure on X. Then for 1 ≤ p < ∞, the complete normed space (Lp(µ), ∥ · ∥p) is separable
and Cc(X,R) is dense in it, where ∥f∥p := (∫X |f |pdµ)1/p. [Hint : We argue for the case where p = 1.
The collection of all µ-integrable simple Borel functions on X is dense in L1(µ) by the definition of
integration. Now let U be a countable base of open balls with compact closures in X and let V be all
finite unions of members of U . Consider a µ-integrable simple Borel function g =∑k
i=1 ci 1Bi . Since
µ is regular by [165], we can approximate each Bi in measure by an open supersetW , and then there
is V ∈ V such that µ[W∆V ] is sufficiently small. Also each ci can be approximated by a rational.
Hence the countable collection F = ∑k
i=1 ci 1Vi : k ∈ N, ci ∈ Q, Vi ∈ V is dense in L1(µ). Now
to show Cc(X,R) is dense in L1(µ), it is enough to show that 1V , V ⊂ X open with µ[V ] < ∞,
can be approximated by members of Cc(X,R) since Cc(X,R) is a vector subspace and since F is
dense in L1(µ). Let Kn’s be compact with Kn ⊂ Kn+1 and V =∪∞
n=1Kn. Let fn ∈ Cc(X,R) be
such that 0 ≤ fn ≤ 1, fn ≡ 1 on Kn, and fn ≡ 0 on X \ V . Then, ∥1V − fn∥1 ≤ µ[V \Kn] → 0.]
17. Measure Theory and Functional Analysis
Definition: Let µ, β be measures on an abstract measurable space (X,A). We say β is absolutely
continuous with respect to µ if µ[A] = 0 ⇒ β[A] = 0 for every A ∈ A (notation: β << µ); and we
say µ, β are orthogonal (or singular) if there is A ∈ A with β[A] = 0 = µ[X \A] (notation: µ ⊥ β).
Examples: (i) Let µ be the Lebesgue measure on [0, 1], β1[A] := β[A ∩ [0, 1/2]], and β2[A] :=
[A ∩ [1/2, 1]] for Lebesgue measurable A ⊂ [0, 1]. Then β1, β2 << µ, β1 ⊥ β2 and µ = β1 + β2.
(ii) Let µ be the Lebesgue measure on [0, 1], and let β1, β2 be Borel measures on [0, 1] defined by:
β1[A] = 1 if 0 ∈ A, β1[A] = 0 if 0 /∈ A, and β2[A] =∫A fdµ where f : [0, 1] → R is µ-integrable and
≥ 0 (here, the countable additivity of β2 is checked using the Monotone Convergence Theorem).
Then β1 ⊥ µ and β2 << µ.
(iii) Let f : R → R be a Lipschitz continuous homeomorphism, µ be the Lebesgue measure on R
and define β[A] = µ[f(A)]. Then β is a Borel measure on R. Let λ > 1 be a Lipschitz constant for
f . If B ⊂ R is Borel with µ[B] = 0 and ε > 0, then there are countably many intervals Jn such that
B ⊂∪∞
n=1 Jn and∑∞
n=1 µ[Jn] < ε/λ. If Kn = f(Jn), then µ[Kn] ≤ λµ[Jn] so that∑∞
n=1 µ[Kn] < ε.
Also f(B) ⊂∪∞
n=1Kn. Hence β[B] = µ[f(B)] = 0. Thus β << µ.
Exercise-65: [Reason for the name ‘absolutely continuous’] Let µ, β be finite measures on (X,A).
Then β << µ iff for every ε > 0 there is δ > 0 such that β[A] < ε for every A ∈ A with µ[A] < δ.
[Hint : Let ε > 0. If there is no δ > 0 with the required property, then there are An ∈ A such that
µ[An] < n−2 and β[An] ≥ ε. If B =∩∞
k=1
∪∞n=k An, then µ[B] = 0 and β[B] ≥ ε > 0. To obtain
‘⇐’, note that if β[A] > ε > 0 and if δ > 0 is chosen for this ε, then µ[A] ≥ δ > 0.]
