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Chapter 3: Integration

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Chapter 3: Integration Chapter 3 Integration 3.1 Measurable Functions Definition 3.1.1. Let (X; A) be a measurable space. We say that f : X ! R is measurable if f −1((α; 1; )) 2 A for every α 2 R. Let M(X; A) denote the set of measurable functions. Lemma 3.1.2. The following are equivalent: 1) f is measurable. 2) f −1((−∞; α]) 2 A for all α 2 R. 3) f −1([α; 1)) 2 A for all α 2 R. 4) f −1((−∞; α)) 2 A for all α 2 R. Proof. 1) , 2) This follows since (f −1((−∞; α]))c = f −1((α; 1)): 1) ) 3) Observe that 1 \ 1 f −1([α; 1)) = f −1((α − ; 1)) 2 A: n n=1 3) ) 1) Observe that 1 [ 1 f −1((α; 1)) = f −1([α + ; 1)) 2 A: n n=1 3) , 4) This follows since (f −1((−∞; α)))c = f −1([α; 1)): Example 3.1.3. 1) Given any (X; A), the constant function f(x) = c for all x 2 X is measurable for any c 2 R. 2) If A ⊆ X, the characteristic function χA(x) of A is measurable if and only if A 2 A. 37 3) If X = R and B(R) ⊆ A, then the continuous functions are measurable since f −1((α; 1)) is open and hence measurable. Proposition 3.1.4. [Arithmetic of Measurable Functions] Assume that f; g 2 M(X; A). 1) cf 2 M(X; A) for every c 2 R 2) f 2 2 M(X; A) 3) f + g 2 M(X; A) 4) fg 2 M(X; A) 5) jfj 2 M(X; A) 6) maxff; gg 2 M(X; A) 7) minff; gg 2 M(X; A) Proof. 1) Assume that c = 0. Then ( ; if α ≥ 0 (cf)−1((α; 1)) = X if α < 0 Hence clearly f is measurable. If c 6= 0, then ( −1 α −1 f (( c ; 1)) if c ≥ 0 (cf) ((α; 1)) = −1 α f ((−∞; c )) if c < 0: 2) ( p f −1(( α; 1)) if α ≥ 0 (f 2)−1((α; 1)) = X if α < 0: 3) Note that [ fx 2 Xjf + g(x) > αg = fx 2 Xjf(x) > rg \ fx 2 Xjg(x) > α − rg: r2Q It follows that f + g is measurable. 4) Note that (f + g)2 − (f 2 + g2) fg = : 2 5) ( f −1((α; 1)) [ f −1((−∞; −α)) if α ≥ 0 (jfj)−1((α; 1)) = X if α < 0: 6) Observe that (f + g) + jf − gj maxff; gg = : 2 7) We have that minff; gg = − max{−f; −gg: 38 Remark 3.1.5. Let (X; A) be a measure space and let f : X ! R. Let f +(x) := supff(x); 0g and f −(x) := sup{−f(x); 0g. + − + − + 1 − 1 + It follows that f = f − f and jfj = f + f , so f = 2 (jfj + f) and f = 2 (jfj − f). Hence f and f − are measurable if and only if f is measurable. Definition 3.1.6. Let (X; A) be a measure space. An extended real-valued function is a function f : X ! R∗ = R [ {±∞}. It is measurable if f −1((α; 1]) 2 A for each α 2 R. Notice that if f is measurable, then both 1 \ f −1(f1g) = f −1((n; 1]) n=1 and 1 !c [ f −1({−∞}) = f −1((−n; 1]) n=1 are in A. The following proposition is straight forward. The proof is left as an exercise. Proposition 3.1.7. An extended real-valued function f : R ! R∗ is measurable if and only if f −1(f1g) and f −1({−∞}) are in A and the real-valued function defined by 0 x 2 f −1(f1g) [ f −1({−∞}) f (x) = 1 f(x) otherwise is measurable. Remark 3.1.8. The collection of all extended real-valued measurable functions will be denoted M(X; A). With the convention that 0(±∞) = 0, cf, f 2, f +, f − 2 M(X; A) if and only if f 2 M(X; A). For f + g, we use the convention that 1 − 1 = 0. Then f + g 2 M(X; A) if f, g 2 M(X; A). Definition 3.1.9. Given fang ⊆ R, define lim sup an := inf supfakg and lim inf an := sup inf fakg n!1 n k≥n n!1 n k≥n Note: In general, lim inf an ≤ lim sup an for any sequence fang. Moreover lim an = L if and only if n!1 n!1 n!1 lim inf an = L = lim sup an. n!1 n!1 Proposition 3.1.10. If ffng 2 M(X; A) then let f(x) = infffn(x)g F (x) = supffn(x)g n n ∗ ∗ f (x) = lim infffn(x)g F (x) = lim supffn(x)g n!1 n!1 All of these functions are measurable. Proof. We will show first that F (x) = supnffn(x)g is measurable. This follows since 1 [ fx 2 XjF (x) > αg = fx 2 Xjfn(x) > αg: n=1 39 That f(x) is measurable follows since infffn(x)g = − sup{−fn(x)g n n It is now immediate that f ∗ and F ∗ are also measurable. The following corollary is immediate from the previous proposition. Corollary 3.1.11. Let ffng 2 M(X; A). Assume also that ffng converges pointwise on X to f0. Then f0 2 M(X; A). Corollary 3.1.12. Let f; g 2 M(X; A). Then fg 2 M(X; A). Proof. For each n; m 2 define N 8 < f(x) if jf(x)j ≤ n fn(x) = n if f(x) > n : −n if f(x) < −n and 8 < g(x) if jg(x)j ≤ m gm(x) = m if g(x) > m : −m if g(x) < −m It is straight forward to show that each fn and gm is measurable. Consequently, we have that fngm is also measurable. From this we conclude that for each m 2 N that f(x)gm(x) = lim fn(x)gm(x) n!1 is measurable. And as such so is f(x)g(x) = lim f(x)gm(x): m!1 Definition 3.1.13. A simple function is a function ' : X ! R such that '(R) is finite. If the range of ' −1 Pn is fa1; : : : ; ang and Ei = ' (faig) then ' = i=1 aiχEi . We say that ' is in standard form if the ai's are distinct and the Ei's partition R. Pn Up to order, the standard form of a simple function is unique. ' = i=1 aiχEi (in standard form) is measurable if and only if each Ei 2 A. + + Theorem 3.1.14. Let f 2 M (X; A). Then there exists f'ng ⊆ M (X; A) such that 1) 'n is simple for each n. 2) 0 ≤ 'n(x) ≤ 'n+1(x) for all x 2 R and each n. 3) f(x) = lim 'n(x) for all x 2 . n!1 R n k k+1 n Proof. Let n 2 N. For k = 0; 1; : : : ; n2 − 1, let En;k := fx 2 R j 2n ≤ f(x) < 2n g and if k = n2 then Pn2n k let En;k := fx 2 R j f(x) ≥ ng. Define 'n = k=0 2n χEn;k . It is easy to see that f'ng is the desired sequence. 40 3.2 The Generalized Lebesgue Integral In this section we will introduce the basic theory of integration for a measure space (X; A; µ). Definition 3.2.1. Let (X; A; µ) be a measure space. Assume that ' 2 M+(X; A) is simple and suppose Pn that ' = i=1 aiχEi is the standard form of '. We define n Z X ' dµ := aiµ(Ei): i=1 Remark 3.2.2. It is often inconvenient to have to assume that a simple function is always in standard form. Pn Indeed it is a relatively simple exercise to show that if that ' = i=1 aiχEi , where Ei 2 A, but this is not the standard form for ', then we still have n Z X ' dµ := aiµ(Ei): i=1 Lemma 3.2.3. Let ', 2 M+(X; A) be simple functions and c ≥ 0. Then 1) R c' dµ = c R ' dµ 2) R (' + ) dµ = R ' dµ + R dµ ∗ R Furthermore, if we define λ : A! R by λ(E) = ϕχE dµ, then λ is a measure on A. Pn Pm Proof. Let ' = i=1 aiχEi and = j=1 bjχFj . Since n X c' = caiχEi i=1 and n m X X ' + = aiχEi + bjχFj ; i=1 j=1 then the previous remark allows us to immediately show that Z c' dm = c ' dµ and Z Z Z (' + ) dµ = ' dµ + dµ: Definition 3.2.4. If f 2 M+(X; A), we define Z Z f dµ := sup ' dµ 0≤'≤f where the supremum is taken over all simple function ' 2 M+(R; A) with 0 ≤ ' ≤ f. R R If E 2 A, then f dµ := fχE dµ. E Pn To establish the last statement observe that if ' = i=1 aiχEi and if we let γi(E) = µ(E \ Ei), then n X λ = aiγi: i=1 41 The following proposition is immediate from the definition of the integral. Proposition 3.2.5. 1) If f, g 2 M+(X; A) and f ≤ g then Z Z f dµ ≤ g dµ. R R + 2) If E ⊆ F 2 A then E f dµ ≤ F f dµ for all f 2 M (X; A).
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