Chapter 6. Integration. 1. Integrals of Nonnegative Functions. a j µ(e j ) (ca j )µ(e j ) = c X. and ψ =

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1 Chapter 6. Integration 1. Integrals of Nonnegative Functions Let (, S, µ) be a measure space. We denote by L + the set of all measurable functions from to [0, ]. Let φ be a simple function in L +. Suppose f() = {a 1,..., a m }. Then φ has the standard representation φ = m a jχ Ej, where E j := f 1 ({a j }), j = 1,..., m. We define the integral of φ with respect to µ to be φ dµ := a j µ(e j ). For c 0, we have cφ = m (ca j)χ Ej. It follows that cφ dµ = (ca j )µ(e j ) = c φ dµ. Suppose that φ = m a jχ Ej, where a j 0 for j = 1,..., m and E 1,..., E m are mutually disjoint measurable sets such that m E j =. Suppose that ψ = n b kχ Fk, where b k 0 for k = 1,..., n and F 1,..., F n are mutually disjoint measurable sets such that n F k =. Then we have φ = a j χ Ej F k and ψ = If φ ψ, then a j b k, provided E j F k. Consequently, a j µ(e j ) = In particular, if φ = ψ, then In general, if φ ψ, then a j µ(e j F k ) φ dµ = a j µ(e j ) = a j µ(e j ) 1 b k χ Fk E j. b k µ(e j F k ) = b k µ(f k ). b k µ(f k ) = ψ dµ. b k µ(f k ).

2 Furthermore, (φ + ψ) dµ = (a j + b k )µ(e j F k ) = φ dµ + ψ dµ. Suppose φ is a simple function in L + and m a jχ Ej is its standard representation. If S, then φχ = m a jχ Ej. We define φ dµ := φχ dµ = a j µ(e j ). The set function ν from S to [0, ] given by ν() := φ dµ is a measure. Indeed, ν( ) = 0. To prove σ-additivity of ν, we let ( k ),2,... be a sequence of mutually disjoint measurable sets and := k. Then we have ν() = = φ dµ = a j µ(e j ) = a j µ(e j k ) = a j µ(e j k ) k φ dµ = ν( k ). The Lebesgue integral of a function f L + is defined to be { } f dµ = sup φ dµ : 0 φ f, φ simple. When f is a simple function, this definition agrees with the previous one. If f, g L + and f g, then it follows from the definition that f dµ g dµ. If f L+ and S, we define f dµ := (fχ ) dµ. Since fχ f, we have f dµ f dµ. Theorem 1.1. (The Monotone Convergence Theorem) Let (f n ),2,... be an increasing sequence of functions in L +. If f = lim f n, then f dµ = lim f n dµ. Proof. Since each f n is measurable and f = lim f n, the function f is measurable. Moreover, f n f n+1 f for every n IN. Hence, f n dµ f n+1 dµ 2 f dµ n IN.

3 Thus, there exists an element α [0, ] such that α = lim f n dµ f dµ. Let φ be any simple function on (, S) such that 0 φ f. Fix c (0, 1) for the time being and define E n := {x : f n (x) cφ(x)}, n IN. Each E n is measurable. For n IN, we have f n dµ f n dµ E n (cφ) dµ = c E n φ dµ. E n But E n E n+1 for n IN and = E n. Hence, lim E n φ dµ = φ dµ. It follows that α = lim f n dµ lim c φ dµ = c φ dµ. E n Thus, α c φ dµ for every c (0, 1). Consequently, α φ dµ for every simple function satisfying 0 φ f. Therefore, α f dµ and This completes the proof. f dµ α = lim f n dµ f dµ. The monotone convergence theorem is still valid if the conditions f n (x) f n+1 (x) for all n IN and lim f n (x) = f(x) hold for almost every x. Indeed, under the stated conditions, there exists a null set N such that for every x \ N, f n (x) f n+1 (x) and lim f n (x) = f(x). Note that N g dµ = 0 for all g L+. With E := \ N we have f dµ = fχ E dµ = lim f n χ E dµ = lim f n dµ. The hypothesis that the sequence (f n ),2,... be increasing almost everywhere is essential for the monotone convergence theorem. For example, let µ be the Lebesgue measure on = IR, and let f n := χ (n,n+1) for n IN. Then (f n ),2,... converges to 0 pointwise, but IR f n dµ = 1 for every n IN. 3