66 T.K.SUBRAHMONIAN MOOTHATHU
Exercise-66: Let µ, β be measures on (X,A) with β << µ and β ⊥ µ. Then β ≡ 0.
Recall the following elementary geometric fact: if u, v are two vectors in the plane, then v can
be written uniquely as v = v1 + v2 so that the vector v1 is parallel to u and v2 ⊥ u. A result of the
same spirit for measures is:
[166] [Lebesgue Decomposition Theorem] (true for σ- finite measures also) Let β, µ be two finite
measures on a measurable space (X,A). Then there are unique measures βa, βs on (X,A) such
that β = βa + βs, βa << µ and βs ⊥ µ.
Proof. Note that β << β + µ. The idea is to obtain a suitable g such that β[A] =∫A gd(β + µ) for
every A ∈ A. Then βa, βs can be defined with the help of g. We have (β + µ)[X] < ∞ and hence
L2(β + µ) ⊂ L1(β + µ). Consider the linear functional ϕ : L2(β + µ) → R given by ϕ(h) =∫X hdβ.
By Cauchy-Schwarz,we have
|ϕ(h)| ≤∫X|h|dβ ≤
∫X|h|d(β + µ) = ∥h∥1 = ∥h · 1∥1 ≤ ∥h∥2∥1∥2 = ∥h∥2[(β + µ)[X]]1/2,
and hence ϕ is bounded. By Riesz representation theorem, there is g ∈ L2(β + µ) such that
ϕ(h) = ⟨h, g⟩. That is,
(i)∫X hdβ =
∫X hgd(β + µ) for every h ∈ L2(β + µ), and
(ii) β[A] =∫A gd(β + µ) for every A ∈ A (taking h = 1A in (i)).
We use (ii) for the rest of the proof; (i) will be used to prove the next theorem. First we claim
that 0 ≤ g ≤ 1. Let An = x ∈ X : g(x) ≤ −n−1 and Bn = x ∈ X : g(x) ≥ 1 + n−1. Using
A = An in (ii), β[An] ≤ −n−1(β+µ)[An] and hence (β+µ)[An] = 0. Similarly, A = Bn in (ii) gives
β[Bn] ≥ (1 + n−1)(β[Bn] + µ[Bn]) which implies β[Bn] = 0 and then µ[Bn] = 0. Thus 0 ≤ g ≤ 1
(β + µ)-almost everywhere.
Let S = x ∈ X : g(x) = 1, T = X \ S. Then, T =∪∞
n=1 Tn, Tn = x ∈ X : g(x) ≤ 1 − n−1.
Taking A = S in (ii), we get µ[S] = 0. For Y ∈ A with µ[Y ] = 0, taking A = Y ∩ Tn in (ii), we
have β[Y ∩ Tn] ≤ (1− n−1)β[Y ∩ Tn] implying β[Y ∩ T ] ≤∑∞
n=1 β[Y ∩ Tn] = 0. Thus if we define
βs[Y ] = β[Y ∩ S], βa[Y ] = β[Y ∩ T ], then βs ⊥ µ, βa << µ and β = βa + βs.
Uniqueness: Suppose β = αa + αs is another decomposition with respect to µ. Then αa[S] = 0
since µ[S] = 0. Hence βs[Y ] = β[Y ∩ S] = αa[Y ∩ S] + αs[Y ∩ S] = αs[Y ∩ S] ≤ αs[Y ]. Thus
βs ≤ αs and hence βa ≥ αa. Then the measure ν = αs − βs = βa − αa satisfies ν ⊥ µ and ν << µ.
Therefore ν ≡ 0.
[167] [Radon-Nikodym Theorem for finite measures] Let β, µ be finite measures on a measurable
space (X,A) and suppose that β << µ. Then there exists a non-negative µ-integrable function
f : X → R such that β[A] =∫A fdµ for every A ∈ A. Moreover, f is unique µ-almost everywhere.
MEASURE THEORY 67
(Thus β << µ⇔ β is obtained by integrating against µ.)
Proof. Let ϕ, g, S, T be as in the proof of [166]. Let β = βa+βs be the Lebesgue decomposition of β
with respect to µ. Since β << µ, we have βa = β and βs = 0 by the uniqueness of the decomposition.
