Chapter 6. Integration §1. Integrals of Nonnegative Functions Let (X, S, Μ

Chapter 6. Integration

§1. Integrals of Nonnegative Functions

Let (X, S, µ) be a measure space. We denote by L+ the set of all measurable functions from X to [0, ∞]. + Let φ be a simple function in L . Suppose f(X) = {a1, . . . , am}. Then φ has the Pm −1 standard representation φ = j=1 ajχEj , where Ej := f ({aj}), j = 1, . . . , m. We deﬁne the integral of φ with respect to µ to be

m Z X φ dµ := ajµ(Ej). X j=1

Pm For c ≥ 0, we have cφ = j=1(caj)χEj . It follows that

m Z X Z cφ dµ = (caj)µ(Ej) = c φ dµ. X j=1 X

Pm Suppose that φ = j=1 ajχEj , where aj ≥ 0 for j = 1, . . . , m and E1,...,Em are m Pn mutually disjoint measurable sets such that ∪j=1Ej = X. Suppose that ψ = k=1 bkχFk , where bk ≥ 0 for k = 1, . . . , n and F1,...,Fn are mutually disjoint measurable sets such n that ∪k=1Fk = X. Then we have

m n n m X X X X φ = ajχEj ∩Fk and ψ = bkχFk∩Ej . j=1 k=1 k=1 j=1

If φ ≤ ψ, then aj ≤ bk, provided Ej ∩ Fk 6= ∅. Consequently,

m m n m n n X X X X X X ajµ(Ej) = ajµ(Ej ∩ Fk) ≤ bkµ(Ej ∩ Fk) = bkµ(Fk). j=1 j=1 k=1 j=1 k=1 k=1

In particular, if φ = ψ, then

m n X X ajµ(Ej) = bkµ(Fk). j=1 k=1

In general, if φ ≤ ψ, then

m n Z X X Z φ dµ = ajµ(Ej) ≤ bkµ(Fk) = ψ dµ. X j=1 k=1 X

1 Furthermore,

m n Z X X Z Z (φ + ψ) dµ = (aj + bk)µ(Ej ∩ Fk) = φ dµ + ψ dµ. X j=1 k=1 X X

+ Pm Suppose φ is a simple function in L and j=1 ajχEj is its standard representation. Pm If A ∈ S, then φχA = j=1 ajχEj ∩A. We deﬁne

m Z Z X φ dµ := φχA dµ = ajµ(Ej ∩ A). A X j=1

R The set function ν from S to [0, ∞] given by ν(A) := A φ dµ is a measure. Indeed, ν(∅) = 0. To prove σ-additivity of ν, we let (Ak)k=1,2,... be a sequence of mutually disjoint ∞ measurable sets and A := ∪k=1Ak. Then we have

m m ∞ Z X X X ν(A) = φ dµ = ajµ(Ej ∩ A) = ajµ(Ej ∩ Ak) A j=1 j=1 k=1 ∞ m ∞ ∞ X X X Z X = ajµ(Ej ∩ Ak) = φ dµ = ν(Ak). k=1 j=1 k=1 Ak k=1

The Lebesgue integral of a function f ∈ L+ is deﬁned to be

Z nZ o f dµ = sup φ dµ : 0 ≤ φ ≤ f, φ simple . X X

When f is a simple function, this definition agrees with the previous one. If f, g ∈ L+ and R R + f ≤ g, then it follows from the definition that X f dµ ≤ X g dµ. If f ∈ L and A ∈ S, R R R R we define A f dµ := X (fχA) dµ. Since fχA ≤ f, we have A f dµ ≤ X f dµ.

Theorem 1.1. (The Monotone Convergence Theorem) Let (fn)n=1,2,... be an increasing + sequence of functions in L . If f = limn→∞ fn, then Z Z f dµ = lim fn dµ. X n→∞ X

Proof. Since each fn is measurable and f = limn→∞ fn, the function f is measurable.

Moreover, fn ≤ fn+1 ≤ f for every n ∈ IN. Hence, Z Z Z fn dµ ≤ fn+1 dµ ≤ f dµ ∀ n ∈ IN. X X X

2 Thus, there exists an element α ∈ [0, ∞] such that

Z Z α = lim fn dµ ≤ f dµ. n→∞ X X

Let φ be any simple function on (X,S) such that 0 ≤ φ ≤ f. Fix c ∈ (0, 1) for the time being and deﬁne

En := {x ∈ X : fn(x) ≥ cφ(x)}, n ∈ IN.

