a primer of lebesgue integration with a …math.uchicago.edu/~may/reu2013/reupapers/skenderi.pdfa...

A PRIMER OF LEBESGUE INTEGRATION WITH A VIEW TO

THE LEBESGUE-RADON-NIKODYM THEOREM

MISHEL SKENDERI

Abstract. In this paper, we begin by introducing some fundamental con-cepts and results in measure theory and in the Lebesgue theory of integration.We then introduce some functional-analytic concepts and results that will benecessary for the proof of the Lebesgue-Radon-Nikodym Theorem. In the Ap-pendix, we construct the Lebesgue measure on Rk and present an importantexample that features the Lebesgue measure. We assume a background in in-troductory mathematical analysis and point-set topology; however, we assumeno background in measure theory.

Contents

1. Introduction 12. Basic Measure-Theoretic Concepts, the Lebesgue Integral, and Related

Results 23. The Uniqueness of Bounded Linear Functionals De�ned on Hilbert

Spaces 124. A Construction of Lp on Arbitrary Measure Spaces 165. The Lebesgue-Radon-Nikodym Theorem 196. Appendix: A Construction of the Lebesgue Measure on Rk and a

Relevant Example Involving the Lebesgue Measure 26Acknowledgments 30References 31

1. Introduction

The Lebesgue theory of integration is of great importance in mathematics. TheLebesgue integral has several advantages over its historical predecessor, the Rie-mann integral. It allows for greater variety and �exibility than does the Riemannintegral in several avenues, including the available classes of integrable functionsand the available classes of domains of integration.The Lebesgue-Radon-Nikodym Theorem, which we will prove in this paper, is aremarkable result in measure theory that is of interest in several disciplines, amongthem measure theory, functional analysis, probability theory, �nancial mathemat-ics, statistics, economics, physics, and engineering. It is also a noteworthy result inand of itself.We begin with some elementary de�nitions.

Date: August 31, 2013.

1

2 MISHEL SKENDERI

2. Basic Measure-Theoretic Concepts, the Lebesgue Integral, and

Related Results

De�nitions 2.1. Let X be a set. A σ-algebra M in X is a collection of subsets ofX that satis�es the following properties:

(i) X ∈M.(ii) If A ∈M, then Ac := (X −A) ∈M.(iii) Suppose that for every n ∈ N, An ∈M. Then A := ∪n∈NAn ∈M.

If M is a σ-algebra in X, then the pair (X,M) is said to be a measurable space.The elements of M are said to be the measurable sets in X.

Remarks 2.2. (a) If X is a measurable space, then by (i) and (ii) in the abovede�nition, ∅ = Xc ∈ M.

(b) Due to (ii) and (iii), M is closed under complements and countable unions.Note that if M is closed under countable unions, then it is also closed under�nite unions, since for every n su�ciently large, we can take An = ∅.

(c) By DeMorgan's Law, ∩n∈NAn = (∪n∈NAcn)c, so M is closed under count-able intersections and thus �nite ones.

(d) Because A − B = A ∩ Bc, it follows that if A ∈ M and B ∈ M, thenA−B ∈M.

De�nition 2.3. Suppose X is a measurable space and Y is a topological space.If f is a function from X into Y , f is measurable if given any open set V in Y ,fpre(V ) is a measurable set in X.

Proposition 2.4. Suppose Y and Z are topological spaces, and let g : Y → Z becontinuous. Suppose X is a measurable space. If f : X → Y is measurable and ifh : X → Z is given by h = g ◦ f , then h is measurable.

Proof. Let V be open in Z. Since g is continuous, gpre(V ) is open in Y, andsince f is measurable, fpre(gpre(V )) is a measurable set in X. Because hpre(V ) =fpre(gpre(V )), the proof is complete. �

This conclusion is often stated as continuity preserves measurability.

Proposition 2.5. Suppose X is a measurable space and Y is a topological space.Let u and v be measurable functions from X into R, and let Φ : R2 → Y be acontinuous function. If h : X → Y is given by h(x) = Φ(u(x), v(x)), then h ismeasurable.

Proof. De�ne f : X → R2 by f(x) = (u(x), v(x)). Then h = Φ ◦ f , and as Φ iscontinuous, it su�ces to prove (by the previous result) that f is measurable. Now,let V in R2 be open. If A is any open rectangle in R2 whose sides are parallel to thecoordinate axes, then there exist two open intervals α1 and α2 in R such that A =α1×α2 , and f

pre(A) = upre(α1)∩vpre(α2), which is measurable, by measurabilityof u and v. Since there exist such open rectangles An with V = ∪n∈NAn, fpre(V )is measurable, for fpre(V ) = fpre(∪n∈NAn) = ∪n∈Nfpre(An). �

Corollary 2.6. Let X be a measurable space.

(a) If u and v are measurable functions that map X into R and f = u +iv, then f : X → C is measurable.

(b) If f : X → C is measurable, then Re(f), Im(f), and |f | are measurable.

A PRIMER OF LEBESGUE INTEGRATION WITH A VIEW TO THE LEBESGUE-RADON-NIKODYM THEOREM3

(c) If f : X → C and g : X → C are measurable, then f + g and fg aremeasurable.

Proof. (a) This follows by taking Φ to be the identity function on C.(b) For z ∈ C, take g(z) = Re(z), Im(z), and |z|, respectively, in Proposition

2.4. (Any projection is continuous and so is the absolute value function.)(c) For real-valued f and g, set Φ(u, v) = u + v and Φ(u, v) = uv, respectively.

If f and g are complex-valued, then Re(f), Im(f), Re(g), and Im(g) aremeasurable, by (b). Since f + g = (Re(f) + Re(g)) + (Im(f) + Im(g)), thedesired result follows from Proposition 2.5 and part (a) of this Corollary.

�

De�nition 2.7. Suppose (X,M) is a measurable space. If E ⊆ X, then thefunction 1E : X → R given by

f(x) =

{1, x ∈ E0, x /∈ E

is called the indicator function of E. If E is measurable, then 1E is measurable,for let A ⊆ R be open. If 0 ∈ A and 1 ∈ A, then fpre(A) = X. If 0 ∈ A and 1 /∈ A,then fpre(A) = Ec. If 0 /∈ A and 1 ∈ A, then fpre(A) = E. If 0 /∈ A and 1 /∈ A,then fpre(A) = ∅. It follows 1E is measurable.

Proposition 2.8. Let X be a measurable space. If f : X → C is measurable, thenthere exists g : X → C such that |g| = 1 and f = g|f |.

Proof. Let E = fpre(0), and let Y = C − {0}. Note that Y is open in C. Forz ∈ Y , de�ne h(z) = z

|z| , and for x ∈ X, de�ne g(x) = h(f(x) + 1E(x)). Note that

g is well-de�ned, and if x ∈ E, then g(x) = 1, and if x /∈ E, then g(x) = f(x)|f(x)| ,

which implies |g| = 1 on X. Since f is measurable, E = (fpre(Y ))c is measurable.It follows 1E is measurable. Therefore, g is measurable, from Corollary 2.6(c) andProposition 2.4, since h is continuous on Y . �

Theorem 2.9. If F ⊆ P (X), then there exists a smallest σ-algebra M∗ in X suchthat F ⊆M∗. By "smallest", we mean that if M is any σ-algebra such that F ⊆M,then M∗ ⊆M. M∗ is said to be the σ-algebra generated by F.

Proof. Let Ψ = {M : F ⊆ M and M is a σ-algebra in X}. Notice Ψ 6= ∅, asP (X) ∈ Ψ. De�ne M∗ = ∩M∈Ψ M. Clearly, F ⊆ M∗ and if M∗ were a σ-algebra, then it would be the smallest one containing F. We now show M∗ is aσ-algebra. Obviously, X ∈ M∗. Let A ∈ M∗. Then given any M ∈ Ψ, it followsA ∈M, which implies Ac ∈M, which shows Ac ∈M∗. Finally, let {An}n∈N ⊆M∗.Then given any M ∈ Ψ, {An}n∈N ⊆ M. This implies ∪n∈NAn ∈ M, which shows∪n∈NAn ∈M∗. Thus, M∗ is indeed a σ-algebra. �

De�nitions 2.10. Suppose (X,T(X)) is a topological space. By the above theo-rem, there exists a smallest σ-algebra that contains T(X). We denote this σ-algebrabyB(X). The elements ofB(X) are known as the Borel sets ofX. As particular ex-amples of noteworthy Borel sets, we introduce the Fσ and Gδ sets. An Fσ is a count-able union of closed sets, and a Gδ is a countable intersection of open sets. Now,consider the measurable space (X,B(X)) and let (Y,T(Y ) be a topological space. If

4 MISHEL SKENDERI

f : X → Y is continuous, then V ∈ T(Y ) implies fpre(V ) ∈ T(X) ⊆ B(X). There-fore, every continuous function is Borel measurable; that is to say, the preimage ofany open set is a Borel set. We call Borel measurable functions Borel functions.

Before proceeding any further, it is necessary to make a small, but very impor-tant, de�nition.

De�nition 2.11. We assume the reader is familiar with the extended real numbersystem (see pages 11-12 of [1] for a discussion), and we now declare 0·∞ =∞·0 = 0,a convention that we will use frequently.

Proposition 2.12. Suppose (X,M) is a measurable space, Y is a topological space,and f : X → Y .

(a) If Y = [−∞,∞] and for every α ∈ R, fpre((α,∞]) ∈M, then f is measur-able.

(b) Suppose Z is a topological space, f is measurable, and g : Y → Z is a Borelfunction. If h : X → Z is given by h = g ◦ f , then h is measurable.

Proof. (a) Given any α ∈ R, if fpre((α,∞]) ∈ M, then (fpre((α,∞]))c =fpre([−∞, α]) ∈ M. Thus, fpre([−∞, α)) = (∪n∈Nfpre([−∞, α − 1

n ])) ∈M. Now, let β and γ ∈ R be such that β < γ. Then fpre((β, γ)) =(fpre([−∞, γ)) ∩ fpre((β,∞])) ∈ M. Now if V ⊆ [−∞,∞] is open, thenthere exists a sequence of open intervals (βn, γn) such that V = ∪n∈N(βn, γn).Thus, fpre(V ) ∈M, as desired.

(b) First, we show if N = {E ⊆ Y : fpre(E) ∈ M}, then N is a σ-algebrain Y that contains all the Borel sets of Y . Since X ∈ M, Y ∈ N. If Ein N, then fpre(E) ∈ M, which implies (fpre(E))c = fpre(Ec) ∈ M, sothat Ec ∈ N. Now, suppose {En}n∈N ⊆ N. Then for each n, fpre(En) ∈M, which implies fpre((∪n∈NEn)) = (∪n∈Nfpre(En)) ∈ M. This, in turn,implies (∪n∈NEn) ∈ N. Thus, N is a σ-algebra. Clearly, if U ⊆ Y is open,then U ∈ N. By minimality of B(Y ), B(Y ) ⊆ N. Now, let V ⊆ Z beopen. Then gpre(V ) ∈ B(Y ) ⊆ N. Thus, fpre(gpre(V )) ∈ M, and sincehpre(V ) = fpre(gpre(V )), the result follows.

�

Remarks 2.13. Inspecting the proof of (a), we see that we have also proved thefollowing three statements: (1) If the preimage of any open interval is a mea-surable set, then f is measurable. (2) If Y = [−∞,∞] and for every α ∈ R,fpre([−∞, α]) ∈ M, then f is measurable. (3) If Y = [−∞,∞] and for everyα ∈ R, fpre([−∞, α)) ∈M, then f is measurable. It is also true, and easy to show,that if Y = [−∞,∞] and for every α ∈ R, fpre([α,∞]) ∈M, then f is measurable.Conclusion (b) is often phrased as Borel measurability preserves measurability. Itis a generalization of the fact that continuity preserves measurability.

De�nitions 2.14. For completeness, we give the following de�nitions: If {αn} isa sequence in [−∞,∞], then lim supαn = infn∈N supm≥n αm, and lim inf αn =supn∈N infm≥n αm. lim supαn and lim inf αn are called the limit superior andthe limit inferior, respectively, of {αn}. Recall that if limn→∞ αn exists, thenlimn→∞ αn = lim supαn = lim inf αn.If necessary, consult any introductory real analysis textbook for a discussion of thelimit superior and the limit inferior, whose properties we shall frequently use with-out necessarily saying so.


If X is a set and {fn} is a sequence of functions such that for every n ∈ N, fn : X →[−∞,∞], then if x ∈ X is given, we de�ne the functions sup fn : X → [−∞,∞]and lim sup fn : X → [−∞,∞] by x 7→ sup(fn(x)) and x 7→ lim sup(fn(x)), respec-tively.inf fn : X → [−∞,∞] and lim inf fn : X → [−∞,∞] are de�ned analogously.

Proposition 2.15. Let X be a measurable space. If fn : X → [−∞,∞] is mea-surable for every n ∈ N, g = supn∈N fn, and h = lim sup fn, then g and h aremeasurable.

Proof. Given any α ∈ R, it is easy to verify gpre((α,∞]) = ∪n∈Nfnpre((α,∞]).This implies g is measurable. Analogously, it can be shown that if g∗ = infn∈N fn,then g∗ is measurable. This implies h = infn∈N supm≥n fm is measurable. We cangive a similar proof to show that if h∗ = lim inf fn, then h

∗ is measurable. �

Corollary 2.16. Suppose X is a measurable space.

(a) If fn : X → C is measurable for every n ∈ N, and f is the pointwise limitof {fn}, then f is measurable.

(b) If f : X → [−∞,∞] and g : X → [−∞,∞] are measurable, then so aremax{f, g} and min{f, g}.

Proof. (a) There exist real-valued functions r and s such that f = r + is.Moreover, for every n, there exist real-valued functions rn and sn such thatfn = rn + isn. Thus, {rn} and {sn} converge pointwise to r and s, respec-tively. By Corollary 2.6(b), for every n, rn and sn are measurable. By theabove Proposition, r and s are measurable, and by Corollary 2.6(a), f ismeasurable.

(b) This follows by taking f1 = f , f2 = g, and fn = 0 for n > 2, in the aboveProposition.