Now, let us make a guess about f . Taking h = f 1A in (i) of the proof of [166] we see that the
required f must satisfy∫A fdβ =
∫A fgd(β+µ), or (adding
∫A fdµ to both sides and rearranging,)∫
A f(1 − g)d(β + µ) =∫A fdµ. On the other hand, we must have
∫A fdµ = β[A] =
∫A gd(β + µ)
by (ii) from the proof of [166], so that we anticipate the identity f(1 − g) = g. Define f : X → R
as f ≡ 0 on S and f = g/(1 − g) on T . Then∫A f(1 − g)d(β + µ) =
∫A∩T f(1 − g)d(β + µ) =∫
A∩T gd(β+µ) = (by (ii) from [166] )β[A∩ T ] = βa[A] = β[A]. But we have already derived above
that∫A f(1− g)d(β + µ) =
∫A fdµ. Thus β[A] =
∫A fdµ for every A ∈ A.
Uniqueness µ-almost everywhere: If h is another such function and if An = x ∈ X : f(x)−h(x) ≥
n−1, then∫Anfdµ = β[An] =
∫Anhdµ and hence 0 =
∫An
(f−h)dµ ≥ n−1µ[An] implying µ[An] = 0
and therefore µ[x ∈ X : f(x) > h(x)] = 0. Similarly µ[x ∈ X : h(x) > f(x)] = 0 also.
[167′] [Radon-Nikodym Theorem] Let β, µ be σ-finite measures on a measurable space (X,A)
and let β << µ. Then there exists a non-negative measurable function f : X → R such that
β[A] =∫A fdµ for every A ∈ A. Moreover, f is unique µ-almost everywhere.
Proof. Write X =∪∞
i=1Xi, a disjoint union of measurable sets Xi with (β + µ)[Xi] < ∞. Get
fi : Xi → R by [167] and let f : X → R be f |Xi = fi.
We know that locally finite Borel measures on locally compact second countable metric spaces
are σ-finite. Therefore, we have:
[167′′] [Radon-Nikodym Theorem for locally compact second countable metric spaces] Let β, µ be
locally finite Borel measures on a locally compact second countable metric space X and let β << µ.
Then there exists a non-negative Borel function f : X → R such that β[A] =∫A fdµ for every Borel
A ⊂ X. Moreover, f is unique µ-almost everywhere.
Remark: The function f given by the Radon-Nikodym Theorem is called the Radon-Nikodym
derivative of β with respect to µ and is denoted by dβdµ . This notation is justified by the following
observation. If g : (X,A) → R is a non-negative measurable function, then∫X gdβ =
∫X g dβ
dµdµ [∵if g = 1A, then
∫X gdβ = β[A] =
∫A
dβdµµ =
∫X g dβ
dµµ; extend this to nonnegative simple functions
by linearity and then use approximation].
Exercise-67: Let µ be the Lebesgue measure on [0, 1], let f : [0, 1] → [0, 1] be a twice differentiable
homeomorphism with f ′ > 0 on (0, 1) (example: f(x) = xn) and define β[A] = µ[f(A)] for Borel
A ⊂ [0, 1]. We already know that β << µ (since f is Lipschitz by Mean value theorem). Show that
68 T.K.SUBRAHMONIAN MOOTHATHU
dβdµ = f ′. [Hint : Since f ′ > 0, f is increasing on (0, 1) and hence on [0, 1]. If [a, b] ⊂ [0, 1], then∫[a,b] f
′dµ =∫ ab f
′(t)dt = f(b) − f(a) = µ[f([a, b])] = β[[a, b]] =∫[a,b]
dβdµdµ. Since closed intervals
generate the Borel σ-algebra on [0, 1], we deduce∫A f
′dµ =∫A
dβdµdµ for every Borel A ⊂ [0, 1]. By
uniqueness, dβdµ = f ′. ]
Exercise-68: Let Y,X be measurable spaces, g : Y → X be a measurable surjection and let β
be a finite measure on Y . Define µ[A] = β[g−1(A)] for measurable A ⊂ X. Then µ is a finite
measure on X and∫Y f gdβ =
∫X fdµ for integrable f : X → R. [Hint : If f = 1A, then∫
Y f gdβ =∫g−1(A) 1dβ = β[g−1(A)] = µ[A] =
∫X fdµ.]