Each En is measurable. For n ∈ IN, we have

Z Z Z Z fn dµ ≥ fn dµ ≥ (cφ) dµ = c φ dµ. X En En En

∞ R R But En ⊆ En+1 for n ∈ IN and X = ∪ En. Hence, limn→∞ φ dµ = φ dµ. It n=1 En X follows that Z Z Z α = lim fn dµ ≥ lim c φ dµ = c φ dµ. n→∞ n→∞ X En X R R Thus, α ≥ c X φ dµ for every c ∈ (0, 1). Consequently, α ≥ X φ dµ for every simple R function satisfying 0 ≤ φ ≤ f. Therefore, α ≥ X f dµ and

Z Z Z f dµ ≤ α = lim fn dµ ≤ f dµ. X n→∞ X X

This completes the proof.

The monotone convergence theorem is still valid if the conditions fn(x) ≤ fn+1(x) for all n ∈ IN and limn→∞ fn(x) = f(x) hold for almost every x ∈ X. Indeed, under the stated conditions, there exists a null set N such that for every x ∈ X \ N, fn(x) ≤ fn+1(x) R + and limn→∞ fn(x) = f(x). Note that N g dµ = 0 for all g ∈ L . With E := X \ N we have Z Z Z Z f dµ = fχE dµ = lim fnχE dµ = lim fn dµ. X X n→∞ X n→∞ X

The hypothesis that the sequence (fn)n=1,2,... be increasing almost everywhere is es- sential for the monotone convergence theorem. For example, let µ be the Lebesgue measure on X = IR, and let fn := χ(n,n+1) for n ∈ IN. Then (fn)n=1,2,... converges to 0 pointwise, R but IR fn dµ = 1 for every n ∈ IN.

3 Theorem 1.2. Let f, g ∈ L+ and c ≥ 0. Then Z Z Z Z Z (cf) dµ = c f dµ and (f + g) dµ = f dµ + g dµ. X X X X X

+ Proof. For f, g ∈ L , we can ﬁnd increasing sequences (fn)n=1,2,... and (gn)n=1,2,... of + simple functions in L such that f = limn→∞ fn and g = limn→∞ gn. By the monotone convergence theorem we have Z Z Z Z (cf) dµ = lim (cfn) dµ = lim c fn dµ = c f dµ X n→∞ X n→∞ X X

and Z Z Z Z Z Z (f +g) dµ = lim (fn +gn) dµ = lim fn dµ+ lim gn dµ = f dµ+ g dµ. X n→∞ X n→∞ X n→∞ X X X

+ P∞ Theorem 1.3. If (fk)k=1,2,... is a sequence of functions in L and f = k=1 fk, then

∞ Z X Z f dµ = fk dµ. X k=1 X

Pn Proof. For n ∈ IN, let gn := k=1 fk. Then (gn)n=1,2,... is an increasing sequence of + functions in L such that limn→∞ gn = f. By the monotone convergence theorem we have

n ∞ Z Z Z X X Z f dµ = lim gn dµ = lim gk dµ = fn dµ. n→∞ n→∞ X X X k=1 k=1 X

Theorem 1.4. Let f be a function in L+, and let ν be the function from S to [0, ∞] given R by ν(A) := A f dµ, A ∈ S. Then ν is a measure on S.

Proof. Clearly, ν(∅) = 0. Suppose that (En)n=1,2,... is a sequence of disjoint sets in S and ∞ P∞ E = ∪n=1En. Then we have fχE = n=1 fχEn . By Theorem 1.3 we obtain

∞ ∞ Z Z X Z X Z f dµ = fχE dµ = fχEn dµ = f dµ. E X n=1 X n=1 En

This proves that ν is σ-additive.

+ Theorem 1.5. Let (fn)n=1,2,... be a sequence in L . Then Z Z lim inf fn dµ ≤ lim inf fn dµ. X n→∞ n→∞ X

4 Proof. Let f := lim infn→∞ fn and gk := infn≥k fn for k ∈ IN. Then (gk)k=1,2,... is an

increasing sequence and limk→∞ gk = f. By the monotone convergence theorem, we have Z Z f dµ = lim gk dµ. X k→∞ X R R For ﬁxed k, gk ≤ fn for all n ≥ k; hence, X gk dµ ≤ X fn dµ for all n ≥ k. It follows that Z Z gk dµ ≤ inf fn dµ. X n≥k X We conclude that Z Z Z Z f dµ = lim gk dµ ≤ lim inf fn dµ = lim inf fn dµ. X k→∞ X k→∞ n≥k X n→∞ X The above theorem is often called Fatou’s lemma.