�

De�nitions 2.17. We now give some de�nitions related to the above Corollarythat are important in integration theory. If X is a set, and f : X → [−∞,∞], wede�ne

f+ = max{f, 0}, f− = −min{f, 0}.Notice

|f | = f+ + f−, f = f+ − f−.(If x is such that f(x) is �nite, then the above decomposition can be checked usingthe absolute value formulae for max{f, 0} and min{f, 0}. If x is such that f(x) isnot �nite, then simple computations show that the above decomposition remainstrue.)

De�nitions 2.18. Suppose (X,M) is a measurable space. If s : X → C is afunction such that s(X) is �nite, then s is a simple function. The nonnegativesimple functions are those simple functions whose image lies in [0,∞). If s(X) ={α1, . . . , αn}, and if we de�ne Aj = fpre(αj), then s =

∑nj=1 αj1Aj

, as is easy tocheck. We now show s is measurable if and only if each Aj is measurable. Suppose�rst that s is measurable. Then given any αj , s

pre(αj) = (spre({αj}c))c, whichproves Aj is measurable, since {αj}c is open and M is closed under complements.Suppose, conversely, that each Aj is measurable. Then each 1Aj

is measurable.

6 MISHEL SKENDERI

Now, consider the function h : X → C given by h(x) = αj , and suppose V ⊆ Cis open. If αj ∈ V , then hpre(V ) = X. If αj /∈ V , then hpre(V ) = ∅. Thus, h ismeasurable. Corollary 2.6(c) now implies s is measurable.

Theorem 2.19. Suppose X is a measurable space, and let f : X → [0,∞] bemeasurable. Then there exists a sequence {sn} of measurable simple functions suchthat for any n, 0 ≤ sn ≤ sn+1 ≤ f , and for any x in X, sn(x) converges to f(x).

Proof. For any n, de�ne δn = 2−n. Then for any x ∈ [0,∞) and for any n, thereexists a unique nonnegative integer kn(x) such that kn(x) ≤ x

δn< 1 + kn(x). Given

any n, de�ne φn : [0,∞]→ [0,∞) by

φn(x) =

{kn(x)δn, x ∈ [0, n)n, x ∈ [n,∞]

It is clear that each φn is nonnegative. We now show that each φn is simple. Letn be given and let x ∈ [0,∞]. If x ∈ [n,∞], then φn(x) = n. If x ∈ [0, n), thenφn(x) = kn(x)δn ≤ x < n. Clearly, there exist only �nitely many nonnegativeintegers k such that kδn < n. This proves φn is simple. We now show that for eachx ∈ [0,∞], φn(x) converges to x: If x = ∞, then φn(x) = n converges to ∞. Ifx ∈ [0,∞), then for any n > x, φn(x) = kn(x)δn, which implies 0 ≤ x−φn(x) < δn.This gives the desired result. It is not di�cult to show that for each x ∈ [0,∞]and for each n, φn(x) ≤ φn+1(x), and it is also not di�cult to show that each φnis a Borel function. However, the veri�cation is rather lengthy on account of thevarious cases that have to be considered; we therefore omit it. Now, since Borelmeasurability preserves measurability, if, for each n, we de�ne sn = φn ◦ f , then itfollows that sn : X → [0,∞) is a simple measurable function such that for each n,0 ≤ sn ≤ sn+1 ≤ f and such that f is the pointwise limit of the sn. �

This theorem is often summarized as, "Every nonnegative measurable functioncan be approximated from below by nonnegative measurable simple functions."

De�nitions 2.20. Let (X,M) be a measurable space. If µ : M → [0,∞] isa function that satis�es the following properties, then µ is a measure on M, orsimply a measure, when the σ-algebra is clear from the context.

(a) There exists E ∈M such that µ(E) <∞.(b) µ is countably additive: If {Ai}i∈N ⊆M is pairwise disjoint (that is, if i 6= j

implies Ai ∩ Aj = ∅), then µ(∪∞i=1Ai) =∑∞i=1 µ(Ai). (The nonnegativity

of µ implies∑∞i=1 µ(Ai) is either a nonnegative �nite number or ∞; that

is,∑∞i=1 µ(Ai) is always a well-de�ned element of [0,∞]. Notice also that

the order of summation does not matter.)

If µ(X) <∞, then µ is a �nite measure.A complex measure on M is a countably additive function that maps M into C.

If A ⊆ P (X), and µ : A→ [0,∞] or µ : A→ C, then µ is said to be a set function.

De�nition 2.21. A measure space is a triple (X,M, µ), where µ is a measure onM. Frequently, we will simply say "X is a measure space".

Proposition 2.22. Let (X,M, µ) be a measure space. Then the following hold:

(a) µ(∅) = 0.(b) µ is �nitely additive: If A1, . . . , Am are pairwise disjoint elements of M,

then µ(∪mi=1Ai) =∑mi=1 µ(Ai).


(c) µ is monotonic: If A and B are elements of M such that A ⊆ B, thenµ(A) ≤ µ(B).

(d) µ is continuous from below: If {An}n∈N ⊆M, An ⊆ An+1 for every n, andA = ∪n∈NAn, then limn→∞ µ(An) = µ(A).

(e) µ is continuous from above: If {An}n∈N ⊆M, µ(A1) <∞, An+1 ⊆ An forevery n, and A = ∩n∈NAn, then limn→∞ µ(An) = µ(A).

(f) µ is countably subadditive: If {An}n∈N ⊆M and A = ∪n∈NAn, then µ(A) ≤∑∞n=1 µ(An).

(g) µ is �nitely subadditive: If A1, . . . , Am are elements ofM, then µ(∪mi=1Ai) ≤∑mi=1 µ(Ai).

(h) µ is �nite if and only if µ is bounded.

Proof. (a) There exists A such that µ(A) <∞. Set A1 = A and for n > 1, setAn = ∅. The result now follows from countable additivity.

(b) For n > m, set An = ∅. The result now follows from (a) and countableadditivity.

(c) We have (B − A) ∈ M and µ((B − A)) ≥ 0. By �nite addivitity, µ((B −A)) + µ(A) = µ(B). If µ(A) = ∞, then clearly, µ(B) = ∞. If µ(A) < ∞,then we may subtract it to obtain µ(B) − µ(A) ≥ 0. Since µ(A) is �nite,we may add it to obtain µ(B) ≥ µ(A).

(d) De�ne B1 = A1 and for every n > 1, de�ne Bn = (An − An−1). Then{Bn}n∈N ⊆ M is pairwise disjoint, An = ∪ni=1Bn for every n, and A =∪n∈NBn. By �nite and countable additivity, respectively, µ(An) =

∑ni=1 µ(Bi)

and µ(A) =∑∞i=1 µ(Bi). This completes the proof of (d).

(e) For every n ∈ N, de�ne Cn = A1 − An. Then for each n, Cn ⊆ Cn+1,and by �nite addivity, µ(Cn) +µ(An) = µ(A1). Since monotonicity impliesµ(An) ≤ µ(A1) < ∞, subtraction yields µ(Cn) = µ(A1) − µ(An). Now,∪n∈NCn = (A1 −A), so by continuity from above, µ(A1)− µ(A) = µ(A1 −A) = limn→∞ µ(Cn) = limn→∞(µ(A1)−µ(An)) = µ(A1)− limn→∞ µ(An).(Notice that we used the fact µ(A) ≤ µ(A1) < ∞, which is true by mono-tonicity.) Since µ(A1) <∞, we may subtract it. This completes the proofof (e).

(f) De�ne B1 = A1, and for n ≥ 2, de�ne Bn = (An − (∪n−1i=1 Ai)). Then

{Bn}n∈N ⊆M, and {Bn}n∈N is a pairwise disjoint collection. Since ∪n∈NBn =A, and for each n, Bn ⊆ An, monotonicity and countable additivity implyµ(A) = µ(∪n∈NBn) =

∑∞n=1 µ(Bn) ≤

∑∞n=1 µ(An).

(g) This follows from countable subadditivity in the same way �nite additivityfollows from countable additivity.

(h) One implication is obvious; the other one follows from monotonicity.�

Example 2.23. Consider the measurable space (X,P (X)).Given any A ⊆ X, de�ne

µ(A) =

{∞, A is infinite|A|, A is finite

µ is called the counting measure.

De�nitions 2.24. Let (X,M, µ) be a measure space. If s is a measurable nonneg-ative simple function de�ned on X, then there exist distinct α1, . . . , αn ∈ [0,∞)

8 MISHEL SKENDERI

and there exist A1, . . . , An ∈M such that

s =

n∑i=1

αi1Ai .

If E ∈M, then de�ne ∫E

s dµ =

n∑i=1

αiµ(E ∩Ai).

If f : X → [0,∞] is measurable and E ∈M, then de�ne∫E

f dµ = sup0≤s≤f

∫E

s dµ.

Notice that although there exist two de�nitions of∫Es dµ for measurable nonneg-

ative simple s, they are equivalent.∫Ef dµ is known as the Lebesgue integral of f

over E, with respect to µ. Clearly, it is an element of [0,∞]. Furthermore, we em-phasize that all throughout these de�nitions, we use the convention 0·∞ =∞·0 = 0.

We record the following properties of Lebesgue integrals of nonnegative measur-able functions as a Corollary. As this Corollary follows directly from the de�nitions,we omit the proof.

Corollary 2.25. Suppose f : X → [0,∞] and g : X → [0,∞] are measurable, andsuppose B, C, and E ∈M. Then the following results hold:

(a) If f ≤ g on E, then∫Ef dµ ≤

∫Eg dµ.

(b) If B ⊆ C, then∫Bf dµ ≤

∫Cf dµ.

(c) If k is a nonnegative �nite real number, then∫Ekf dµ is well-de�ned, and∫

Ekf dµ = k

∫Ef dµ.

(d) If f = 0 on E, then∫Ef dµ = 0.

(e) If µ(E) = 0, then∫Ef dµ = 0.

(f)∫Ef dµ =

∫X1Ef dµ. (Notice that the right-hand side is well-de�ned.)

Proposition 2.26. Let (X,M, µ) be a measure space. Suppose s : X → [0,∞) andt : X → [0,∞) are simple and measurable. If Φs : M→ [0,∞] is given by Φs(A) =∫As dµ, then Φs is a measure on M. Furthermore,

∫X

(s+t) dµ =∫Xs dµ+

∫Xt dµ.

Proof. Let s be as in De�nition 2.24. If {Bj}j∈N is a pairwise disjoint collectionof measurable sets whose union we call B, then by the countable additivity of µ,Φs(B) =

∑ni=1 αiµ(B ∩ Ai) =

∑ni=1 αi

∑∞j=1 µ(Bj ∩ Ai) =

∑ni=1

∑∞j=1 αiµ(Bj ∩

Ai) =∑∞j=1

∑ni=1 αiµ(Bj ∩Ai) =

∑∞j=1 Φs(Bj), which proves that Φs is countably

additive on M. Notice that by (e) of the above Corollary, Φs(∅) = 0 <∞, so thatΦs is indeed a measure on M.If s is as above, t(X) = {β1, . . . , βm}, and for each j ∈ {1, . . . ,m}, Bj = tpre(βj),then Ei,j := (s+t)pre(αi+βj) = (Ai∩Bj), so that

∫Ei,j

(s+t) dµ = (αi+βj)µ(Ei,j),

and∫Ei,j

s dµ+∫Ei,j

t dµ = αiµ(Ei,j)+βjµ(Ei,j). Since the Ei,j are pairwise disjoint

sets whose union is X and Φs, Φt, and Φs+t are measures on M, it follows that∫X

(s+ t) dµ =∫Xs dµ+

∫Xt dµ. �

We now develop some of the most important results in the Lebesgue theoryof integration, namely those which concern the interchangeability of limiting pro-cesses. In Riemann integration, uniform convergence was necessary to successfully


exchange the order of limit operations. As we shall see, Lebesgue integration be-haves much better.

Theorem 2.27. Lebesgue's Monotone Convergence Theorem

Suppose X is a measure space and let {fn} be a sequence of functions de�ned onX such that for every n, 0 ≤ fn ≤ fn+1 ≤ ∞. Then there exists a measurable f :X → [0,∞] such that for every x, limn→∞ fn(x) = f(x), and limn→∞

∫Xfn dµ =∫

Xf dµ.

Proof. Fix any x ∈ X. If {fn(x)} is bounded above, then it converges to somenonnegative real number, which we call f(x). If {fn(x)} is unbounded above,then it converges, by monotonicity, to ∞, and we set f(x) = ∞. In this way, wede�ne a measurable function f (Proposition 2.15), and for every n, fn ≤ f , andthus

∫Xfn dµ ≤

∫Xf dµ (Corollary 2.25(a)). Since for every n, 0 ≤

∫Xfn dµ ≤∫

Xfn+1 dµ, there exists k ∈ [0,∞] such that limn→∞

∫Xfn dµ = k�this follows

by an argument similar to the one we used in order to de�ne f�and because limitspreserve inequalities, k ≤

∫Xf dµ. Now, let s be any measurable simple function

on X that satis�es 0 ≤ s ≤ f , let β ∈ (0, 1) be given, and set En = {x : fn(x) ≥βs(x)}. Notice that for every n, En ⊆ En+1, and we claim ∪n∈NEn = X. Forif x0 ∈ X, then f(x0) = 0 implies f1(x0) = 0 and s(x0) = 0, so that x0 ∈ E1.If f(x0) = ∞, then clearly there exists some n such that fn(x0) ≥ βs(x0). If0 < f(x0) < ∞, then by monotone convergence for numerical sequences, thereexists some n such that fn(x0) ≥ βs(x0). Note that each En is measurable, so thatfor each n,

∫Xfn dµ ≥

∫Enfn dµ ≥ β

∫Ens dµ, by Corollary 2.25 ((a), (b), and

(c)). Letting n approach in�nity, Proposition 2.26 and continuity from below showthat k ≥ β

∫Xs dµ. Since this holds for every β ∈ (0, 1), k ≥

∫Xs dµ. Now, by the

de�nition of∫Xf dµ, k ≥

∫Xf dµ, which completes the proof. �

Example 2.28. Consider the measure space (N,P (N), µ), where µ is the countingmeasure. Let f : N → [0,∞]. Then it is obvious that f is measurable, and wehave

∫N f dµ =

∑∞n=1 f(n). The latter assertion can be shown using the Monotone

Convergence Theorem and the fact that f can be approximated from below bynonnegative measurable simple functions.What is interesting about this example is that whereas in Riemann integration,in�nite series were merely analogous to integrals, in Lebesgue integration, in�niteseries are integrals.