4 Theorem 1.2. Let f, g L + and c 0. Then (cf) dµ = c f dµ and (f + g) dµ = f dµ + g dµ. Proof. For f, g L +, we can find increasing sequences (f n ),2,... and (g n ),2,... of simple functions in L + such that f = lim f n and g = lim g n. By the monotone convergence theorem we have and (f +g) dµ = lim (cf) dµ = lim (cf n ) dµ = lim c f n dµ = c f dµ (f n +g n ) dµ = lim f n dµ+ lim g n dµ = f dµ+ g dµ. Theorem 1.3. If (f k ),2,... is a sequence of functions in L + and f = f k, then f dµ = f k dµ. Proof. For n IN, let g n := n f k. Then (g n ),2,... is an increasing sequence of functions in L + such that lim g n = f. By the monotone convergence theorem we have f dµ = lim g n dµ = lim g k dµ = f n dµ. Theorem 1.4. Let f be a function in L +, and let ν be the function from S to [0, ] given by ν() := f dµ, S. Then ν is a measure on S. Proof. Clearly, ν( ) = 0. Suppose that (E n ),2,... is a sequence of disjoint sets in S and E = E n. Then we have fχ E = fχ E n. By Theorem 1.3 we obtain E f dµ = This proves that ν is σ-additive. fχ E dµ = fχ En dµ = Theorem 1.5. Let (f n ),2,... be a sequence in L +. Then lim inf f n dµ lim inf f n dµ. 4 E n f dµ.

5 Proof. Let f := lim inf f n and g k := inf n k f n for k IN. Then (g k ),2,... is an increasing sequence and lim k g k = f. By the monotone convergence theorem, we have f dµ = lim g k dµ. k For fixed k, g k f n for all n k; hence, g k dµ f n dµ for all n k. It follows that We conclude that f dµ = lim k g k dµ inf f n dµ. n k The above theorem is often called Fatou s lemma. g k dµ lim inf f n dµ = lim inf f n dµ. k n k 2. Integrable Functions Let (, S, µ) be a measure space. measurable function f from to IR is said to be integrable over a set E S if both E f + dµ and E f dµ are finite. In this case, we define E f dµ := Clearly, f is integrable if and only if f is. E f + dµ E f dµ. Theorem 2.1. Let f be an integrable function on a measure space (, S, µ). Then f is finite almost everywhere. Moreover, if 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 E n := {x : f(x) n}. For E := E n we have f dµ f dµ nµ(e n ) nµ(e). E n It follows that µ(e) f dµ/n for all n IN. Since 0 f dµ <, we conclude that µ(e) = 0. If x \ E, then x / E n for some n IN, that is, f(x) < n. This shows that f(x) < for all x \ E. In other words, f is finite almost everywhere. For the second statement, let K n := {x : f(x) 1/n} for n IN, and let K := K n. Then we have K = {x : f(x) > 0} and 0 = f dµ f dµ µ(k n )/n. K n 5

6 It follows that µ(k n ) = 0 for all n IN. Hence, µ(k) = lim µ(k n ) = 0. Thus, f = 0 almost everywhere. Theorem 2.2. Let f and g be integrable functions on a measure space (, S, µ). Then the following statements are true. (1) For each c IR, cf is integrable and (cf) dµ = c f dµ. (2) The function f + g is integrable, and (f + g) dµ = f dµ + g dµ. (3) If f g almost everywhere, then f dµ g dµ. (4) If = n, where n (n IN) are mutually disjoint sets in S, then f dµ = n f dµ. Proof. (1) If c 0, then (cf) + = cf + and (cf) = cf. If c < 0, then (cf) + = cf and (cf) = cf +. In both cases, cf is integrable and (cf) dµ = c 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 \ E. Let h(x) := f(x) + g(x) for x \ E and h(x) := 0 for x E. On \ E we have h + h = f + f + g + g. It follows that h + + f + g = h + f + + g +. Hence, h + dµ + f dµ + g dµ = Consequently, (f + g) dµ = f dµ + g dµ. h dµ + f + dµ + g + dµ. (3) Let h := g f. Then h 0 almost everywhere. Hence, h dµ 0. It follows that f dµ g dµ. (4) By Theorem 1.4 we have f dµ = f + dµ f dµ = n f + dµ n f dµ. Since f = f + + f is integrable, we have ( n f + dµ + n f dµ ) <. Consequently, n f + dµ n f dµ = This completes the proof of the theorem. ( ) f + dµ f dµ = n n 6 n f dµ.