§2. Integrable Functions

Let (X, S, µ) be a measure space. A measurable function f from X to IR is said to R + R − be integrable over a set E ∈ S if both E f dµ and E f dµ are ﬁnite. In this case, we deﬁne Z Z Z f dµ := f + dµ − f − dµ. E E E Clearly, f is integrable if and only if |f| is.

Theorem 2.1. Let f be an integrable function on a measure space (X, S, µ). Then f is R finite almost everywhere. Moreover, if X |f| dµ = 0, then f = 0 almost everywhere. Proof. In order to prove the theorem, it suffices to consider the case f ≥ 0. For n ∈ IN, ∞ let En := {x ∈ X : f(x) ≥ n}. For E := ∩n=1En we have Z Z f dµ ≥ f dµ ≥ nµ(En) ≥ nµ(E). X En R R It follows that µ(E) ≤ X f dµ/n for all n ∈ IN. Since 0 ≤ X f dµ < ∞, we conclude that µ(E) = 0. If x ∈ X \ E, then x∈ / En for some n ∈ IN, that is, f(x) < n. This shows that f(x) < ∞ for all x ∈ X \ E. In other words, f is finite almost everywhere.

For the second statement, let Kn := {x ∈ X : f(x) ≥ 1/n} for n ∈ IN, and let ∞ K := ∪n=1Kn. Then we have K = {x ∈ X : f(x) > 0} and Z Z 0 = f dµ ≥ f dµ ≥ µ(Kn)/n. X Kn

5 It follows that µ(Kn) = 0 for all n ∈ IN. Hence, µ(K) = limn→∞ µ(Kn) = 0. Thus, f = 0 almost everywhere.

Theorem 2.2. Let f and g be integrable functions on a measure space (X, S, µ). Then the following statements are true. R R (1) For each c ∈ IR, cf is integrable and X (cf) dµ = c X f dµ. R R R (2) The function f + g is integrable, and X (f + g) dµ = X f dµ + X g dµ. R R (3) If f ≤ g almost everywhere, then X f dµ ≤ X g dµ. ∞ (4) If A = ∪n=1An, where An (n ∈ IN) are mutually disjoint sets in S, then

∞ Z X Z f dµ = f dµ. A n=1 An

Proof. (1) If c ≥ 0, then (cf)+ = cf + and (cf)− = cf −. If c < 0, then (cf)+ = −cf − − + R R and (cf) = −cf . In both cases, cf is integrable and X (cf) dµ = c X f dµ. (2) Since |f + g| ≤ |f| + |g|, f + g is integrable. Moreover, by Theorem 2.1, both f and g are ﬁnite almost everywhere. Hence, there exists a null set E such that both f(x) and g(x) are ﬁnite for all x ∈ X \ E. Let h(x) := f(x) + g(x) for x ∈ X \ E and h(x) := 0 for x ∈ E. On X \ E we have h+ − h− = f + − f − + g+ − g−. It follows that h+ + f − + g− = h− + f + + g+. Hence, Z Z Z Z Z Z h+ dµ + f − dµ + g− dµ = h− dµ + f + dµ + g+ dµ. X X X X X X R R R Consequently, X (f + g) dµ = X f dµ + X g dµ. R (3) Let h := g − f. Then h ≥ 0 almost everywhere. Hence, X h dµ ≥ 0. It follows R R that X f dµ ≤ X g dµ. (4) By Theorem 1.4 we have

∞ ∞ Z Z Z X Z X Z f dµ = f + dµ − f − dµ = f + dµ − f − dµ. A A A n=1 An n=1 An

Since |f| = f + + f − is integrable, we have P∞ R f + dµ + R f − dµ < ∞. Conse- n=1 An An quently,

∞ ∞ ∞ ∞ X Z X Z XZ Z X Z f + dµ − f − dµ = f + dµ − f − dµ = f dµ. n=1 An n=1 An n=1 An An n=1 An

This completes the proof of the theorem.