Theorem 2.29. Let X be a measure space. If for every n ∈ N, fn : X →[0,∞] is measurable and for every x ∈ X, f(x) =

∑∞n=1 fn(x), then

∫Xf dµ =∑∞

n=1

∫Xfn dµ.

Proof. By Theorem 2.19, there exist sequences {s′j} and {s′′j } of nonnegative mea-surable simple functions that converge pointwise to f1 and f2, respectively. Forevery j, set sj = s′j + s′′j . Then {sj} converges pointwise to f1 + f2. By Proposition

2.26,∫Xsj dµ =

∫Xs′j dµ+

∫Xs′′j dµ. The Monotone Convergence Theorem there-

fore shows∫Xf1 + f2 dµ =

∫Xf1 dµ+

∫Xf2 dµ. Now, for any N ∈ N, de�ne gN =∑N

i=1 fi. By induction,∫XgN dµ =

∑Ni=1

∫Xfi dµ. It is easy to see that for every

x, {gN (x)} converges monotonically, from below, tof(x), so by the Monotone Con-

vergence Theorem,∫Xf dµ = limN→∞

∫XgN dµ = limN→∞

∑Ni=1

∫Xfn dµ. �

Note that we have now generalized the second result of Proposition 2.26.

10 MISHEL SKENDERI

Lemma 2.30. Fatou's Lemma

Let X be a measure space. If for every n ∈ N, fn : X → [0,∞] is measurable, then∫X

lim inf fn dµ ≤ lim inf∫Xfn dµ.

Proof. Note �rst that as lim inf fn : X → [0,∞] is measurable,∫X

lim inf fn dµ iswell-de�ned. Now, �x any k ∈ N and de�ne, for any x ∈ X, gk(x) = infi≥k fi(x).gk is thus measurable, and gk ≤ fk, which implies

∫Xgk dµ ≤

∫Xfk dµ. Thus,

lim inf∫Xgn dµ ≤ lim inf

∫Xfn dµ. Notice that for each n, 0 ≤ gn ≤ gn+1. Also, for

any x, lim inf fn(x) = supk gk(x) = limk→∞ gk(x). By the Monotone ConvergenceTheorem,

∫X

lim inf fn dµ = limk→∞∫Xgk dµ = lim inf

∫Xgn dµ. �

Theorem 2.31. Suppose (X,M, µ) is a measure space, and let f : X → [0,∞] bemeasurable. If Φ : M→ [0,∞] is given by Φ(A) =

∫Af dµ, then Φ is a measure on

M. Moreover, if g : X → [0,∞] is measurable, then∫Xg dΦ =

∫Xgf dµ.

Proof. For any A ∈M, clearly Φ(A) ≥ 0. Also, Φ(∅) = 0. We now prove countableadditivity. Let {Bn}n∈N ⊆M be a collection of pairwise disjoint sets whose unionwe call B. It follows that Φ(B) =

∫X1Bf dµ, for each n, Φ(Bn) =

∫X1Bn

f dµ,

and for any x ∈ X, (1Bf)(x) =∑∞n=1(1Bn

f)(x). By Theorem 2.29, Φ(B) =∑∞n=1 Φ(Bn). Now, if A ∈ M, then

∫X1A dΦ = Φ(A) =

∫Af dµ =

∫X1Af dµ.

Suppose now that s is as in De�nition 2.24. Then∫Xs dΦ =

∑ni=1 αiΦ(Ai) =∑n

i=1 αi∫Aif dµ =

∑ni=1 αi

∫X1Ai

f dµ =∫X

(∑ni=1 αi1ai)f dµ =

∫Xsf dµ, us-

ing Theorem 2.29. Now let g : X → [0,∞] be measurable. Then there exists asequence {sn} of measurable nonnegative simple functions that converges point-wise to g, monotonically and from below. Since f is nonnegative, the sequence{snf} converges pointwise to gf monotonically and from below. The MonotoneConvergence Theorem, applied to {snf}, shows

∫Xg dΦ =

∫Xgf dµ. �

Notice that the �rst conclusion of this Theorem generalizes the �rst conclusionof Proposition 2.26. The second conclusion of this theorem is sometimes written asdΦ = f dµ, but we do not assign any independent meaning to dΦ, dµ, or f dµ.The second conclusion of this Theorem has a converse known as the Radon-NikodymTheorem, which we shall prove in this paper.

We will now de�ne integration for functions that map into C and [−∞,∞].

De�nitions 2.32. Suppose (X,M, µ) is a measure space. Let f : X → C be ameasurable function such that

∫X|f | dµ <∞. Such a function f is called summable

(with respect to µ). (Recall that if f is measurable, then so is |f | (Corollary 2.6(b)), so that the above integral is well-de�ned.) If f is summable and A ∈M, thende�ne∫A

f dµ =

∫A

(Re(f))+ dµ −∫A

(Re(f))− dµ + i

∫A

(Im(f))+ dµ − i∫A

(Im(f))− dµ.

Notice that (Re(f))+, (Re(f))−, (Im(f))+, and (Im(f))− are measurable, by Corol-lary 2.6 (b) and Corollary 2.16 (b); thus, our de�nition makes sense. Note also that∫Af dµ ∈ C, for (Re(f))+ ≤ |Re(f)| ≤ |f |, so that

∫A

(Re(f))+ dµ ≤∫A|f | dµ <∞,

and analogous statements hold for (Re(f))−, (Im(f))+, and (Im(f))−.Suppose f : X → [−∞,∞] is measurable and A ∈ M. If

∫Af+ dµ < ∞ or

A PRIMER OF LEBESGUE INTEGRATION WITH A VIEW TO THE LEBESGUE-RADON-NIKODYM THEOREM11∫Af− dµ <∞, then de�ne∫

A

f dµ =

∫A

f+ dµ−∫A

f− dµ.

Notice that∫Af dµ ∈ [−∞,∞].

Remark 2.33. Recall that for real-valued f , f = f+ − f−, and notice that forcomplex-valued f , f = (Re(f))+ − (Re(f))− + i(Im(f))+ − i(Im(f))−. These factsare what motivate the above de�nitions.

Proposition 2.34. Suppose X is a measure space. If f and g are summable andα and β are complex numbers, then αf + βg is summable, and

∫Xαf + βg dµ =

α∫Xf dµ+ β

∫Xg dµ.

Proof. We omit the proof because it consists only of simple algebraic manipulations.�

Remarks 2.35. The above Proposition is frequently expressed by saying the summa-ble functions form a vector space over C and the function I : {f : f is summable} →C given by I(f) =

∫Xf dµ is a linear functional on the complex vector space of

summable functions, that is a linear function that maps the vector space in questioninto its underlying scalar �eld.

Proposition 2.36. Suppose X is a measure space. If f is summable, then |∫Xf dµ| ≤∫

X|f | dµ.

Proof. Set z =∫Xf dµ. If z = 0, then the proof is complete. If z 6= 0, then de�ne

c = |z|z . Then |c| = 1, and cz = |z|. Thus, |z| = c

∫Xf dµ =

∫Xcf dµ. Now,

notice that Re(cf) ≤ |cf | = |f |. In conjunction with the fact |z| ∈ R+, this implies∫Xcf dµ =

∫XRe(cf) dµ ≤

∫X|f | dµ, as desired. �

Theorem 2.37. Lebesgue's Dominated Convergence Theorem

Let X be a measure space. Suppose {fn} is a sequence of complex-valued measur-able functions de�ned on X, and suppose also that given any x ∈ X, f(x) =limn→∞ fn(x) exists. If there exists a summable function g on X that domi-nates the fn in the sense that for every n, |fn| ≤ g, then f is summable andlimn→∞

∫X|fn − f | dµ = 0, which also shows that limn→∞

∫Xfn dµ =

∫Xf dµ.

Proof. Continuity implies |fn| converges pointwise to |f |, and thus passing to thelimit shows |f | ≤ g; thus, f is summable. Now, 2g − |fn − f | ≥ 0 for everyn, and notice lim inf(2g − |fn − f |) = 2g, so by Fatou's Lemma,

∫X

2g dµ ≤lim inf

∫X

2g − |fn − f | dµ = lim inf[(∫X

2g dµ) + (∫X−|fn − f | dµ)] =

∫X

2g dµ+

lim inf(∫X−|fn−f | dµ) =

∫X

2g dµ− lim sup(∫X|fn−f | dµ). As

∫X

2g dµ <∞, we

may subtract it, which implies lim sup(∫X|fn− f | dµ) ≤ 0. Since 0 ≤ |fn− f | ≤ 2g

for every n, 0 ≤∫X|fn − f | dµ < ∞. Hence, lim sup(

∫X|fn − f | dµ) = 0. Clearly,

lim inf(∫X|fn − f | dµ) = 0, so limn→∞(

∫X|fn − f | dµ) = 0. Now, the above

Proposition shows that for any n, 0 ≤ |∫Xfn dµ −

∫Xf dµ| ≤

∫X|fn − f |, so

limn→∞∫Xfn dµ =

∫Xf dµ. (Notice that several previously established results

were used freely throughout this proof.) �

De�nition 2.38. Suppose (X,M, µ) is a measure space. Let P be a property thata point x ∈ X may or may not possess. If A ∈ M, then we say that P holdsalmost everywhere (with respect to µ) on A if there exists B ∈M such that B ⊆ A,

12 MISHEL SKENDERI

µ(B) = 0, and every point that is an element of (A−B) possesses the property P .We also say "P holds for almost every x ∈ A" to describe the same situation.

We now introduce some important results regarding sets of measure zero.

Theorem 2.39. Let (X,M, µ) be a measure space, and let f : X → [0,∞] bemeasurable. If E ∈M, then

∫Ef dµ = 0 if and only if f = 0 almost everywhere on

E.

Proof. Suppose∫Ef dµ = 0. We wish to show A = {x ∈ E : f(x) > 0} has

measure zero. Notice A = ∪n∈NAn, where An = {x ∈ E : f(x) > 1n}. It is clear

that µ(A) = 0 if and only if µ(An) = 0 for every n. Suppose, for contradiction,that there exists some m such that µ(Am) > 0. Then as 1

m1Am≤ f on E, we have

0 < µ(Am)m = 1

m

∫E1Am

dµ ≤∫Ef dµ, a contradiction. The converse is almost

obvious. �

Corollary 2.40. Let (X,M, µ) be a measure space, and let f be summable. Thenf = 0 almost everywhere on X if and only if for every E ∈M,

∫Ef dµ = 0.

Proof. Suppose �rst that for any E ∈ M,∫Ef dµ = 0. Let A = {x ∈ X :

Re(f(x)) ≥ 0}. As Re(f) is measurable, A is measurable. By hypothesis, Re(∫Af dµ) =

0. Clearly, (Re(f))− = 0 on A. Therefore, 0 = Re(∫Af dµ) =

∫A

(Re(f))+ dµ. The

above Theorem therefore implies (Re(f))+ = 0 almost everywhere on A, and there-fore (Re(f))+ = 0 almost everywhere on X. Similary, we can show (Re(f))−,(Im(f))+, and (Im(f))− equal zero almost everywhere on X. The result follows.Conversely, suppose f = 0 almost everywhere on X. Then |f | = 0 almost every-where on X, which implies

∫X|f | dµ = 0, by the above Theorem. Now, let E ∈M.

As 0 ≤ |∫Ef dµ| ≤

∫E|f | dµ ≤

∫X|f | dµ = 0, the result follows. �

Theorem 2.41. Let (X,M, µ) be a measure space with �nite measure, let f be asummable function, and let S ⊆ C be a closed set. Suppose further that for anyE ∈M with positive measure, AE(f) = ( 1

µ(E)

∫Ef dµ) ∈ S. Then for almost every

x ∈ X, f(x) ∈ S.

Proof. Notice �rst that AE(f) is well-de�ned, as f is summable, and 0 < µ(E) ≤µ(X) < ∞. If S = C, then we are done. If not, then let S∗ ⊆ Sc be a closeddisc with center at z and radius r > 0. As Sc is equal to a countable union ofsuch discs, it su�ces to show E = fpre(S∗) has measure zero. If µ(E) > 0, then|AE(f) − z| = 1

µ(E) |∫E

(f − z) dµ| ≤ 1µ(E)

∫E|(f − z)| dµ. Now, if x ∈ E, then

|f(x) − z| ≤ r, so 1µ(E)

∫E|(f − z)| dµ ≤ 1

µ(E) · rµ(E) = r. But since AE(f) ∈ S,|AE(f)− z| > r, a contradiction. �

3. The Uniqueness of Bounded Linear Functionals Defined on

Hilbert Spaces

We now present a brief introduction to Hilbert spaces with a view to proving acertain uniqueness result regarding bounded linear functionals, a result of immenseimportance in the proof of the Lebesgue-Radon-Nikodym Theorem.

De�nitions 3.1. Let J be a vector space over C. If there exists a function 〈·, ·〉 :J ×J → C (called the inner product) such that the following hold for any x, y, andz ∈ J , and any α ∈ C, then J is called a unitary space.


(a) 〈y, x〉 = 〈x, y〉(b) 〈x+ y, z〉 = 〈x, z〉+ 〈y, z〉(c) 〈αx, y〉 = α〈x, y〉(d) 〈x, x〉 ≥ 0(e) If 〈x, x〉 = 0, then x = 0.

We now discuss some consequences of these de�nitions. (a) and (c) show thatfor any x ∈ J , 〈0, x〉 = 0 = 〈x, 0〉. Thus, by (e), 〈x, x〉 = 0 if and only if x = 0. (b)and (c) say that for any �xed z ∈ H, the function y 7→ 〈y, z〉 is a linear functionalon J . (a) and (c) show that for any x and y ∈ J and any α ∈ C, 〈x, αy〉 = α〈x, y〉(a) and (b) show that for any x, y, and z ∈ J , 〈z, x+ y〉 = 〈z, x〉+ 〈z, y〉.

De�nitions 3.2. De�ne the function || · || : J → (R+∪{0}) by x 7→ 〈x, x〉12 . Notice

that this is well-de�ned by (d) of the above de�nitions. || · || is the norm of J , and||x|| is the norm of x.

Theorem 3.3. The Schwarz Inequality

If J is a unitary space, then for any x and y ∈ J , |〈x, y〉| ≤ ||x||||y||.