Exercise-69: Let X be a compact metric space and consider C(X,R) with the supremum norm. If
ϕ : C(X,R) → R is linear, then the following are equivalent:
(i) ϕ is positive (i.e., ϕ(f) ≥ 0 for every f ∈ C(X,R with f ≥ 0).
(ii) ϕ(f) ≤ ϕ(g) for f, g ∈ C(X,R) with f ≤ g.
(iii) ϕ is bounded and ∥ϕ∥ = ϕ(1).
[Hint : (i) ⇔ (ii) is easy. To see (ii) ⇒ (iii), consider f ∈ C(X,R) with |f | ≤ 1. Then f ≤ |f | ≤ 1
and −f ≤ |f | ≤ 1 so that ϕ(f) ≤ ϕ(1) and −ϕ(f) = ϕ(−f) ≤ ϕ(1). Hence |ϕ(f)| ≤ ϕ(1), and
equality holds when f = 1. To prove (iii) ⇒ (i), it suffices to show ϕ(f) ≥ 0 for f ∈ C(X,R) with
0 ≤ f ≤ 1. Since 0 ≤ 1− f ≤ 1, ϕ(1)− ϕ(f) = ϕ(1− f) ≤ ∥ϕ∥ = ϕ(1) and so ϕ(f) ≥ 0.]
Remarks: (i) The above result is true for linear ϕ : C(X,C) → C, but the proof is more involved.
(ii) If X is only a locally compact second countable metric space, a positive linear functional
ϕ : (Cc(X,R), sup norm) → R need not be bounded. For instance, consider ϕ : Cc(X,R) → R
given by ϕ(f) =∫R f(x)dx.
Next we show how positive linear functionals on certain function spaces can be identified with
locally finite Borel measures. If µ is a locally finite Borel measure on a locally compact second
countable metric space X and if we define ϕ : Cc(X,R) as ϕ(f) =∫X fdµ, then it is easy to see
that ϕ is a positive linear functional. Our aim is to prove the converse that every positive linear
functional on Cc(X,R) has this form. We need some preparation.
[168] If X is a compact metric space, then there is a continuous surjection f : 0, 1N → X.
Proof. Let A(1), . . . , A(n1) ⊂ X be finitely many nonempty compact sets with diam[A(i)] < 1 and
X =∪n1
i=1A(i). At the (k+1)th step, for each (i1, . . . , ik) ∈∏k
j=11, 2, . . . , nj, choose finitely many
nonempty compact sets A(i1, . . . , ik, 1), . . . , A(i1, . . . , ik, n(k+1)) having diameter < 1/(k + 1) such
that their union equals A(i1, . . . , ik). Since 0, 1N =∏∞
k=10, 1nk , it is easy to find a continuous
surjection g : 0, 1N →∏∞
k=11, 2, . . . , nk. Hence it suffices to find a continuous surjection h :
MEASURE THEORY 69∏∞k=11, 2, . . . , nk → X (for then we may take f = hg). For y = (y1, y2, . . .) ∈
∏∞k=11, 2, . . . , nk
define h(y) to be the unique point in∩∞
k=1A(y1, . . . , yk). Verify that this does the job.
We prove Riesz Representation Theorem for Measures (RRTM) in three steps2: first for 0, 1N,
then for compact metric spaces, and then for locally compact second countable metric spaces.
[169] [RRTM for 0, 1N] Let X = 0, 1N and let ϕ : C(X,R) → R be a positive linear functional.
Then there is a unique finite Borel measure µ on X such that ϕ(f) =∫X fdµ for every f ∈ C(X,R).
