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.