Proof. Let A = ||x||2, B = |〈x, y〉|, and C = ||y||2. If B = 0, then the resultis obvious. If B 6= 0, then de�ne α ∈ C by α = B

〈y,x〉 . Then α〈y, x〉 = B, and

|α| = 1. Given any r ∈ R, simple computations show 0 ≤ 〈x − rαy, x − rαy〉 =〈x, x〉 − rα〈y, x〉 − rα〈x, y〉+ r2〈y, y〉. It follows that A− 2Br+Cr2 ≥ 0. If C = 0,then this forces B = 0, for otherwise, the above quadratic form would be negativefor su�ciently large r > 0. This gives the desired result when C = 0. If C > 0,then let r = B

C , which implies B2 ≤ AC with little e�ort. �

Corollary 3.4. The Triangle Inequality

If x and y ∈ J , then ||x+ y|| ≤ ||x||+ ||y||.

Proof. Simple computations and the Schwarz inequality show that 0 ≤ 〈x+ y, x+y〉 = 〈x, x〉+ 〈x, y〉+ 〈y, x〉+ 〈y, y〉 ≤ ||x||2 + 2|(x, y)|+ ||y||2 ≤ ||x||2 + 2||x||||y||+||y||2 = (||x||+ ||y||)2. �

De�nitions 3.5. For any x, y, and z ∈ J , the triangle inequality implies ||x−z|| ≤||x − y|| + ||y − z||. If we de�ne a function ρ on J by ρ(x, y) = ||x − y||, then itis easy to check that ρ satis�es all the requirements of a metric, and we call ρ themetric induced by the norm of J. The pair (J, ρ) is thus a metric space. If thismetric space is complete in the sense that every Cauchy sequence in J convergesto some point of J , then J is a Hilbert space. Usually Hilbert spaces are denotedby H, and we will do so. In the remainder of this section, H will denote a Hilbertspace and nothing but a Hilbert space.

Proposition 3.6. Fix any y ∈ H. The functions given by x 7→ 〈x, y〉, x 7→ 〈y, x〉,and x 7→ ||x|| are then uniformly continuous on H.

Proof. Let ε > 0 be given. Choose δ1 = ε1+||y|| . If x1 and x2 are such that ||x1 −

x2|| < δ1, then the Schwarz inequality shows |〈x1, y〉 − 〈x2, y〉| = |〈x1 − x2, y〉| ≤||x1 − x2||||y|| < ε, as desired. A very similar argument works for the functionx 7→ 〈y, x〉. Now, choose δ2 = ε. If x1 and x2 satisfy ||x1 − x2|| < δ2, then bythe triangle inequality ||x1|| − ||x2|| ≤ ||x1 − x2|| < ε. By symmetry, we conclude| ||x1|| − ||x2|| | ≤ ||x1 − x2|| < ε. �

14 MISHEL SKENDERI

De�nitions 3.7. If V is a vector space, then recall that A ⊆ V is convex if forany x ∈ A, y ∈ A, and t ∈ [0, 1], zt = (x+ t(y− x)) ∈ A. Geometrically, this meansthat given any two points in A, the straight line segment between those points liesentirely within A. Clearly, any subspace of V is convex. Also, if A is convex andx ∈ V , then A + x = {y + x : y ∈ A}, the translate of A by x, is convex. This isclear geometrically (well, up to three dimensions, at any rate) and is not di�cultto verify analytically.

Theorem 3.8. Suppose A ⊆ H is nonempty, closed, and convex. Then there existsone and only one x0 ∈ A such that for every x ∈ A, ||x0|| ≤ ||x||.

Proof. It is not di�cult to check that for any x and y ∈ H, the following identity,known as the parallelogram law, is true: ||x+y||2 + ||x−y||2 = 2||x||2 +2||y||2. Now,de�ne α = inf {||z|| : z ∈ A}. Given any x and y ∈ A, applying the parallelogram

law to x2 and y

2 shows that 14 ||x− y||

2 = 12 ||x||

2 + 12 ||y||

2 − || (x+y)2 ||2. By convexity

of A, (x+y)2 ∈ A. It follows that for any x and y ∈ A, ||x− y||2 = 2||x||2 + 2||y||2 −

4|| (x+y)2 ||2 ≤ 2||x||2 + 2||y||2 − 4α2. If ||x|| = ||y|| = α, then ||x − y||2 ≤ 0, which

implies x = y. This proves uniqueness. We now prove existence. Clearly, there existsa sequence {yk} in A that converges to α.We claim {yk} is Cauchy: For, given ε > 0,

there existsK ∈ N such that for every k > K, 0 ≤ 2||yk||2−2α2 < ε2

2 . Hence, for any

n > K and for any m > K, ||ym− yn||2 ≤ 2||ym||2 + 2||yn||2− 4α2 < ε2. Since H iscomplete and A ⊆ H is closed, there exists x0 ∈ A such that limk→∞ yk = x0. Sincefor any z ∈ H, the function z 7→ ||z|| is continuous, ||x0|| = limk→∞ ||yk|| = α. �

We now prove a simple, but important, result regarding continuity.

Proposition 3.9. Let Y be a metric space. If f : Y → C is continuous at c ∈ Yand f(c) 6= 0, then there exists δ > 0 such that for any x, if ρ(x, c) < δ, thenf(x) 6= 0.

Proof. Since f is continuous at c, there exists δ > 0 such that if ρ(x, c) < δ, then

|f(x)− f(c)| < |f(c)|2 . The triangle inequality then implies |f(x)| ≥ |f(c)|

2 > 0. �

De�nitions 3.10. Let x and y ∈ H be given. If 〈x, y〉 = 0, then x is orthogonalto y, and we write x ⊥ y. Note that x ⊥ y implies y ⊥ x. Given any x ∈ H, de�nex⊥ = {y ∈ H : y ⊥ x}. The properties of the inner product imply x⊥ is a subspaceof H. Notice that x⊥ is the zero set of the continuous function, with domain H,given by y 7→ 〈x, y〉. The above Proposition thus implies x⊥ is a closed subspace ofH. If M is a subspace of H, then de�ne M⊥ = {y ∈ H : for every x ∈M, y ⊥ x}.Notice M⊥ = ∩x∈M x⊥. Since the arbitrary intersection of subspaces is a subspaceand the arbitrary intersection of closed sets is a closed set, M⊥ is a closed subspaceof H. Thus, (M⊥)⊥ is well-de�ned. Moreover, it is easy to prove that if M is aclosed subspace, then M = (M⊥)⊥.

We now prove an important result regarding the representation of elements of aHilbert space.

Theorem 3.11. If M ⊆ H is a closed subspace of H, then the following hold:

(a) If x ∈ H, then there exist unique rx ∈ M and sx ∈ M⊥ such that x = rx + sx,and rx and sx are the points closest to x in M and M⊥, respectively.


(b) x 7→ rx and x 7→ sx are linear transformations. These transformations areknown as the orthogonal projections of H onto M and M⊥, respectively.(c) ||x||2 = ||rx||2 + ||sx||2.

Proof. (a) Let x ∈ H be given. To prove uniqueness, let rx ∈M and sx ∈M⊥and r′x ∈ M and s′x ∈ M⊥ be such that x = rx + sx = r′x + s′x. Then(rx − r′x) = (s′x − sx), and (rx − r′x) ⊥ (s′x − sx), as is easy to check.Hence, (rx − r′x) = (sx − s′x) = 0. To prove existence, notice that becauseM is closed and convex, x + M = {x + y : y ∈ M} is also closed andconvex. Using the previous Theorem, de�ne sx as the unique element ofsmallest norm in x + M. De�ne rx = x − sx. Clearly, rx ∈ M. We nowprove sx ∈ M⊥. Let y ∈ M be given. If y = 0, then sx ⊥ y. If not,then suppose, without loss of generality, ||y|| = 1. If λ ∈ C, then because(sx − λy) ∈ x+M , the minimality of ||sx|| implies ||sx||2 ≤ ||sx − λy||2 =||sx||2−λ〈y, sx〉−λ〈sx, y〉+ ||λy||2, which implies 0 ≤ −λ〈y, sx〉−λ〈sx, y〉+λλ. Letting λ = 〈sx, y〉 shows 0 ≤ −|〈sx, y〉|2, which implies 〈sx, y〉 = 0,which means sx ∈ M⊥. Now, if z ∈ M , then (rx − z) ∈ M , which implies||x− z||2 = ||sx + (rx − z)||2 = ||sx||2 + ||rx − z||2, as sx ⊥ (rx − z). Thus,||x− z||2 is clearly minimized when z = rx. An analogous argument showssx is the point in M⊥ nearest to x.

(b) Given any λ and θ ∈ C and any x and y ∈ H, (a) implies r(λx+θy) − λrx −θry = λsx+θsy−s(λx+θy). The left-hand side of this equation is an element

of M , whereas the right-hand side is an element of M⊥. Hence, both sidesare equal to 0, and linearity follows easily.

(c) This is a straightforward consequence of the fact rx ⊥ sx.�

Corollary 3.12. If M is a closed subspace of H and M 6= H, then there existsz ∈M⊥ such that z 6= 0.

Proof. Since M 6= H, there exists y ∈ (H −M). Now, sy ∈ M⊥ and sy 6= 0, forthen we would have y = ry ∈M. Set z = sy. �

We already know that given any �xed y ∈ H, the mapping x 7→ 〈x, y〉 is acontinuous linear functional on H. The following theorem asserts the converse.

Theorem 3.13. Suppose L is a continuous linear functional on H. Then thereexists a unique y ∈ H such that for every x ∈ H, L(x) = 〈x, y〉.

Proof. We �rst prove existence. If L = 0, then let y = 0. If not, then de�neM = {x : L(x) = 0}. It is easy to see that M is a subspace. Moreover, M is closed,because the continuity of L implies (H −M) is open. By hypothesis, M 6= H,so the above Corollary implies the existence of w ∈ M⊥ such that w 6= 0. De�nez = w

||w|| ; then z ∈M⊥ and ||z|| = 1. Given any x ∈ H, let x∗ = (L(x))z− (L(z))x.

The linearity of L implies L(x∗) = 0. Thus, x∗ ∈ M , which implies 〈x∗, z〉 = 0,which shows 〈(L(x))z, z〉 = 〈(L(z))x, z〉. Hence, L(x) = L(x)〈z, z〉 = L(z)〈x, z〉 =

〈x, (L(z))z〉. Finally, set y = (L(z))z. To prove uniqueness, let y′ ∈ H be such thatfor any x ∈ H, 〈x, (y − y′)〉 = 0. Choosing x = (y − y′), we conclude y = y′. �

Remark 3.14. We therefore see that if L is a linear functional on H, then L iscontinuous on H if and only if there exists a unique y ∈ H such that for everyx ∈ H, L(x) = 〈x, y〉.

16 MISHEL SKENDERI

We now prove a basic fact about linear functionals on Hilbert spaces that will beused in the proof of the Lebesgue-Radon-Nikodym Theorem; but �rst, we introducesome de�nitions and related facts.

De�nitions 3.15. If L is a linear functional on H, then de�ne ||L|| = sup {|L(x)| :x ∈ H such that ||x|| = 1}. Notice that ||L|| = sup {|L(x)| : x ∈ H such that ||x|| ≤1}, for given any x such that ||x|| ∈ (0, 1], if y = x

||x|| , then |L(y)| = 1||x|| |L(x)| ≥

|L(x)|. Moreover, for every x ∈ H, |L(x)| ≤ ||L|| ||x||. If ||x|| = 0 or ||x|| = 1,then this is obvious. If not, then naturally |L( x

||x|| )| ≤ ||L||, so the result follows

from the linearity of L. Notice also that ||L|| is the smallest b such that for everyx ∈ H, |L(x)| ≤ b||x||. For if there were some b such that for every x ∈ H,|L(x)| ≤ b||x|| < ||L|| ||x||, then for every z with ||z|| = 1, |L(z)| ≤ b < ||L||, acontradiction. Lastly, if ||L|| <∞, then L is bounded.

Proposition 3.16. Let L be a linear functional on H. Then L is bounded if andonly if L is uniformly continuous on H, and L is uniformly continuous on H if andonly if L is continuous at some point of H.

Proof. If L is bounded, then by linearity of L, for any x and y ∈ H, |L(y)−L(x)| ≤||L|| ||y − x||, so L is uniformly continuous (choose δ = ε

1+||L|| , which is possible,

as ||L|| is �nite). Clearly, if f is uniformly continuous on X, then f is continuousat each point of X. Now, suppose f is continuous at some x0 ∈ H. Then thereexists δ > 0 such that for any x with ||x − x0|| ≤ δ, |L(x − x0)| ≤ 1. Let y with||y|| ≤ 1 be given. Then ||δy|| ≤ δ. Choosing x = x0 + δy, we see ||x− x0|| ≤ δ, so|L(x− x0)| = |L(δy)| ≤ 1. Hence, |L(y)| ≤ 1

δ . Thus, ||L|| ≤1δ <∞. �

4. A Construction of Lp on Arbitrary Measure Spaces

In this section, we will construct the Lp spaces, where p ∈ [1,∞). We willdo so because L2, by way of its being a Hilbert space, plays a powerful role inthe proof of the Lebesgue-Radon-Nikodym Theorem. Although we shall need onlyknowledge of L2 for said proof, the construction would not be simpli�ed signi�cantlyif we were to restrict ourselves to p = 2. We will therefore construct Lp for everyp ∈ [1,∞). However, before doing so, we will �rst state and prove some importantinequalities, indispensable in the study of Lp spaces, whose proofs rely stronglyon the notion of convex functions. Recall from calculus that if f : (a, b) → (c, d),then f is convex on (α, β) ⊆ (a, b) if for any x and y ∈ (α, β) and any t ∈ [0, 1],f(tx + (1 − t)y) ≤ tf(x) + (1 − t)f(y). We allow a = −∞, α = −∞, b = ∞, andβ = ∞. Recall also that if f ′′ exists on (α, β), then f is convex on (α, β) if andonly if f ′′ > 0 on (α, β). We start with a simple de�nition.

De�nition 4.1. If p and q are positive real numbers such that 1p + 1

q = 1, then p

and q are a pair of conjugate exponents. Notice that this implies p and q ∈ (1,∞).