Proof. We know from [132] that A := ∅ ∪ all cylinder sets Cw is a semi-algebra on X, where
Cw = (xn) ∈ X : x1 · · ·xk = w for w ∈ 0, 1k. By Exercise-31, the algebra A′ generated by A is
equal to the collection of all finte unions of pairwise disjoint members of A. Also recall that every
member of A, and hence every member of A′, is both open and compact in X. Hence, 1A ∈ C(X,R)
for every A ∈ A′. If µ : A′ → [0,∞) is defined as µ[A] = ϕ(1A), then µ[A ∪ B] = ϕ(1A∪B) =
ϕ(1A + 1B) = ϕ(1A) + ϕ(1B) = µ[A] + µ[B] for disjoint A,B ∈ A′, and thus µ is finitely additive
on A′. Since every member of A′ is both compact and open,∪∞
n=1An /∈ A′ if An ∈ A′ are pairwise
disjoint nonempty sets. Thus trivially, µ is countably additive on A′. Also µ[X] = ϕ(1) ∈ [0,∞).
So µ is a finite pre-measure on A′. Since A′ is also a base for the topology of X, the algebra
generated by A′ is the Borel σ-algebra. So µ extends to a finite Borel measure µ on X by [131].
Let G be the collection of all functions of the form g =∑
w∈0,1k aw1Cw , where k ∈ N and
aw ∈ R. If f ∈ C(X,R), then for every ε > 0, we may find k ∈ N by the uniform continuity of
f such that |f((xn)) − f((yn))| < ε for every (xn), (yn) ∈ X with x1 · · ·xk = y1 · · · yk. From this
observation, it follows that F is dense in C(X,R) with respect to the supremum norm. Define
ψ : C(X,R) → R as ψ(f) =∫X fdµ. Then ψ is linear and positive, and ψ(g) = ϕ(g) for every g ∈ G
by the definition of µ and linearity. As both ϕ and ψ are continuous by Exercise-69, and since G is
dense in C(X,R), it follows that ϕ = ψ, which means ϕ(f) =∫X fdµ for every f ∈ C(X,R).
Uniqueness: If β is another such measure, then β[A] = ϕ(1A) = µ[A] for every A ∈ A′ and then
β[U ] = µ[U ] for every open U ⊂ X by approximation. Then β = µ by Exercise-63.
[169′] [RRTM for compact metric spaces] LetX be a compact metric space and let ϕ : C(X,R) → R
be a positive linear functional. Then there is a unique finite Borel measure µ on X such that
ϕ(f) =∫X fdµ for every f ∈ C(X,R).
Proof. Let Y = 0, 1N and let p : Y → X be a continuous surjection given by [168]. Define
T : C(X,R) → C(Y,R) as T (f) = f p. Then T is a linear isometry and T (f) ≥ 0 whenever
f ≥ 0. So C(X,R) may be identified with a vector subspace of C(Y,R) and ϕ may be considered
as a positive linear functional on this subspace. By Exercise-69, ϕ is bounded and ∥ϕ∥ = ϕ(1). By
2We adapt the proof given by V.S. Sunder, who also mentions an earlier proof by V.S. Varadarajan.
70 T.K.SUBRAHMONIAN MOOTHATHU
Hahn-Banach Theorem, there is a bounded linear functional ϕ : C(Y,R) → R such that ϕ(T (f)) =
ϕ(f) for every f ∈ C(X,R) and ∥ϕ∥ = ∥ϕ∥ = ϕ(1X) = ϕ(T (1X)) = ϕ(1Y ). So ϕ is positive
by Exercise-69. Therefore, by [169] there is a unique finite Borel measure β on Y such that
ϕ(g) =∫Y gdβ for every g ∈ C(Y,R). Define µ[A] = β[p−1(A)] for Borel A ⊂ X. Then µ is a
finite Borel measure on X and∫Y f pdβ =
∫X fdµ for every f ∈ C(X,R) by Exercise-68. Hence
ϕ(f) = ϕ(T (f)) = ϕ(f p) =∫Y f pdβ =
∫X fdµ for every f ∈ C(X,R). Uniqueness of µ follows
from Exercise-63.
Remark: It may be noted that ∥ϕ∥ = ϕ(1) =∫X 1dµ = µ[X], so that positive linear functionals of
norm 1 on C(X,R) correspond to Borel probability measures on X, when X is a compact metric
space.
