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FUNDAMENTALS of REAL ANALYSIS by Do˘Gan C¸Ömez III

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FUNDAMENTALS of REAL ANALYSIS by Do˘Gan C¸Ömez III FUNDAMENTALS OF REAL ANALYSIS by Do˘ganC¸¨omez III. MEASURABLE FUNCTIONS AND LEBESGUE INTEGRAL III.1. Measurable functions Having the Lebesgue measure define, in this chapter, we will identify the collection of functions that are compatible with Lebesgue measure and then define Lebesgue integral on these functions. Recall: f is continuous if and only if f −1(O) is open for all O open. Also, recall that 1 1 [ −1 [ −1 O = Ik , f (O) = f (Ik): k=1 k=1 Continuous functions are well behaved for even Riemann-integral; hence we will naturally be compatible with measurable sets. We will consider a larger class of functions: Definition. Let f : E ! R#, where E 2 F. f is called (Lebesgue) measurable if for all a 2 R, the set fx 2 E : f(x) > ag 2 F. Fact.1 Let f : E ! R#, E 2 F. The following are equivalent: i) f is measurable. ii) fx 2 E : f(x) ≥ ag 2 F for all a 2 R. iii) fx 2 E : f(x) < ag 2 F for all a 2 R. iv) fx 2 E : f(x) ≤ ag 2 F for all a 2 R. Furthermore i)-iv) imply v) fx 2 E : f(x) = ag 2 F for all a 2 R#. Proof. (i) ) (ii): First, note that fx 2 E : f(x) ≥ ag = f −1([a; 1]). Also, since \ 1 [a; 1] = (a − ; 1]; n n it follows that 1 ! 1 \ 1 \ 1 fx 2 E : f(x) ≥ ag = f −1 (a − ; 1] = f −1 (a − ; 1] 2 F: n n n=1 n=1 (ii) ) (iii): −1 −1 −1 fx 2 E : f(x) < ag = f ((−∞; a)) = f (R n [a; 1]) = E n f ([a; 1]) 2 F: (iii) ) (iv) and (iv) ) (i) are similar and left as an exercise. For (v), if a 2 R, then ff(x) = ag = f −1(fag) = f −1((−∞; a]) \ f −1([a; 1]) 2 F. If a = 1, 1 \ ff(x) = 1g = ff(x) > ng 2 F: n=1 1 2 Examples. 1. Let f : [0; 1] ! R be a function such that 0 if x 2 f(x) = Q : 1 if x 2 Qc Then, −1 1 1 c f ( 2 ; 1) = fx 2 [0; 1] : f(x) > 2 g = Q \ [0; 1] 2 F; if a = 1, f −1(1; 1) = ; 2 F; if a = −1, then f −1(−1; 1) = [0; 1] 2 F. It follows that, for all a 2 R; fx 2 [0; 1] : f(x) > ag 2 F; hence, f is measurable. 2. Let f(x) be the piecewise function defined as 8 > 0 if x ≤ −1 <> x2 if − 1 < x < 1 f(x) = > 2 if x = 1 :> 2 − x if x > 1 If a < 0, then f −1(a; 1) = fx : f(x) > ag = (−∞; 2 − a) 2 F. −1 If a = 0, then f (0; 1) = pfx : f(x) > 0gp= (−1p; 0) [ (0; 2) 2 F. If 0 < a < 1, then f −1(−1; a) = (−1; − a) [ ( a; 2 − a) 2 F: If 2 ≤ a, then f −1(a; 1) = ; 2 F. Hence, f is measurable. Fact.2 f : E ! R, E 2 F, is measurable if and only if for all O 2 B, f −1(O) 2 F. Proof. Exercise. Corollary.1 Continuous functions are measurable. Proof. Exercise. Corollary.2 Let f : E ! R be a measurable function on E: i) If D ⊂ E, D 2 F, then fjD and fjEnD are measurable on E. ii) If f = g almost everywhere on E, then g is measurable on E. Proof. (i) Since f : E ! R be measurable on E, −1 −1 fx 2 D : f(x) > ag = (fjD) (a; 1) = D \ f (a; 1) 2 F (ii) Let A = fx 2 E : f(x) 6= g(x)g. Note that m(A) = 0; hence A 2 F: Then, fx 2 E : g(x) > ag = fx 2 A : g(x) > ag [ fx 2 E n A : g(x) > ag = fx 2 A : g(x) > ag [ fx 2 E n A : f(x) > ag 2 F: | {z } | −{z1 } 2F (fjEnA) (a;1)2F Fact.3 Let E 2 F and f; g : E ! R be measurable functions and finite almost everywhere. Then: i) αf + βg is measurable on E for all α; β 2 R. ii) f 2 is measurable on E (hence, fg is measurable on E). 