Theorem 4.2. Let X be a measure space, and let p and q be conjugate exponents.If f : X → [0,∞] and g : X → [0,∞] are measurable, then∫

X

fg dµ ≤ (

∫X

fp dµ)1p (

∫X

gq dµ)1q

and

(

∫X

(f + g)p dµ)1p ≤ (

∫X

fp dµ)1p + (

∫X

gp dµ)1p .

These are Hölder's and Minkowski's inequalities, respectively.


Proof. We �rst prove Hölder's inequality. For notational ease, let A =∫Xfg dµ,

let B = (∫Xfp dµ)

1p , and let C = (

∫Xgq dµ)

1q . If B = 0, then fp = 0 al-

most everywhere, which implies f = 0 almost everywhere. This implies fg = 0almost everywhere, which implies A = 0. In this case, the inequality holds. IfB > 0 and C = ∞, then the inequality clearly holds, so we now consider thecase in which both B and C are �nite and positive. De�ne measurable func-tions F : X → [0,∞] and G : X → [0,∞] by F = f

B and G = gC . Then∫

XF p dµ =

∫X

fp∫Xfp dµ

dµ =∫Xfp dµ∫

Xfp dµ

= 1. Similarly,∫XGq = 1. Now, if x0 ∈ X

is such that F (x0) and G(x0) ∈ (0,∞), since the exponential function maps Ronto R+, there exists s∗ ∈ R such that F (x0) = exp(s∗). If s = p(s∗), thenF (x0) = exp( sp ). Similarly, there exists t ∈ R such that G(x0) = exp( tq ). As p and q

are conjugate exponents and the exponential function is convex, exp( sp ) · exp( tq ) =

exp( sp + tq ) ≤ 1

p exp(s) + 1q exp(t). Thus, F (x0)G(x0) ≤ 1

p (F (x0))p + 1q (G(x0))q.

Notice that this last inequality persists if x0 is such that F (x0) = 0 or F (x0) =∞.Therefore, for every x ∈ X, F (x)G(x) ≤ 1

p (F (x))p + 1q (G(x))q. Integrating over X,

we have∫XFG dµ ≤ 1

p + 1q = 1. Rearrangement yields A ≤ BC, as desired.

We now prove Minkowski's inequality. Notice that the following holds: (f + g)p =f · (f + g)p−1 + g · (f + g)p−1. Hölder's inequality shows

∫Xf · (f + g)p−1 dµ ≤

(∫Xfp dµ)

1p · (∫X

(f+g)q(p−1) dµ)1q . By symmetry,

∫Xg·(f+g)p−1 dµ ≤ (

∫Xgp dµ)

1p ·

(∫X

(f + g)q(p−1) dµ)1q . Adding these two inequalities and using the fact (p −

1)q = p, we see∫X

(f + g)p dµ ≤ (∫X

(f + g)p dµ)1q [(∫Xfp dµ)

1p + (

∫Xgp dµ)

1p ].

Now, notice that it su�ces to prove the Minkowski inequality under the condi-

tions (∫X

(f + g)p) dµ)1p > 0 and [(

∫Xfp dµ)

1p + (

∫Xgp dµ)

1p ] < ∞, since the

failure of either condition implies the Minkowski inequality immediately. Underthese constraints, we have f < ∞ almost everywhere and g < ∞ almost every-

where. If (f + g)(x) = 0, then obviously, [ f(x)+g(x)2 ]p ≤ 1

2 [(f(x))p + (g(x))p].If 0 < (f + g)(x) < ∞, then the convexity of t 7→ tp on (0,∞) shows that

[ f(x)+g(x)2 ]p ≤ 1

2 [(f(x))p+(g(x))p]. Now, as f and g are �nite almost everywhere, so

is (fp+gp), which implies∫X

( f+g2 )p dµ ≤ 1

2

∫X

(fp+gp) dµ <∞.Multiplying by 2p

and keeping our earlier constraints in mind, we have 0 <∫X

(f+g)p dµ <∞. Hence,0 < (

∫X

(f+g)p dµ)1q <∞. Thus,

∫X

(f+g)p dµ

(∫X

(f+g)p dµ)1qis well-de�ned, positive, and �nite.

Furthermore, an above inequality shows (∫X

(f +g)p dµ)1p = (

∫X

(f +g)p dµ)1− 1q =∫

X(f+g)p dµ

(∫X

(f+g)p dµ)1q≤ (∫Xfp dµ)

1p + (

∫Xgp dµ)

1p , which completes the proof. �

We now proceed with the construction of Lp(µ).

De�nition 4.3. Let X be a measure space. If p ∈ [1,∞) and f : X → C is

measurable, then de�ne ||f ||p = (∫X|f |p dµ)

1p and de�ne Lp(µ) to be the set of all

f such that ||f ||p < ∞. Notice that ||f ||p is well-de�ned, as |f |p : X → [0,∞] ismeasurable. Notice also that L1(µ) is precisely the class of all summable functions.

Proposition 4.4. Let p and q be conjugate exponents such that p ∈ (1,∞), andsuppose f ∈ Lp(µ) and g ∈ Lq(µ). Then fg is summable, and ||fg||1 ≤ ||f ||p ||g||q.

Proof. Apply Hölder's inequality to |f | and |g| to obtain the desired result. �

18 MISHEL SKENDERI

Proposition 4.5. Let p ∈ [1,∞), and suppose f and g ∈ Lp(µ). Then (f + g) ∈Lp(µ), and ||f + g||p ≤ ||f ||p + ||g||p.

Proof. As |f+g| ≤ |f |+|g|, (∫X|f+g|p dµ)

1p ≤ (

∫X

(|f |+|g|)p dµ)1p . By Minkowski's

inequality, (∫X

(|f |+ |g|)p dµ)1p ≤ (

∫X|f |p dµ)

1p + (

∫X|g|p dµ)

1p <∞. �

Proposition 4.6. Let p ∈ [1,∞), let α ∈ C, and suppose f ∈ Lp(µ). Thenαf ∈ Lp(µ), and ||αf ||p = |α| ||f ||p.

Proof. This is clear. �

Remark 4.7. The two Propositions above may be rephrased as follows: If p ∈ [1,∞),then Lp(µ) is a vector space over C.

We now �nally de�ne Lp(µ).

De�nitions 4.8. Let X be a measure space, and let p ∈ [1,∞). We de�ne anequivalence relation ∼ on Lp(µ) as follows: If f and g ∈ Lp(µ), then call f andg equivalent and write f ∼ g if f = g almost everywhere on X. (Notice that fand g are equivalent if and only if ||f − g||p = 0.) It is not di�cult to verify thatthis is indeed an equivalence relation. Denote the set of all equivalence classes ofelements of Lp(µ) by Lp(µ). Now, let f∗ and g∗ ∈ Lp(µ), and let f1 and g1 ∈ Lp(µ)be arbitrary representatives of f∗ and g∗, respectively. De�ne ρ : Lp(µ)×Lp(µ)→(R∪ {0}) by ρ(f∗, g∗) = ||f1− g1||p. It is easy to check that if f2 ∼ f1 and g2 ∼ g1,then ||f1 − g1||p = ||f2 − g2||p, which implies ρ is well-de�ned. Notice that ρ issymmetric. It satis�es the triangle inequality because if f∗, g∗, and h∗ ∈ Lp(µ) andf , g, and h ∈ Lp(µ) are their respective arbitrary representatives, then Proposition4.5 implies ||f − g||p ≤ ||f −h||p + ||h− g||p. Lastly, note that ρ is positive-de�nite:It is clearly nonnegative. If f∗ = g∗, then clearly, ρ(f∗, g∗) = 0. If ρ(f∗, g∗) = 0,then because of the equivalence relation we have de�ned, f∗ = g∗.Thus, we have turned Lp(µ) into a metric space. Notice that Lp(µ) is also a vectorspace over C. This follows from Remark 4.7 and the following facts: If f1 ∼ f2 andg1 ∼ g2, then f1 + g1 ∼ f2 + g2; if α ∈ C and f1 ∼ f2, then αf1 ∼ αf2.

Remarks 4.9. Although the elements of Lp(µ) are equivalence classes of functions,lest communication become overly cumbersome, we shall frequently treat the ele-ments of Lp(µ) as functions in their own right. Therefore, we will often say thingssuch as the following: "There exists f ∈ Lp(µ) such that

∫X|f | dµ = 0." We will

say this even though the above expression is, in fact, unde�ned. (Obviously, the in-tegrand is actually the absolute value of a representative element of the equivalenceclass under consideration; it is of no consequence which representative is chosen.)

With this convention in mind, if f ∈ Lp(µ), then we de�ne ||f ||p, the Lp(µ)-norm

of f , by ||f ||p = (∫X|f |p dµ)

1p . It is clear that the metric induced by this norm

is the metric ρ de�ned above. We have already seen that Lp(µ) is a metric space.Happily, it is a complete metric space, as we now show.

Theorem 4.10. Let (X,M, µ) be a measure space. If p ∈ [1,∞), then Lp(µ) is acomplete metric space.

Proof. Let {fn} be a Cauchy sequence in Lp(µ). As {fn} is Cauchy, there existsa subsequence {fni

} such that for every i ∈ N, ||fni+1− fni

||p < 3−i. (To seewhy this is so, notice that for every i ∈ N, there exists Ni ∈ N such that if


m ≥ Ni and n ≥ Ni, then ||fn − fm||p < 3−i.) Let k ∈ N, and de�ne measurable

nonnegative functions gk =∑ki=1 |fni+1

−fni| and g =

∑∞i=1 |fni+1

−fni|.A repeated

application of the Minkowski Inequality shows ||gk||p ≤∑ki=1 ||fni+1 − fni ||p ≤

12 . As g

p is the pointwise limit of {(gk)p}, Fatou's Lemma implies

∫Xgp dµ ≤

lim infk∈N∫X

(gk)p dµ ≤ ( 12 )p. This, in turn, implies ||g||p ≤ 1

2 . Thus, g < ∞almost everywhere on X. Hence, there exists E ⊆ X such that µ((X−E)) = 0 and[fn1(x) +

∑∞i=1(fni+1(x)− fni(x))] converges absolutely on E. If x ∈ E, denote the

above sum by f(x) and for x ∈ (X −E), de�ne f(x) = 0. Now, for any k ≥ 2, it is

clear fnk= fn1

+∑k−1i=1 (fni+1

− fni). Thus, if x ∈ E, then f(x) = limi→∞ fni

(x).We now show limn→∞ ||f − fn||p = 0. Let ε > 0 be given. There exists N ∈ Nsuch that if n > N and m > N , then ||fn − fm||p < ε

2 . Fix any l > N. Fatou'sLemma then implies (||f − fl||p)p ≤ lim infi∈N(||fni

− fl||p)p ≤ ( ε2 )p. This implies||f−fl||p < ε. Lastly, f ∈ Lp(µ), for Lp(µ) is a vector space, f = (f−fN+1)+fN+1,and (f − fN+1) ∈ Lp(µ). �

We now know that if p ∈ [1,∞), then Lp(µ) is a complex vector space that is alsoa complete metric space in the metric induced by its norm. We now show that L2(µ)is a unitary space. Given any f and g ∈ L2(µ), de�ne 〈·, ·〉 : L2(µ) × L2(µ) → Cby 〈f, g〉 =

∫Xfg dµ. We now check that

∫Xfg dµ is indeed a complex number.

Notice |Re(g)| ≤ |g| and |Im(g)| ≤ |g|, which implies Re(g) and Im(g) ∈ L2(µ). AsL2(µ) is a vector space, g = (Re(g)− Im(g)) ∈ L2(µ). Then Proposition 4.4 (withp = q = 1

2 ) implies fg ∈ L1(µ), which implies |∫Xfg dµ| <∞, as desired. It is not

di�cult to check that 〈·, ·〉 satis�es the remaining requirements for an inner product.Now, notice that for any f ∈ L2(µ), (||f ||2)2 =

∫X|f |2 dµ =

∫Xff dµ = 〈f, f〉.

Thus, ||f ||2 = 〈f, f〉12 . This shows that the L2(µ)-norm does indeed arise from the

inner product we have just de�ned, so that L2(µ) is a unitary space. Combinedwith the preceding discussion, this proves that L2(µ) is indeed a Hilbert space,which will be crucial in the proof of the Lebesgue-Radon-Nikodym Theorem, whichwe present in the next section.

5. The Lebesgue-Radon-Nikodym Theorem

In this section, we will �nally prove the Lebesgue-Radon Nikodym Theorem, aseminal result in measure theory and the Lebesgue theory of integration.

De�nitions 5.1. Let (X,M) be a measurable space. If E ∈M, then {Ei}i∈N ⊆Mis said to be a partition of E if {Ei}i∈N is a pairwise disjoint collection andE = ∪i∈NEi. (Notice that a partition always exists; take E1 = E, and Ei = ∅for i ≥ 2.) Recall, then that if µ : M → C is a complex measure on M, thenµ(E) =

∑∞i=1 µ(Ei).We make note of the following observations: First,

∑∞i=1 µ(Ei)

converges in the traditional sense; that is, it converges to a complex number. Sec-ond, notice that the order of summation is irrelevant. A classical result from el-ementary analysis therefore says that

∑∞i=1 µ(Ei) converges absolutely; that is,∑∞

i=1 |µ(Ei)| is equal to a �nite nonnegative number. This fact justi�es the follow-ing de�nition. De�ne |µ| : M → [0,∞] by |µ|(E) = sup

∑∞i=1 |µ(Ei)|, where the

supremum is taken across all partitions {Ei} of E. We call |µ| the total variationmeasure of µ. This name suggests that µ is a measure on M; it is indeed, as weshow after these brief remarks.

20 MISHEL SKENDERI

Remarks 5.2. Notice �rst that as the collection {Ei}i∈N given by E1 = E andEi = ∅ for i ≥ 2 is a partition of E, it follows that µ(∅) = 0. This therefore implies|µ|(E) ≥ |µ(E)|. The fact µ(∅) = 0 also makes it easy to show that µ is �nitelyadditive. Notice also that if µ is a measure, then it is not necessarily a complexmeasure; therefore |µ| is not always de�ned if µ is a measure. However, if µ is a�nite measure, then it is also a complex measure, and therefore |µ| is de�ned; andin this case, it is easy to see |µ| = µ.

Theorem 5.3. Let (X,M) be a measurable space. If µ is a complex measure onM, then |µ| is a measure on M.