6 Theorem 2.3. (The Lebesgue Dominated Convergence Theorem) Let (fn)n=1,2,... be a sequence of integrable functions such that the sequence converges to f a.e. and there exists a nonnegative integrable function g such that |fn| ≤ g a.e. for all n ∈ IN. Then Z Z lim fn dµ = f dµ. n→∞ X X

Proof. The functions g − fn (n ∈ IN) are nonnegative, and so by Fatou’s lemma, Z Z (g − f) dµ ≤ lim inf (f − gn) dµ. X n→∞ X Since |f| ≤ g, f is integrable, and we have Z Z Z Z g dµ − f dµ ≤ g dµ − lim sup fn dµ. X X X n→∞ X It follows that Z Z f dµ ≥ lim sup fn dµ. X n→∞ X

Similarly, by considering g + fn for n ∈ IN, we get Z Z f dµ ≤ lim inf fn dµ. X n→∞ X This establishes the desired result.

§3. The Riemann and Lebesgue Integrals

We start by reviewing the deﬁnition of the Riemann integral. Suppose that a, b ∈ IR and a < b. Let f be a bounded function from the interval [a, b] to IR. A collection of points P = {x0, x1, . . . , xn} is called a partition of [a, b] if a = x0 < x1 < ··· < xn = b. The norm of the partition is deﬁned to be

kP k := max{xi − xi−1 : 1 ≤ i ≤ n}.

For i = 1, . . . , n, let

mi := inf{f(x): x ∈ [xi−1, xi]} and Mi := sup{f(x): x ∈ [xi−1, xi]}.

Then the lower sum S∗(f, P ) of f corresponding to the partition P is deﬁned by

n X S∗(f, P ) = mi(xi − xi−1), i=1

7 and similarly, the upper sum S∗(f, P ) of f by

n ∗ X S (f, P ) = Mi(xi − xi−1). i=1

∗ Clearly, S∗(f, P ) ≤ S (f, P ) holds for every partition P of [a, b]. Let P and Q be two partitions of [a, b]. We say that P is ﬁner than Q if Q ⊆ P . If this is the case, then

∗ ∗ S∗(f, Q) ≤ S∗(f, P ) and S (f, P ) ≤ S (f, Q).

To establish the inequalities, it suﬃces to assume that P has only one more point than

Q. Suppose that Q = {x0, x1, . . . , xn} and P = Q ∪ {t}. Then xk−1 < t < xk for some k,

1 ≤ k ≤ n. Set s1 := inf{f(x): xk−1 ≤ x ≤ t} and s2 := inf{f(x): t ≤ x ≤ xk}. Observe

that mk ≤ s1 and mk ≤ s2. Therefore,

n X X S∗(f, Q) = mi(xi − xi−1) = mi(xi − xi−1) + mk(xk − xk−1) i=1 i6=k X ≤ mi(xi − xi−1) + s1(t − xk−1) + s2(xk − t) = S∗(f, P ). i6=k

The proof for S∗(f, P ) ≤ S∗(f, Q) is similar. Consequently, for two arbitrary partitions P ∗ and Q of [a, b] we have S∗(f, P ) ≤ S (f, Q). Indeed,

∗ ∗ S∗(f, P ) ≤ S∗(f, P ∪ Q) ≤ S (f, P ∪ Q) ≤ S (f, Q).

The lower Riemann integral of f is deﬁned by

I∗(f) := sup{S∗(f, P ): P is a partition of [a, b]}, and the upper Riemann integral of f is deﬁned by

I∗(f) := inf{S∗(f, P ): P is a partition of [a, b]}.

∗ Clearly, I∗(f) ≤ I (f). ∗ A bounded function f :[a, b] → IR is called Riemann integrable if I∗(f) = I (f). In this case, the common value is called the Riemann integral of f and is denoted by

Z b f(x) dx. a

8 A Lebesgue measurable function from [a, b] to IR is said to be Lebesgue integrable if f is integrable with respect to the Lebesgue measure λ on [a, b]. In this case, the Lebesgue integral is denoted by Z f dλ. [a,b] Theorem 3.1. If f :[a, b] → IR is a bounded Riemann integrable function, then f is Lebesgue integrable and Z b Z f(x) dx = f dλ. a [a,b] R b Proof. Let I := a f(x) dx. Since f is Riemann integrable, for each k ∈ IN there exists a partition Qk of [a, b] such that

∗ I − 1/k ≤ S∗(f, Qk) ≤ S (f, Qk) ≤ I + 1/k.

Let Pk := Q1 ∪ · · · ∪ Qk. Then Pk ⊆ Pk+1 for k ∈ IN and

∗ ∗ I − 1/k ≤ S∗(f, Qk) ≤ S∗(f, Pk) ≤ S (f, Pk) ≤ S (f, Qk) ≤ I + 1/k.

∗ It follows that limk→∞(S (f, Pk) − S∗(f, Pk)) = 0.