1 iii) f is measurable on E. 3 iv) jfj is measurable on E. Proof. (i) First, we show that αf is measurable on E for all α 2 R. If α = 0, then αf = 0. Hence αf is measurable. If α > 0, then n a o fx 2 E :(αf)(x) > ag = x 2 E : f(x) > 2 F: α Similarly for α < 0: Hence αf is measurable, and so is βg. Therefore, it is enough to show that f + g is measurable on E. Notice that, fx 2 E : f(x) + g(x) > ag = fx 2 E : f(x) > a − g(x)g Now, there exists rx 2 Q such that f(x) > rx > a − g(x). Hence [ fx 2 E : f(x) > a − g(x)g = (fx 2 E : f(x) > rg \ fx 2 E : r > a − g(x)g) 2 F: | {z } | {z } r2Q 2F 2F 1 2 2 2 (ii) First, observe that fg = 2 [(f + g) − f − g ]. Therefore it is enough to show that f measurable implies f 2 is measurable. Now, we have fx 2 E :(f 2)(x) > ag = fx 2 E :[f(x)]2 > ag p p = fx 2 E : f(x) > ag [ fx 2 E : f(x) < − ag 2 F: | {z } | {z } 2F 2F 1 (iii) If a = 0, then fx 2 E : f (x) > ag = fx 2 E : f(x) > 0g 2 F. So assume that a > 0. Then 1 1 x 2 E : (x) > a = x 2 E : f(x) < 2 F: f a (iv) Follows from fx 2 E : jfj(x) > ag = fx 2 E : jf(x)j > ag = fx 2 E : f(x) > ag [ fx 2 E : f(x) < −ag 2 F: Definition. For functions f; g : E ! R, define (f _ g)(x) = maxff(x); g(x)g (f ^ g)(x) = minff(x); g(x)g f +(x) = f(x) _ 0 −f −(x) = f(x) ^ 0 Notice that, with the definitions above, we have that jfj = f + + f − and f = f + − f −. Corollary. If f; g : E ! R, E 2 F is measurable on E, then: i) f +, f − are measurable on E. ii) (f _ g)(x) and (f ^ g)(x) are measurable on E. Proof. Exercise. Convention. For the rest of this chapter f : E ! R. E will always be measurable. Definition. Let A ⊂ R be a set, then then the function 1 if x 2 A χ (x) = A 0 otherwise 4 is called the characteristic function of the set A. Easy Fact. E 2 F if and only if χE is a measurable function. For, observe that, ; if a ≥ 1 χ−1(a; 1) = E E otherwise Hence χE is measurable (on E). If E 62 F, E 2 R, then χE is not measurable. Question: If f and g are measurable functions. What about f ◦ g? The answer is \No" in general. Example. Recall that f : [0; 1] ! [0; 2] defined by f(x) = x + '(x); where φ is the Cantor- Lebesgue function, is strictly increasing. Then, by Fact 9 in Chapter.II, there exists A 2 F such that f(A) 62 F: Extend f continuously in a strictly increasing manner onto R. Say g : R ! R such that −1 −1 gj[0;1] = f. Then g is continuous. Next, consider χA and define h(x) := (χA ◦ g )(x). Now 1 1 1 h−1 ; 1 = (χ ◦ g−1)−1 ; 1 = (g ◦ χ−1) ; 1 2 A 2 A 2 1 = g χ−1 ; 1 = g(A) = f(A) 62 F: A 2 Theorem.1 Let f; g be measurable functions on their respective domains. If g is continuous, then g ◦ f is measurable. Proof. Define h(x) := (g ◦f)(x): If O ⊂ R is a Borel set, then we have h−1(O) = (g ◦f)−1(O) = −1 −1 −1 −1 −1 f (g (O)). Since g is continuous, g (O) 2 B. Hence f (g (O)) 2 F. Recall that, for a given collection of functions ffng; f : E ! R, we say that fn ! f pointwise if and only if for all x 2 E, for all > 0, there exists N(, x) such that whenever n ≥ N we have jfn(x) − f(x)j < . Also, fn ! f uniformly if and only if for all > 0, there exists N() such that whenever n ≥ N we have jfn(x) − f(x)j < for all x 2 E. Definition. Let ffng be a sequence of real-valued functions on a measurable set E: We say that ffng converges almost everywhere (a.e.) to f : E ! R if there exists N ⊂ E with m(N) = 0 such that fn ! f pointwise on E n N. Remarks. Uniform convergence ) pointwise convergence ) almost everywhere convergence. Note that the converse is not always true (Exercise: Give examples). Observation. Let fn ! f pointwise, where each fn is measurable (on E). For x 2 E; if f(x) > a; 1 then 9m ≥ 1 such that f(x) > a + m : Thus, since fn ! f pointwise, for large enough n; we will 1 have fn(x) > a + m : This means that, 1 9m 9N such that, 8n ≥ N; f (x) > a + : (∗) n m 5 1 Conversely, if (*) is satisfied, then f(x) = limn fn(x) ≥ a + m : Hence, we have fx : f(x) > ag = fx 2 E :(9m ≥ 1)(9N ≥ 1)(8n ≥ N)(fn(x) > a + 1=m)g !! [ [ \ 1 = x : f (x) > a + 2 F: n m m≥1 N≥1 n≥N Theorem.2 Let E 2 F and fn : E ! R be a measurable function for all n ≥ 1.
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