Proof. Let {Ei}i∈N ⊆ M be a pairwise disjoint collection, and set E = ∪i∈NEi.Given any i, if |µ|(Ei) = 0, then let ti = 0. If |µ|(Ei) > 0, let ti be an arbitrarilychosen nonnegative number such that ti < |µ|(Ei). (In this manner, we create{ti}i∈N, a sequence of nonnegative real numbers.) There exists a partition {Ai,j}j∈Nof {Ei} such that ti ≤

∑∞j=1 |µ(Ai,j)|. Now, because {Ai,j}i,j∈N is a partition of E,

we have∑∞i=1 ti ≤

∑∞i=1

∑∞j=1 |µ(Ai,j)| ≤ |µ|(E). Taking the supremum of

∑∞i=1 ti

across all such sequences {ti}, we see∑∞i=1 |µ|(Ei) ≤ |µ|(E).

We now prove the opposite inequality. Let {Bj}j∈N be a partition of E. Thenfor any j, {(Bj ∩ Ei)}i∈N is a partition of Bj , and for any i, {(Bj ∩ Ei)}j∈N isa partition of Ei. Thus, µ(Bj) =

∑∞i=1 µ((Bj ∩ Ei)), and hence,

∑∞j=1 |µ(Bj)| =∑∞

j=1 |∑∞i=1 µ((Bj ∩ Ei))| ≤

∑∞j=1

∑∞i=1 |µ((Bj ∩ Ei))| =

∑∞i=1

∑∞j=1 |µ((Bj ∩

Ei))| ≤∑∞i=1 |µ|(Ei), since for any i,

∑∞j=1 |µ((Bj ∩ Ei))| ≤ |µ|(Ei). Taking the

supremum across all partitions {Bj}j∈N, we have |µ|(E) ≤∑∞i=1 |µ|(Ei). Hence,

|µ| is countably additive on M. As µ(∅) = 0, |µ|(∅) = 0 <∞, which completes theproof. �

We will soon prove a useful, and perhaps unexpected, fact regarding the totalvariation measure, namely its �niteness. The proof will require a Lemma, whichwe prove below.

Lemma 5.4. If {1, . . . , N} ⊆ N and {z1, . . . , zN} ⊆ C, then there exists R ⊆{1, . . . , N} such that 1

π

∑Nk=1 |zk| ≤ |

∑k∈R zk|.

Proof. If f is de�ned on R by x 7→ exp(ix), then f maps R onto the boundaryof the unit disc. Hence, for each k ∈ {1, . . . , N}, there exists αk ∈ R such thatzk = |zk| exp(iαk). For any θ ∈ [−π, π], set R(θ) = {k : cos(αk − θ) > 0}. Then

|∑R(θ)

zk| = |∑R(θ)

exp(−iθ)zk| ≥ Re(∑R(θ)

exp(−iθ)zk) = Re(∑R(θ)

exp(−iθ) exp(iαk)|zk|)

= Re(∑R(θ)

exp(i(αk − θ))|zk|) =∑R(θ)

cos(αk − θ)|zk| =N∑k=1

cos+(αk − θ)|zk|.

(The �rst equality holds because | exp(−iθ)| = 1; notice also that further downwe have used the fact Re(exp(ix)) = cosx.) Now, as cos+ is a continuous func-tion and [−π, π] is compact, it follows that there exists θ∗ such that for every

θ ∈ [−π, π],∑Nk=1 cos+(αk − θ)|zk| ≤

∑Nk=1 cos+(αk − θ∗)|zk|. Continuity im-

plies Riemann integrability, and the above inequality makes it easy to see that∫ π−π(∑Nk=1 |zk| cos+(αk−θ)) dθ ≤ 2π(

∑Nk=1 |zk| cos+(αk−θ∗)). Now, for any β ∈ R,

a simple computation shows∫ π−π cos+(β − θ) dθ = 2. Therefore, 2

∑Nk=1 |zk| ≤


2π(∑Nk=1 |zk| cos+(αk−θ∗)), which implies 1

π

∑Nk=1 |zk| ≤

∑Nk=1 |zk| cos+(αk−θ∗).

Set R = R(θ∗). Then∑Nk=1 |zk| cos+(αk − θ∗) = |

∑R(θ∗) zk|, and this completes

the proof. �

At this point, we emphasize that no results concerning the interplay betweenRiemann and Lebesgue integration were used in the above proof; we used onlyproperties of Riemann integrals.

Theorem 5.5. Let (X,M) be a measurable space, and let µ be a complex measureon M. Then |µ|(X) <∞.

Proof. Suppose there exists E ∈M such that |µ|(E) =∞. De�ne t = π(2+ |µ(E)|).As |µ|(E) = ∞, there exists a partition {Ei} of E such that 2t <

∑∞i=1 |µ(Ei)|.

Therefore, there exists some N ∈ N such that t <∑Ni=1 |µ(Ei)|. Applying the above

Lemma, we see there exists R ⊆ {1, . . . , N} such that 1 < tπ < 1

π

∑Ni=1 |µ(Ei)| ≤

|∑i∈R µ(Ei)|. By �nite additivity, if we set A = ∪i∈REi, it follows that 1 <

tπ < |µ(A)|. Now, let B = (E − A). Then µ(B) + µ(A) = µ(E). Hence, µ(B) =

µ(E) − µ(A). Therefore, 2 = tπ − |µ(E)| < |µ(A)| − |µ(E)| ≤ |µ(A) − µ(E)| =

|µ(B)|. Therefore, there exist disjoint sets A and B whose union is E such that|µ(A)| > 1 and |µ(B)| > 1. As A ∈ M, B ∈ M, and |µ| is a measure on M, �niteadditivity implies |µ|(A) = ∞ or |µ|(B) = ∞. This argument therefore impliesthat if |µ|(X) = ∞, there exist disjoint measurable sets A1 and B1 such thatX = (A1 ∪B1), |µ(A1)| > 1 and |µ|(B1) =∞. As |µ|(B1) =∞, there exist disjointmeasurable sets A2 and B2 such that B1 = A2 ∪B2, |µ(A2)| > 1 and |µ|(B2) =∞.Inductively, we may construct a pairwise disjoint collection {Ai}i∈N ⊆ M. Bycountable additivity of µ, it follows that µ(∪i∈NAi) =

∑∞i=1 µ(Ai). The left-hand

side of this equation is a complex number, whereas the right-hand side is a divergentseries. This is a contradiction. �

Remark 5.6. Since |µ|(X) <∞, it follows that |µ| is bounded. Hence, there existsM ≥ 0 such that for every E ∈ M, |µ|(E) ≤ M. Recalling that for every E ∈ M,|µ(E)| ≤ |µ|(E), this implies |µ(E)| ≤M . Complex measures are therefore bounded.

De�nitions 5.7. Let (X,M) be a measurable space, and let µ and λ be complexmeasures on M. We then de�ne µ+ λ : M→ C by E 7→ µ(E) + λ(E), and for anyα ∈ C, we de�ne αµ : M → C by αµ(E) 7→ α · µ(E). It is not di�cult to verifythat µ+λ and αµ are complex measures, from which it follows that the class of allcomplex measures on M is a vector space over C.

De�nitions 5.8. If (X,M) is a measure space, and µ : M → R is a complexmeasure, then µ is a signed measure on M. De�ne µ+ = 1

2 (|µ| + µ), and de�ne

µ− = 12 (|µ| − µ). (We emphasize that despite the symbols µ+ and µ−, |µ| denotes

the total variation measure of µ, not the absolute value function composed withµ.) We call µ+ and µ− the positive variation and the negative variation of µ,respectively.Notice that µ+ and µ− are themselves complex measures, by the remarks in theabove de�nitions. They are therefore bounded. It is, moreover, easy to see thatthey are, in fact, measures, which implies µ+ and µ− map M into bounded subsetsof the nonnegative real line.Lastly, notice that µ = µ+ − µ− and |µ| = µ+ + µ−. This representation of µ isknown as the Jordan decomposition of µ.

22 MISHEL SKENDERI

De�nitions 5.9. Let (X,M) be a measurable space, let µ be a measure on M, letλ be either a measure or complex measure on M, and let ν be either a measure orcomplex measure on M.If for every E ∈ M, µ(E) = 0 implies λ(E) = 0, then λ is absolutely continuouswith respect to µ, which we denote as follows: λ� µ.If there exists A ∈ M such that for every E ∈ M, λ(E) = λ((A ∩ E)), then λ isconcentrated on A.

If there exist B ∈ M and C ∈ M such that (B ∩ C) = ∅, λ is concentrated onB, and ν is concentrated on C, then λ and ν are mutually singular, and we writeλ ⊥ ν or, equivalently, ν ⊥ λ. We also say "λ is singular with respect to ν" or,equivalently, "ν is singular with respect to λ."

We now present some consequences of, and properties pertinent to, these de�ni-tions.

Proposition 5.10. Let (X,M) be a measurable space, let A ∈ M, and let λ beeither a measure or complex measure on M. Then λ is concentrated on A if andonly if for every E ∈M such that (E ∩A) = ∅, λ(E) = 0.

Proof. One direction is obvious, for if λ is concentrated on A, and E ∈M is suchthat (E ∩A) = ∅, then λ(E) = λ(∅) = 0.Now, let B ∈M be given. Then ((B − A) ∩ A) = ∅. Thus, λ((B − A)) = 0. Finiteadditivity now implies λ(B) = λ((B −A)) + λ((B ∩A)). The result follows. �

Proposition 5.11. Let (X,M) be a measurable space, let λ and ν be measures orcomplex measures on M, and let µ be a measure on M. The following then hold:

(a) The set of all complex measures that are absolutely continuous with respectto µ and the set of all complex measures that are singular with respect to µare both complex vector spaces.

(b) If λ is concentrated on A ∈M, then so is |λ|.(c) If λ ⊥ ν, then |λ| ⊥ |ν|.(d) If λ� µ, then |λ| � µ.(e) If λ� µ and ν ⊥ µ, then λ ⊥ ν.(f) If λ� µ and λ ⊥ µ, then λ = 0.

Proof.(a) All throughout the proof of (a), suppose λ and ν are complex measures.If λ� µ, ν � µ and α ∈ C, then it is clear that λ+ ν � µ and αλ� µ.Suppose now that λ ⊥ µ and ν ⊥ µ. Let A1, A2, B, and C ∈ M be such thatA1 ∩ B = ∅, A2 ∩ C = ∅, µ is concentrated on A1 and A2, λ is concentrated onB, and ν is concentrated on C. Now, let E ∈ M; we have µ((E ∩ A1 ∩ A2)) =µ((E ∩ A1)) = µ(E). Now, let F ∈ M be such that (F ∩ (B ∪ C)) = ∅. Then(F ∩B) = ∅ = (F ∩C). Thus, (λ+ ν)(F ) = 0. Since ((A1 ∩A2)∩ (B ∪C)) = ∅, theresult follows. Furthermore, if α ∈ C and S ∈ M, then λ(S) = λ(S ∩ B). Hence,αλ(S) = αλ(S ∩B).(b) Let E ∈M be such that (E∩A) = ∅. Then λ(E) = 0. Let {Ei}i∈N be a partitionof E. For each i, (Ei ∩ A) = ∅, which implies λ(Ei) = 0. Hence,

∑∞i=1 |λ(Ei)| = 0.

Since {Ei} was an arbitrary partition, the result follows.(c) This is a direct consequence of (b).(d) Let E ∈M be such that µ(E) = 0, and let {Ei} be an arbitrary parition of E.For each i, µ(Ei) ≤ µ(E) = 0, which implies λ(Ei) = 0. The result follows.


(e) If ν is concentrated on A ∈ M, then µ(A) = 0. Thus, if E ∈ M is such thatE ⊆ A, then µ(E) = 0. Thus, λ(E) = 0. Now, let B ∈ M. Then λ(B) = λ((B ∩A)) + λ((B ∩Ac)). As (B ∩A) ⊆ A, the result follows.(f) Part (e) above implies λ ⊥ λ. Let A and B ∈ M be disjoint sets such that λis concentrated on A and λ is concentrated on B. Let E ∈ M. Then 0 = λ(∅) =λ((E ∩A ∩B)) = λ((E ∩A)) = λ(E). �

De�nition 5.12. If (X,M, µ) is a measure space, then µ is a σ-�nite measure ifthere exists {En}n∈N ⊆M such that X = ∪n∈NEn and for each n, µ(En) <∞.Lemma 5.13. Let (X,M, µ) be a measure space, where µ is σ-�nite. Then thereexists a function w ∈ L1(µ) such that for every x ∈ X, 0 < w(x) < 1.

Proof. Let {En}n∈N ⊆M be such that X = ∪n∈NEn and for each n, µ(En) < ∞.Let n be given. If x ∈ (X − En), de�ne wn(x) = 0. If x ∈ En, de�ne wn(x) =

13n[1+µ(En)] . De�ne w =

∑∞n=1 wn. Clearly, w(X) ⊆ (0, 1). Now,

∑∞n=1

∫Xwn dµ =∑∞

n=1

∫Enwn dµ ≤

∑∞n=1 3−n <∞, which implies w ∈ L1(µ). �

Remarks 5.14. Let (X,M) be as in the above Lemma. If µ∗ is the measure onM given by dµ∗ = w dµ, then µ∗ is a �nite measure such that for any E ∈ Mµ∗(E) = 0 if and only if µ(E) = 0. (The last assertion follows from the fact w > 0everywhere on X.)

We now �nally prove the Lebesgue-Radon-Nikodym Theorem.

Theorem 5.15. The Lebesgue-Radon-Nikodym Theorem

Let (X,M) be a measurable space. Let µ be a σ-�nite measure on M, and let λ bea complex measure on M. The following then hold:

(a) The Lebesgue Decomposition Theorem

There exists a unique pair (ν, τ) of complex measures onM such that ν � µ,τ ⊥ µ, and λ = ν + τ. (We call (ν, τ) the Lebesgue decomposition of λrelative to µ.) Moreover, if λ is a measure, then ν and τ are measures.

(b) The Radon-Nikodym Theorem

There exists a unique h ∈ L1(µ) such that for any E ∈M, ν(E) =∫Eh dµ.