Exercise-70: Let Y be a compact metric space and ψ : C+(Y,R) → R be such that ψ(f + h) =
ψ(f) + ψ(h) and ψ(αf) = αψ(f) for every f, h ∈ C+(Y,R) and α ≥ 0. Then ψ has a unique
extension to a linear functional on C(Y,R). [Hint : Let f1, f2, h1, h2 ∈ C+(Y,R) and suppose
f1 − f2 = h1 − h2. Then f1 + h2 = h1 + f2 so that by applying ψ and rearranging we obtain
ψ(f1) − ψ(f2) = ψ(h1) − ψ(h2). For f ∈ C(Y,R), define ψ(f) = ψ(h1) − ψ(h2), where h1, h2 ∈
C+(Y,R) are such that f = h1−h2 (for example, f = f+−f−). Then this extension ψ is well-defined
by the above argument. Check other properties.]
[169′′] [Riesz Representation Theorem for Measures for locally compact second countable metric
spaces] Let X be a locally compact second countable metric space and let ϕ : Cc(X,R) → R be
a positive linear functional. Then there is a unique locally finite Borel measure µ on X such that
ϕ(f) =∫X fdµ for every f ∈ Cc(X,R).
Proof. Step-1: Let K ⊂ X be compact and Un = N1/n(K). Note that Un+1 ⊂ Un for all n and Un
is compact for all large n. Let gn ∈ C(X, [0, 1]) be such that gn ≡ 1 on Un+1 and gn ≡ 0 on X \Un.
Then 0 ≤ gn+1 ≤ gn ≤ 1.
Let C+(K,R) = f ∈ C(K,R) : f ≥ 0. If f ∈ C+(K,R), we see that there is f ∈ Cc(X,R) with
f ≥ 0 and f |K = f (∵ choose open U with K ⊂ U , let f = f on K, f ≡ 0 on X \ U and apply
Tietze). Since 0 ≤ fgn+1 ≤ fgn, (ϕ(fgn)) is a decreasing sequence of nonnegative reals and hence
converges. Define ϕK : C+(K,R) → R as ϕK(f) = limn→∞ ϕ(fgn).
To verify that ϕK(f) is independent of the choice of the extension f , consider another extension
f∗ ∈ Cc(X,R) of f with f∗ ≥ 0. Suppose that Un is compact for every n ≥ n0. Given ε > 0
let An = x ∈ Un : |f(x) − f∗(x)| ≥ ε for n ≥ n0. Then An’s are compact, An+1 ⊂ An and∩n≥n0
An ⊂∩
n≥n0Un = K so that
∩n≥n0
An = ∅ (∵ f = f = f∗ on K). So there is n(ε) ≥ n0
such that An(ε) = ∅. Then |f(x) − f∗(x)| < ε for every x ∈ Un(ε). Now gn ≡ 0 on X \ Un(ε) for
MEASURE THEORY 71
n ≥ n(ε). Hence |fgn − f∗gn| ≤ εgn ≤ εgn0 for every n ≥ n(ε) ≥ n0. That is, fgn − f∗gn ≤ εgn0
and f∗gn − fgn ≤ εgn0 for n ≥ n(ε). Applying ϕ, |ϕ(fgn)− ϕ(f∗gn)| ≤ εϕ(gn0) for every n ≥ n(ε)
which implies limn→∞ ϕ(fgn) = limn→∞ ϕ(f∗gn).
Next note that ϕK has the following properties: ϕK(f + h) = ϕK(f) + ϕK(h) and ϕK(αf) =
αϕK(f) for every f, h ∈ C+(K,R) and α ≥ 0. By Exercise-70, ϕK extends uniquely to a linear
functional ϕK on C(K,R). The extended linear functional ϕK : C(K,R) → R is positive (∵ if
f ≥ 0, then ϕK(f) = limn→∞ ϕ(fgn) ≥ 0 since ϕ is positive). So by [169′], there is a unique
finite Borel measure µK on K with ϕK(f) =∫K fdµK . Also observe that ϕK(f |K) = ϕ(f) for
nonnegative f ∈ Cc(X,R) with supp(f) ⊂ K (∵ taking f = f , ϕK(f |K) = limn→∞ ϕ(fgn) = ϕ(f)
since fgn = f).