Suppose Pk = {xk,j : 0 ≤ j ≤ nk}. Let mk,j := inf{f(x): xk,j−1 ≤ x ≤ xk,j} and

Mk,j := sup{f(x): xk,j−1 ≤ x ≤ xk,j} for 1 ≤ k ≤ nk. Deﬁne n n Xk Xk lk := mk,jχ[xk,j−1,xk,j ) and uk := Mk,jχ[xk,j−1,xk,j ). j=1 j=1

Then (lk)k=1,2,... is an increasing sequence of bounded measurable functions on [a, b], and

(uk)k=1,2,... is a decreasing sequence of bounded measurable functions on [a, b]. Moreover, Z Z ∗ lk dλ = S∗(f, Pk) and uk dλ = S (f, Pk). [a,b] [a,b]

Let l and u be the functions deﬁned by l(x) := limk→∞ lk(x) and u(x) := limk→∞ uk(x),

x ∈ [a, b]. Then l and u are Lebesgue measurable. Furthermore, since lk ≤ f ≤ uk a.e. for every k ∈ IN, we have l ≤ f ≤ u a.e. By the monotone convergence theorem, we obtain Z Z ∗ (u − l) dλ = lim (uk − lk) dλ = lim (S (f, Pk) − S∗(f, Pk)) = 0. [a,b] k→∞ [a,b] k→∞ Consequently, u − l = 0 almost everywhere, by Theorem 2.1. It follows that f = u a.e. This shows that f is Lebesgue measurable. Finally, invoking the monotone convergence theorem once more, we arrive at the desired conclusion: Z Z Z b ∗ f dλ = lim uk dλ = lim S (f, Pk) = f(x) dx. [a,b] k→∞ [a,b] k→∞ a

9 Theorem 3.2. A bounded function f :[a, b] → IR is Riemann integrable if and only if there exists a subset N of [a, b] such that λ(N) = 0 and f is continuous at every point x ∈ [a, b] \ N.

Proof. Suppose that f is a bounded Riemann integrable function on [a, b]. There exists a sequence (Pk)k=1,2,... of partitions of [a, b] such that Pk ⊂ Pk+1 for every k ∈ IN and ∗ limk→∞(S (f, Pk) − S∗(f, Pk)) = 0. Let uk and vk (k ∈ IN) be as in the proof of Theorem ∞ 3.1. Set E := ∪k=1Pk. Then λ(E) = 0. Moreover, there exists a subset K of [a, b] such that λ(K) = 0 and limn→∞(uk(x) − lk(x)) = 0 for every x ∈ [a, b] \ K. Let N := E ∪ K. Then λ(N) = 0. We claim that f is continuous at every point x ∈ [a, b] \ N. Let x be a point in [a, b] \ N. For any ε > 0, there exists some k ∈ IN such that uk(x) − lk(x) < ε.

Since x ∈ [a, b] \ Pk, there exists an open interval (α, β) such that x ∈ (α, β) ⊂ [a, b] \ Pk.

For y ∈ (α, β) we have lk(y) = lk(x) and uk(y) = uk(x). Hence,

lk(x) ≤ f(x) ≤ uk(x) and lk(x) = lk(y) ≤ f(y) ≤ uk(y) = uk(x).

Consequently, |f(y) − f(x)| ≤ uk(x) − lk(x) < ε for y ∈ (α, β). This shows that f is continuous at every point x ∈ [a, b] \ N. Now suppose that f is a bounded function on [a, b], N ⊂ [a, b] with λ(N) = 0, and f is continuous at each point in [a, b] \ N. Let (Pk)k=1,2,... be a sequence of partitions of

[a, b] such that limk→∞ kPkk = 0. For k ∈ IN, let uk and vk be deﬁned as in the proof of Theorem 3.1. Suppose x ∈ [a, b] \ N. For any ε > 0, since f is continuous at x, there exists some δ > 0 such that |f(y) − f(x)| < ε whenever y ∈ [a, b] and |y − x| < δ. There exists k0 ∈ IN such that kPkk < δ/2 for all k ≥ k0. Thus,

uk(x) − lk(x) ≤ sup{|f(y) − f(x)| : |y − x| < δ < ε whenever k > k0. Hence, for every point x ∈ [a, b] \ N,

lim uk(x) − vk(x) = 0. k→∞

By the Lebesgue dominated convergence theorem, Z ∗ lim S (f, Pk) − S∗(f, Pk) = lim (uk − lk) dλ = 0. k→∞ k→∞ [a,b]

This shows that f is Riemann integrable.