Moreover, if ν is a measure, then h is nonnegative; if ν is a signed measure,then h is real-valued. h is called the Radon-Nikodym derivative of ν withrespect to µ. As mentioned earlier, we write dν = h dµ. To re�ect thenewly introduced terminology, we now also write h = dν

dµ . (We do not

assign any independent meaning to the symbol dνdµ .)

Proof. We �rst prove the uniqueness of the Lebesgue decomposition and the unique-ness of h ∈ L1(µ). Let (ν, τ) and (ν◦, τ◦) be Lebesgue decompositions of λ rela-tive to µ. Then ν − ν◦ = τ◦ − τ. Since ν − ν◦ � µ and τ◦ − τ ⊥ µ, we haveν − ν◦ = τ◦ − τ = 0. (Notice that we freely used several parts of Proposition 5.11in the above argument.) Suppose now that h and h◦ ∈ L1(µ) are such that forany E ∈M,

∫Eh dµ =

∫Eh◦ dµ. Since L1(µ) is a vector space, (h − h◦) ∈ L1(µ),

and we have∫E

(h−h◦) dµ = 0. An application of Corollary 2.40 yields the desiredresult.

We now prove the existence of the Lebesgue decomposition and the existenceof such h ∈ L1(µ). We �rst consider the special case in which λ is a complexmeasure that maps into [0,∞). (We will prove the general case of the Lebesgue-Radon-Nikodym Theorem after we establish the Theorem in this special case.)

24 MISHEL SKENDERI

As λ is a complex measure, it is bounded. Therefore, dλ is a bounded measure.Letting w be as in the above Lemma, wdµ de�nes a bounded measure. Settingdφ = dλ + wdµ, it is clear that dφ is a bounded measure. Let E ∈ M. Then(dφ)(E) = (dλ)(E)+(wdµ)(E); in other words,

∫X1E dφ =

∫X1E dλ+

∫X1Ew dµ.

By reasoning similar to that in the proof of Theorem 2.31, for any measurablev : X → [0,∞], we have

(5.16)

∫X

v dφ =

∫X

v dλ+

∫X

vw dµ.

(It is clear that if y ∈ L1(φ), then y ∈ L1(dλ) and y ∈ L1(wdµ), for dλ ≤ dφ andwdµ ≤ dφ. Therefore, it is easy to see that the above equation holds for any y ∈L1(φ); simply apply the above equation to y+ : X → [0,∞] and y− : X → [0,∞].We shall use these facts later on in this proof.)

This implies∫Xv dλ ≤

∫Xv dφ. Now, let f ∈ L2(φ). Hölder's Inequality

(with p = q = 12 ), in conjunction with the above inequality, implies |

∫Xf dλ| ≤∫

X|f | dλ ≤

∫X|f | dφ =

∫X

1·|f | dφ ≤ (∫X|f |2 dφ)

12 ·(φ(X))

12 . Therefore, the linear

functionalM : L2(φ)→ C given byM(f) =∫Xf dλ is a bounded linear functional,

for ||M || = sup{|M(f)| : f such that (∫X|f |2 dφ)

12 = 1} ≤ (φ(X))

12 < ∞, as φ is

a bounded measure. As L2(φ) is a Hilbert space, there exists a unique gc ∈ L2(φ)such that

∫Xf dλ =

∫Xfgc dφ. As g := gc ∈ L2(φ), there exists a unique g ∈ L2(φ)

such that

(5.17)

∫X

f dλ =

∫X

fg dφ.

(Notice that although g is unique when considered as an element of L2(φ), it is notnecessarily unique when considered as an element of L2(φ).) Now, let E ∈ M besuch that φ(E) > 0; since φ is a bounded measure, 1E ∈ L2(φ), which implies thatthe above equation holds with 1E in place of f .

(If no such E exists, then both (a) and (b) follow in the general case: For, aswdµ ≤ φ and w > 0 everywhere on X, this implies µ(X) = 0. If λ∗ is an arbitrarycomplex measure on M, then choose ν∗ = µ. Clearly, ν∗ � µ. Choose τ∗ = λ∗. Asµ is concentrated on ∅, τ∗ ⊥ µ. It is obvious that λ∗ = ν∗ + τ∗, and it is also clearthat if λ∗ is a measure, then so are ν∗ and τ∗. This proves (a). To prove (b), sinceµ(X) = 0 and ν∗ = µ, any h ∈ L1(µ) will do; choose h = 1X , for example. Noticethat as µ(X) = 0, h is indeed uniquely determined up to sets of µ-measure zero.)

This therefore implies 0 ≤ λ(E) =∫Eg dφ. As 0 ≤ λ ≤ φ on M, we have

0 ≤ 1φ(E)

∫Eg dφ = λ(E)

φ(E) ≤ 1. Since [0, 1] (or, more formally, ([0, 1] × {0})) is

a closed subset of C, Theorem 2.41 implies that with respect to φ, g(x) ∈ [0, 1]for almost every x ∈ X. It is, therefore, permissible to assume that for everyx ∈ X, g(x) ∈ [0, 1], and we now do so. As f and g ∈ L2(φ), Proposition 4.4implies fg ∈ L1(φ). Using (5.16) (along with the remarks immediately followingthe equation), we re-write (5.17) as

(5.18)

∫X

(1− g)f dλ =

∫X

fgw dµ.

Now, let A = gpre([0, 1)), and let B = gpre(1). It is obvious that A and B ∈ M.De�ne ν : M → [0,∞) and τ : M → [0,∞) by ν(E) = λ((A ∩ E)) and τ(E) =


λ((B ∩ E)). It is easy to check that ν and τ are measures on M. Moreover, as(A∪B) = X, and (A∩B) = ∅, �nite additivity implies λ = ν + τ. We now wish toprove ν � µ and τ ⊥ µ to show that (ν, τ) is the Lebesgue decomposition of λ.

Now, take f = 1B in (5.18). (This is possible because 1B ∈ L2(φ), as φ isa bounded measure.) It follows that

∫Bw dµ = 0. As w > 0 everywhere on

X, µ(B) = 0. Clearly, τ is concentrated on B. Now, for any E ∈ M, µ(E) =µ((E ∩ B)) + µ((E ∩ Bc)). Since µ((E ∩ B)) = 0, µ is concentrated on Bc. Thus,τ ⊥ µ.

Since g is bounded and φ is a �nite measure, it follows that for any n ∈ N andfor any E ∈ M, 1E

∑ni=0 g

i ∈ L2(φ). Therefore, (5.18) (with f = 1E

∑ni=0 g

i)

implies∫E

(1− gn+1) dλ =∫E

∑n+1i=1 g

iw dµ. If G : M→ [0,∞) is given by G(E) =∫E

(1 − gn+1) dλ, then G is a measure on M. This implies∫

(A∩E)(1 − gn+1) dλ +∫

(B∩E)(1− gn+1) dλ =

∫E

(1− gn+1) dλ =∫E

∑n+1i=1 g

iw dµ. Therefore,

(5.19)

∫(A∩E)

(1− gn+1) dλ =

∫E

(1− gn+1) dλ =

∫E

n+1∑i=1

giw dµ.

Now, for each x ∈ (A ∩ E) and for each n ∈ N, (1 − gn+1(x)) > 0; for eachx ∈ (A∩E), (1−gn+1(x)) monotonically increases to 1 as n approaches in�nity. TheMonotone Convergence Theorem applies to show limn→∞

∫(A∩E)

(1 − gn+1) dλ =

λ((A∩E)) = ν(E). Now, for every n and for every x ∈ X,∑n+1i=1 g

i(x)w(x) ≥ 0 and∑n+1i=1 g

i(x)w(x) ≤∑n+2i=1 g

i(x)w(x). Therefore, there exists a measurable h : X →[0,∞] such that for every x ∈ X, limn→∞

∑n+1i=1 g

i(x)w(x) = h(x). The Monotone

Convergence Theorem applies to show∫Eh dµ = limn→∞

∫E

∑n+1i=1 g

iw dµ. Hence,

ν(E) =∫Eh dµ. This shows ν � µ.

Since∫Xh dµ = ν(X) = λ((A ∩X)) ≤ λ(X) <∞, we conclude h ∈ L1(µ).

We now prove the general case of the Lebesgue Decomposition Theorem. Sup-pose λ is an arbitrary complex measure onM, and let λ1 and λ2 be signed measureson M such that λ = λ1 + iλ2. (It is easy to check that such λ1 and λ2 exist.) Itthen follows that λ = λ+

1 − λ−1 + i(λ+

2 − λ−2 ). Since λ+

1 , λ−1 , λ

+2 , and λ

−2 are mea-

sures on M, we may apply to them the special case of the Lebesgue DecompositionTheorem that we have already proved. Let (νλ+

1, τλ+

1), (νλ−

1, τλ−

1), (νλ+

2, τλ+

2), and

(νλ−2, τλ−

2) be the Lebesgue decompositions of λ+

1 , λ−1 , λ

+2 , and λ

−2 , respectively. Set

νλ = νλ+1− νλ−

1+ i(νλ+

2− νλ−

2). Set τλ = τλ+

1− τλ−

1+ i(τλ+

2− τλ−

2). Using Proposi-

tion 5.11, it is not di�cult to check that (νλ, τλ) is the Lebesgue decomposition of λ.

We now prove the general case of the Radon-Nikodym Theorem. Let E ∈ M,and let +h1,

−h1,+h2, and

−h2 ∈ L1(µ) be such that the following hold: νλ+1

=∫E

+h1 dµ, νλ−1

=∫E−h1 dµ, νλ+

2=∫E

+h2 dµ, and νλ−2

=∫E−h2 dµ. Set

h = +h1 − −h1 + i(+h2 − −h2). Note that h ∈ L1(µ), as L1(µ) is a vector spaceover C. It is easy to verify ν(E) =

∫Eh dµ, and this completes the proof. �

Remark 5.20. Suppose that the hypotheses of the Lebesgue-Radon-Nikodym The-orem hold, and suppose further that λ � µ. This implies ν = λ and τ = 0. TheRadon-Nikodym Theorem therefore asserts the existence of h ∈ L1(µ) such that

26 MISHEL SKENDERI

for any E ∈ M, λ(E) =∫Eh dµ. This makes it clear that the Radon-Nikodym

Theorem asserts a converse of Theorem 2.31.

In the Appendix, we construct the Lebesgue measure on Rk. We also present anexample, involving the Lebesgue measure, that is germane to the Radon-NikodymTheorem.

6. Appendix: A Construction of the Lebesgue Measure on Rk and a

Relevant Example Involving the Lebesgue Measure

In this Appendix, we shall construct the Lebesgue measure on Rk, where k ∈ N.We have relegated the construction to the Appendix because the Lebesgue measurewas not required for the ultimate aim of this paper, the Lebesgue-Radon-NikodymTheorem. However, this measure is so ubiquitous that it would be amiss for thispaper, which claims to be a primer of Lebesgue integration, to neglect it. More-over, the Lebesgue measure will enable us to provide an example that illustratesthe necessity of σ-�niteness in the hypotheses of the Radon-Nikodym Theorem.

Our construction will make use of Urysohn's Lemma and the Riesz Representa-tion Theorem (both of which we will soon state). We will also use the de�nitionof∫Rk f(x) dx, for f ∈ Cc(Rk) (we will give the de�nition of Cc(Rk) shortly). The

de�nition of∫Rk f(x) dx for such f can be found in pages 245-247 of [1]. Our con-

struction of the Lebesgue measure is unconventional and so we will prove that it isequivalent to a standard construction, namely the one that is presented in [3].

De�nitions 6.1. If X is a topological space, and f : X → C, then the support off is the closure of the set {x : f(x) 6= 0}, and we de�ne Cc(X) to be the set of allcomplex-valued continuous functions de�ned on X that possess compact support.It is not di�cult to verify that Cc(X) is a vector space over C.

We will now make use of the topological notions of a Hausdor� space and alocally compact space. We say "X is locally compact" to signify the following: Forevery x ∈ X, there exists an open set U such that x ∈ U and U is compact. Weassume the reader is familiar with such spaces and their properties; however, ifnecessary, see pages 35-37 of [2].

Theorem 6.2. Urysohn's Lemma

Suppose X is a locally compact Hausdor� space, V is open in X, K is compact,and K ⊆ V. Then there exists f ∈ Cc(X) such that 0 ≤ f ≤ 1, f = 1 on K, and thesupport of f is contained in V . In terms of indicator functions, this implies thereexists f ∈ Cc(X) such that 1K ≤ f ≤ 1V .

Proof. As Urysohn's Lemma is a standard result in point-set topology, in whichwe assume a background, the proof is omitted. The proof is given on page 39 of[2]. �

Theorem 6.3. The Riesz Representation Theorem

Suppose X is a locally compact Hausdor� space, and let Λ : Cc(X) → C be alinear mapping such that for every f ∈ Cc(X), if f(X) ⊆ [0,∞), then Λ(f) ≥ 0.Then there exists a σ-algebra M in X and a unique measure µ on M for which thefollowing hold:

(a) If f ∈ Cc(X), then Λ(f) =∫Xf dµ. We say therefore that µ represents

Λ.


(b) If E ⊆ X, then µ(E) = inf{µ(V ) : E ⊆ V, V open}. (Notice that thisrelation is well-de�ned, as E ⊆ X, and X is open.)

(c) If K ⊆ X is a compact set, then µ(K) <∞.(d) M contains every Borel set of X, and M is equal to the set of all E ⊆ X

that satisfy the following: For every compact set K ⊆ X, EK := (E ∩K)is such that µ(EK) < ∞ and µ(EK) = sup{µ(J) : J ⊆ EK , J compact}.(Notice that this relation is well-de�ned, as ∅ ⊆ EK , and ∅ is compact.)

(e) If E is an open set or if E ∈ M is such that µ(E) < ∞, then µ(E) =sup{µ(K) : K ⊆ E, K compact}.

(f) If E ∈M, A ⊆ E, and µ(E) = 0, then A ∈M. We say that µ is complete.

We omit the proof of the Riesz Representation Theorem because it is extremelylong and because the only chance which we will have to use it will be in the con-struction of the Lebesgue measure. The proof can be found in pages 41-47 of [2].