7 Theorem 2.3. (The Lebesgue Dominated Convergence Theorem) Let (f n ),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 f n g a.e. for all n IN. Then f n dµ = f dµ. lim Proof. The functions g f n (n IN) are nonnegative, and so by Fatou s lemma, (g f) dµ lim inf (f g n ) dµ. Since f g, f is integrable, and we have g dµ f dµ It follows that g dµ lim sup f n dµ. f dµ lim sup f n dµ. Similarly, by considering g + f n for n IN, we get f dµ lim inf This establishes the desired result. f n dµ. 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. collection of points P = {x 0, x 1,..., x n } is called a partition of [a, b] if a = x 0 < x 1 < < x n = b. The norm of the partition is defined to be P := max{x i x i 1 : 1 i n}. For i = 1,..., n, let m i := inf{f(x) : x [x i 1, x i ]} and M i := sup{f(x) : x [x i 1, x i ]}. Then the lower sum S (f, P ) of f corresponding to the partition P is defined by S (f, P ) = m i (x i x i 1 ), i=1 7

8 and similarly, the upper sum S (f, P ) of f by S (f, P ) = M i (x i x i 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 = {x 0, x 1,..., x n } and P = Q {t}. Then x k 1 < t < x k for some k, 1 k n. Set s 1 := inf{f(x) : x k 1 x t} and s 2 := inf{f(x) : t x x k }. Observe that m k s 1 and m k s 2. Therefore, S (f, Q) = i=1 m i (x i x i 1 ) = i k m i (x i x i 1 ) + m k (x k x k 1 ) i k m i (x i x i 1 ) + s 1 (t x k 1 ) + s 2 (x k t) = S (f, P ). 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). 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 b a f(x) dx. 8

9 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 f dλ. Theorem 3.1. If f : [a, b] IR is a bounded Riemann integrable function, then f is Lebesgue integrable and b a f(x) dx = f dλ. Proof. Let I := b f(x) dx. Since f is Riemann integrable, for each k IN there exists a a partition Q k of [a, b] such that I 1/k S (f, Q k ) S (f, Q k ) I + 1/k. Let P k := Q 1 Q k. Then P k P k+1 for k IN and I 1/k S (f, Q k ) S (f, P k ) S (f, P k ) S (f, Q k ) I + 1/k. It follows that lim k (S (f, P k ) S (f, P k )) = 0. Suppose P k = {x k,j : 0 j n k }. Let m k,j := inf{f(x) : x k,j 1 x x k,j } and M k,j := sup{f(x) : x k,j 1 x x k,j } for 1 k n k. Define n k n k l k := m k,j χ [xk,j 1,x k,j ) and u k := M k,j χ [xk,j 1,x k,j ). Then (l k ),2,... is an increasing sequence of bounded measurable functions on [a, b], and (u k ),2,... is a decreasing sequence of bounded measurable functions on [a, b]. Moreover, l k dλ = S (f, P k ) and u k dλ = S (f, P k ). Let l and u be the functions defined by l(x) := lim k l k (x) and u(x) := lim k u k (x), x [a, b]. Then l and u are Lebesgue measurable. Furthermore, since l k f u k a.e. for every k IN, we have l f u a.e. By the monotone convergence theorem, we obtain (u l) dλ = lim (u k l k ) dλ = lim k k (S (f, P k ) S (f, P k )) = 0. 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: f dλ = lim u k dλ = lim k k S (f, P k ) = 9 b a f(x) dx.

10 Theorem 3.2. 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 (P k ),2,... of partitions of [a, b] such that P k P k+1 for every k IN and lim k (S (f, P k ) S (f, P k )) = 0. Let u k and v k (k IN) be as in the proof of Theorem 3.1. Set E := P k. Then λ(e) = 0. Moreover, there exists a subset K of [a, b] such that λ(k) = 0 and lim (u k (x) l k (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 u k (x) l k (x) < ε. Since x [a, b] \ P k, there exists an open interval (α, β) such that x (α, β) [a, b] \ P k. For y (α, β) we have l k (y) = l k (x) and u k (y) = u k (x). Hence, l k (x) f(x) u k (x) and l k (x) = l k (y) f(y) u k (y) = u k (x). Consequently, f(y) f(x) u k (x) l k (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 (P k ),2,... be a sequence of partitions of [a, b] such that lim k P k = 0. For k IN, let u k and v k 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 k 0 IN such that P k < δ/2 for all k k 0. Thus, u k (x) l k (x) sup{ f(y) f(x) : y x < δ } < ε whenever k > k 0. Hence, for every point x [a, b] \ N, ( lim uk (x) v k (x) ) = 0. k By the Lebesgue dominated convergence theorem, ( lim S (f, P k ) S (f, P k ) ) = lim (u k l k ) dλ = 0. k k This shows that f is Riemann integrable. 10