Step-2: Write X =∪∞
j=1Kj , where Kj ’s are compact and Kj ⊂ int[Kj+1]. Step-1 gives a positive
linear functional ϕj : C(Kj ,R) → R defined using ϕ and let µj be the unique finite Borel measure
on Kj corresponding to ϕj . We claim that µj+1|Kj = µj .
Choose gn’s for Kj as in Step-1 and for f ∈ C+(Kj ,R), let f ∈ Cc(X,R) be an extension with
f ≥ 0 and supp(f) ⊂ Kj+1. Then supp(fgn) ⊂ Kj+1 and hence ϕj+1(fgn|Kj+1) = ϕ(fgn) for all
large n by the observation at the end of Step-1. Hence ϕ(fgn) =∫Kj+1
fgndµj+1 for all large n
and therefore,∫Kjfdµj = ϕj(f) = limn→∞ ϕ(fgn) = limn→∞
∫Kj+1
fgndµj+1, which is equal, by
Dominated convergence theorem, to∫Kj+1
f 1Kjdµj+1 =∫Kjfd(µj+1|Kj ). Hence the claim holds
by the uniqueness part in [169′].
For Borel A ⊂ X, define µ[A] = limj→∞ µj [A ∩ Kj ] ∈ [0,∞]. It may be checked that µ is a
Borel measure. For countable additivity, note the fact that if (xn,j) is a double sequence of reals
increasing in each variable, then limn→∞ limj→∞ xn,j = limj→∞ limn→∞ xn,j , and use this with
xn,j = µj [An ∩ Kj ]. If K ⊂ X is compact, then K ⊂ Kj for some j and this implies µ[K] ≤
µj [K] < ∞ and so µ is locally finite. Also, if f ∈ Cc(X,R) is nonnegative, then supp(f) ⊂ Kj for
some j so that ϕ(f) = ϕj(f |Kj ) =∫Kjfdµj =
∫Kjfdµ =
∫X fdµ, and therefore ϕ(f) =
∫K fdµ for
every f ∈ Cc(X,R) since f = f+ − f−.
Uniqueness: if β is another such measure and if βj = β|Kj , then βj = µj by the uniqueness in
[169′] and hence β = µ.
Remark: RRTM has other formulations when X is a compact metric space: (i) bounded linear
functional on C(X,R) correspond to finite signed real-valued Borel measures, (ii) bounded linear
functional on C(X,C) correspond to ‘finite’ complex-valued Borel measures, etc.
72 T.K.SUBRAHMONIAN MOOTHATHU
Remark: LetX be a compact metric space. Then the dual of the Banach space (C(X,C), sup norm)
can be identified with M(X) := all complex-valued ‘finite’ Borel measures on X by Riesz Theo-
rem. Now recall from Functional Analysis that if Y is a Banach space, then the weak* topology
on the dual Y ∗ is the smallest topology such that the evaluation maps ϕ → ϕ(y) from Y ∗ to C
are continuous for every y ∈ Y ; the closed unit ball of Y ∗ is weak* compact (Alaoglu’s Theorem)
and it is weak* metrizable if Y is separable. Hence the closed unit ball of M(X) = C(X,C)∗ is
weak* compact and weak* metrizable. For µ, µn ∈ M(X), (µn) → µ in the weak* topology iff
(∫X fdµn) →
∫X fdµ in C for every f ∈ C(X,C), since ϕ 7→ ϕ(f) =
∫X fdµ is the evaluation
map if ϕ ∈ C(X,C)∗ corresponds to µ ∈ M(X). If (µn) is a bounded sequence in M(X), then
it is contained in some closed ball (which is compact and metrizable in the weak* topology) and
hence there exist µ ∈ M(X) and a subsequence (µnk) such that (µnk
) weak* converges to µ, or
equivalently (∫X fdµnk
) →∫X fdµ for every f ∈ C(X,C).
Reading assignment: Theorem 6.19 of W. Rudin, Real and Complex Analysis, 3rd Edn.
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