Remarks 6.4. Notice that although µ(E) is de�ned for every E ⊆ X, the RieszTheorem asserts that µ is countably additive on M and not on P (X). Notice,however, that by (b) of the Theorem's conclusion, µ is monotone on P (X), notjust on M. Lastly, notice that if E ⊆ X is such that µ(E) = 0, then E ∈ M. Forif K ⊆ X is compact, then µ(EK) = 0 < ∞ (by monotonicity) and µ(EK) clearlysatis�es conclusion (d) of the Riesz Theorem.

De�nitions 6.5. If k ∈ N, then a k-cell is a set de�ned by∏ki=1〈ai, bi〉, such that,

for every i, we have ai and bi ∈ R, and 〈ai, bi〉 denotes (ai, bi), [ai, bi], (ai, bi], or[ai, bi). If for some j, aj = bj = c, then [aj , bj ] = {c}, and in the other cases,

〈aj , bj〉 = ∅. If B is a k-cell, then we de�ne the volume of B by vol(B) =∏ki=1(bi−

ai).

Now, notice that Rk is a locally compact Hausdor� space. Let Λ : Cc(Rk)→ Cbe given by f 7→

∫Rk f(x) dx. It is clear that Λ satis�es the hypotheses of the Riesz

Representation Theorem. Therefore, there exists a σ-algebraM in Rk that containsevery Borel set in Rk, and there exists a unique measure m on M that satis�es theconclusions of the Riesz Theorem. Very importantly, it satis�es conclusion (a),which means that for every f ∈ Cc(Rk),

∫Rk f dm = Λ(f). We will show that this

m is Lebesgue measure. We now prove our �rst result regarding m.

Theorem 6.6. If K is a compact k-cell, then m(K) = vol(K).

Proof. Let K =∏ki=1[ai, bi]. Given any n ∈ N, de�ne Vn =

∏ki=1(ai − 1

n , bi + 1n ).

Urysohn's Lemma implies there exists fn ∈ Cc(Rk) such that 0 ≤ fn ≤ 1; italso implies the support of fn is a subset of Vn and fn = 1 on K. This, inturn, implies 1K ≤ fn ≤ 1Vn

, as is easy to check. Now for every n, we have∫Rk fn dm =

∫Rk fn(x) dx. By the de�nition of fn, it is clear that vol(K) ≤∫

Rk fn(x) dx ≤ vol(Vn). Subtracting vol(K) shows 0 ≤∫Rk fn(x) dx − vol(K) ≤

vol(Vn) − vol(K) = ( 2n )k. It follows that limn→∞

∫Rk fn(x) dx = vol(K). Now,

notice that 1V1is summable: We have

∫Rk 1V1

dm = m(V1) < ∞, for V1 is con-tained in some compact set J , and the Riesz Theorem tells us m(J) < ∞, whichimplies m(V1) < ∞, by monotonicity of m. Notice also that 1V1

dominates {fn}:For each n, Vn+1 ⊆ Vn, so if x /∈ V1, then for each n, fn(x) = 0 = 1V1

(x), since the

28 MISHEL SKENDERI

support of fn lies in Vn. If x ∈ V1, then for each n, 0 ≤ fn(x) ≤ 1 = 1V1(x).Moreover, 1K is the pointwise limit of {fn}, for if x ∈ K, then for every n,fn(x) = 1. If x /∈ K, then because the support of fn lies in Vn, fn(x) = 0 forevery n su�ciently large; that is, if x /∈ K = ∩n∈NVn, then there exists some Nsuch that x /∈ VN , which implies x /∈ Vn for every n ≥ N . The Dominated Con-vergence Theorem therefore shows limn→∞

∫Rk fn dm =

∫Rk 1K dm = m(K). Since

limn→∞∫Rk fn dm = limn→∞

∫Rk fn(x) dx, m(K) = vol(K), as desired. �

Corollary 6.7. If A is a k-cell, then m(A) = vol(A).

Proof. Let A =∏ki=1〈ai, bi〉. Given any n, de�ne An in the following way: For any

i ∈ {1, . . . , k}, if 〈ai, bi〉 = [ai, bi], then set Ain = [ai, bi]; if 〈ai, bi〉 = (ai, bi), then

set Ain = [ai+

1n , bi−

1n ]; if 〈ai, bi〉 = [ai, bi), then set Ai

n = [ai, bi− 1n ]; if 〈ai, bi〉 =

(ai, bi], then set Ain = [ai + 1

n , bi]. Finally, set An =∏ki=1Ai

n. Then A = ∪n∈NAn,and for each n, An ⊆ An+1. Continuity from below and the above theorem implym(A) = limn→∞m(An) = limn→∞ vol(An). Now, limn→∞ vol(An) = vol(A), since0 ≤vol(A)−vol(An) ≤ ( 2

n )k. This completes the proof. �

Remarks 6.8. If E ⊆ Rk, then denote the interior of E by int(E), and denotethe boundary of E by ∂(E). (Denote the closure of E by E, as usual.) Sinceint(E), E, and ∂E are all Borel sets, they are all elements of M. Now, supposeA is a k-cell. Since vol(A) = vol(int(A)) = vol(A), the above Corollary impliesm(A) = m(int(A)) = m(A). As m(A) <∞, �nite additivity shows m(∂(A)) = 0.

We will shortly employ a standard fact of point-set topology, namely that everyopen set in Rk can be written as the countable union of a family of nearly disjointcompact k-cells. (A family of compact k-cells is said to be nearly disjoint if thefollowing holds: If A and B are any two elements of the family such that A 6= B,then the interiors of A and B are disjoint.) We record this fact below as a Lemma.

Lemma 6.9. If E ⊆ Rk is open, then E can be written as the countable union ofa family of nearly disjoint compact k-cells.

Proof. For the proof, see page 8 of [3]. The proof therein is actually that of astronger conclusion that immediately implies the one we have stated. �

We now present another Lemma that will be quite important.

Lemma 6.10. If {An}n∈N ⊆ Rk is a countable family of nearly disjoint compactk-cells, and A = ∪n∈NAn, then m(A) =

∑∞n=1m(An).

Proof. Let B = ∪n∈N int(An), and let C = ∪n∈N ∂(An). Notice B ∩ C = ∅. Sincem(C) ≤

∑∞n=1m(∂(An)) = 0, this implies m(A) = m(B). Now, as the interiors

are disjoint, this implies m(A) =∑∞n=1m(int(An)) =

∑∞n=1m(An). �

We now proceed to show that our construction of the Lebesgue measure is equiv-alent to that given in [3]. Before doing so, however, it is �tting to explain how theLebesgue measure is constructed in [3]. It is constructed as follows: First, a functionon P(Rk) called outer measure is de�ned, and then the restriction of outer measureto a certain σ-algebra is what is then called Lebesgue measure. As the notationfor outer measure in [3] is rather idiosyncratic, we will denote outer measure by linstead, and we will denote its corresponding σ-algebra by L.We will �rst show that for any E ⊆ Rk, m(E) = l(E). After having shown this, we


will show M = L.

Now, let E ⊆ Rk be given. Then if RE is the class of all countable families ofcompact k-cells whose union covers E, and if {Cn}n∈N ∈ RE , then on page 33 of[3], l(E) is de�ned as follows: l(E) = infRE

∑∞n=1vol(Cn). Thus, 0 ≤ l(E) ≤ ∞.

(Observe that this relation is well-de�ned, as RE is nonempty.)We now proceed to show that the de�nition of l agrees with the de�nition of m

on every open set of Rk.

Theorem 6.11. Let V ⊆ Rk be open. If SV is the class of all countable familiesof nearly disjoint compact k-cells whose union covers V , and if {Dn}n∈N ∈ SV ,then the following is true: m(V ) = infSV

∑∞n=1m(Dn). (Observe that this relation

is well-de�ned, as SV is nonempty.)

Proof. There exists {Dn} ∈ SV such that V = ∪n∈N Dn. Thus,m(V ) = m(∪n∈N Dn) =∑∞n=1m(Dn). The result follows. �

Notice that we may also write m(V ) = infSV

∑∞n=1vol(Dn).

Theorem 6.12. Let V ⊆ Rk be open. If {Cn}n∈N ∈ RV and {Dn}n∈N ∈ SV , thenm(V ) = infRV

∑∞n=1m(Cn) = infSV

∑∞n=1m(Dn).

Proof. Since SV ⊆ RV , infRV

∑∞n=1m(Cn) ≤ infSV

∑∞n=1m(Dn). Now, let

{An} ∈ RV . Then m(V ) ≤∑∞n=1m(An). Also, there exists {Bn} ∈ SV such

that V = ∪n∈N Bn and we know∑∞n=1m(Bn) = m(V ) ≤

∑∞n=1m(An), which

completes the proof. �

It therefore follows that m(V ) = infRV

∑∞n=1vol(Cn).

This shows that l and m do indeed agree on the open sets of Rk. Now, by part(b) of the Riesz Representation Theorem, the following is true: If E ⊆ Rk, thenm(E) = inf{m(V ) : E ⊆ V, V open}. Now, on page 36 of [3], the following isproved: If E ⊆ Rk, then l(E) = inf{l(V ) : E ⊆ V, V open}.

Combined with the agreement of l and m on the open sets of Rk, this shows thatfor every E ⊆ Rk, l(E) = m(E), as desired.

We now prove M = L.A theorem on page 43 of [3] says that E ∈ L if and only if E = A∪Z, where A

is an Fσ set and Z is a set such that l(Z) = 0. It is this criterion that we will use.

Lemma 6.13. If E ⊆ Rk is such that E ∈ M, then there exists an Fσ set A andthere exists a Gδ set B such that A ⊆ E ⊆ B and m(B −A) = 0.

Proof. Notice that there exists a family of compact sets {Kn}n∈N such that Rk =∪n∈NKn. If E ∈M, then for each n, monotonicity and conclusion (c) of the RieszTheorem imply m((E ∩Kn)) ≤ m(Kn) <∞; conclusion (b) of the Riesz Theoremimplies that if ε > 0 and n ∈ N, then there exists an open set Vn such that(E ∩ Kn) ⊆ Vn and m(Vn) − m((E ∩ Kn)) < ε

4n+1 . Since Vn, Kn, and E are allelements of M and m((E ∩Kn)) < ∞, countable additivity implies m(Vn − (E ∩Kn)) = m(Vn)−m(((E ∩Kn)) < ε

4n+1 . Now, set V = ∪n∈NVn. Then (V −E) ∈M,and it is not di�cult to check that (V − E) ⊆ ∪n∈N(Vn − (E ∩Kn)). This impliesm((V −E)) < ε

2 . Now, because V is open and Ec ∈M, the same argument applies

30 MISHEL SKENDERI

to show that there exists an open set W such that Ec ⊆W and m((W −Ec)) < ε2 .

Setting F = W c, we see F ⊆ E, and it is easy to check (E−F ) = (W −Ec). Thus,m((E−F )) < ε

2 . Therefore, m((V −F )) < ε. Notice that F is a closed set. Now, if

i ∈ N, set εi = 1i . Then there exists an open set Vi and there exists a closed set Fi

such that Fi ⊆ E ⊆ Vi and m((Vi−Fi)) < 1i . Set A = ∪i∈NFi, and set B = ∩i∈NVi.

Then A is an Fσ set, B is a Gδ set, and for every i ∈ N, (B −A) ⊆ (Vi − Fi). Thisimplies that for every i ∈ N,m((B−A)) ≤ m((Vi−Fi)) < 1

i . Hence,m((B−A)) = 0.This completes the proof. �

Theorem 6.14. If E ⊆ Rk, then E ∈ M if and only if there exists an Fσ set Aand a set Z with m(Z) = 0 such that E = (A ∪ Z).

Proof. First, suppose E ∈ M. Then there exists a Fσ set A and a Gδ set B suchthat A ⊆ E ⊆ B and m(B−A) = 0. Set Z = (E−A). By monotonicity, m(Z) = 0,and clearly, E = (A∪Z). To prove the converse, notice that if A is an Fσ set, thenA is a Borel set, which implies A ∈M. If Z is such that m(Z) = 0, then Z ∈M, bythe remarks following the Riesz Theorem. If E = (A∪Z), then clearly E ∈M. �

We have therefore shown that the measure m is indeed the Lebesgue measure onRk. Our de�nition of Lebesgue measure suggests there is some connection betweenRiemann integration and integration with respect to the Lebesgue measure. Westate this connection below; its proof can be found on page 323 of [1].

Theorem 6.15. If f : [a, b] → C is Riemann integrable, then f is summable with

respect to the Lebesgue measure m, and∫ baf(x) dx =

∫[a,b]

f dm.

We conclude by presenting an example that shows why the assumption of σ-�niteness is necessary in the hypotheses of the Radon-Nikodym Theorem.

Example 6.16. Consider the measurable space (X,M), where X = (0, 1) andM is the class of all Lebesgue measurable subsets of X. Let m be the Lebesguemeasure on M, and let µ be the counting measure on M. Observe that µ is notσ-�nite, as X is uncountable. Clearly, m � µ. Suppose, for contradiction, thatthere exists h ∈ L1(µ) such that for any E ∈ M, m(E) =

∫Eh dµ. Let x ∈ X.

As {x} is a Borel set, {x} ∈ M. As {x} is a compact 1-cell, Theorem 6.6 impliesm({x}) = vol({x}) = 0. As µ({x}) = 1, this implies h = 0 on X. On the otherhand, Corollary 6.7 implies m(X) = 1 6= 0, a contradiction.

Acknowledgments. It is a great pleasure to thank Hyomin Choi, my mentor,who always listened patiently to, and addressed, any questions and concerns Ihad regarding the material. I particularly wish to thank her for her support andguidance during the most challenging moments of this REU. Furthermore, I thankher for conscientiously proofreading this paper.I would also like to thank my friend and fellow REU participant Simon Lazarus forhis meticulous proofreading of several important parts of this paper.I also thank my friend Joshua Bosshardt for his helpful comments on the �rstcomplete draft of this paper, several of which I took into account when makingrevisions.Finally, I wish to thank Professor J. Peter May: I greatly appreciated his stylisticadvice on this paper, and I am grateful for the opportunity he o�ered me thissummer in his capacity as organizer of this REU.


References

[1] Walter Rudin. Principles of Mathematical Analysis. McGraw Hill, Inc. 1976.[2] Walter Rudin. Real and Complex Analysis. McGraw Hill, Inc. 1987.[3] Richard L. Wheeden and Antoni Zygmund. Measure and Integral: A Introduction to Real

Analysis. Marcel Dekker, Inc. 1977.

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