Homework 06, Math 545 23 September, 2015 Name: 1. Let (X; A; µ) be a measure space. Suppose that s : X ! R is a measurable function n with the property that Range(s)= fai gi=1, where the ai 2 R are distinct. Clearly s is a (measurable) simple function, as it can be written as n X s = ai χAi ;Ai = fx 2 X j s(x) = ai g 2 A; i=1 n where X = [i=1Ai and Ai \ Aj = ;, for i 6= j. This is called the canonical form of s. (a) Suppose that t : X ! R is a measurable simple function, in the sense that it may be written as m X t = bi χBi ; i=1 where Bi 2 A, for each i = 1; ··· ; m, but the bi are not necessarily distinct, the Bi are not necessarily pairwise disjoint and, finally it is not necessarily true m that X = [i=1Bi . (We use this second definition as the definition of a general measurable simple function.) Show that t has a finite range, and, therefore, it can be written in canonical form. (b) Suppose that s; t : X ! R are a measurable simple functions, not necessarily in canonical form: n n X X s = ai χAi ; t = bj χBj : i=1 j=1 Prove that, if s = t on X, then m n X X I(s) = ai µ(Ai ) = bj µ(Bj ) = I(t): i=1 j=1 This number I( · ) we will call the simple function integral. (c) Suppose that f : X ! R is a non-negative µ-measurable function. Define S(f ) := fs : X ! R j s is a µ-measurable simple function and 0 ≤ s ≤ f g ; and recall that Z f dµ := sup fI(s) j s 2 S(f )g : 1 If t : X ! R is a non-negative µ-measurable simple function, prove that R tdµ = I(t). 2. Let (X; A; µ) be a measure space, and suppose that f : X ! R is a non-negative µ-measurable function. Prove that Z Z min(f ; n)dµ ! f dµ, as n ! 1: Do not use the convergence theorems of chapter 7. 3. Let (X; A; µ) be a measure space, with µ(X) < 1, and suppose that fn : X ! R is a sequence of bounded µ-measurable functions that converge to a function f uniformly. Prove that Z Z lim fndµ = f dµ. n!1 Do not use the convergence theorems of chapter 7. 4. Let (X; A; µ) be a measure space, and suppose that fn : X ! R is a sequence of non-negative, µ-measurable, and integrable functions. Suppose that fn ! f point- R R wise a.e., such that f (x) is finite for each x 2 X, fndµ ! f dµ, as n ! 1, and f is integrable. Prove that, for each A 2 A, Z Z fndµ ! f dµ, as n ! 1: A A 5. Let (X; A; µ) be a measure space, and suppose that f ; fn : X ! R are µ-measurable R R and integrable functions, fn ! f point-wise a.e., and jfnjdµ ! jf jdµ. Prove that, as n ! 1, Z jfn − f jdµ ! 0: Hint: Use the result of Exercise 7.2 from the book. 6. Suppose that f : R ! R is Lebesgue integrable, a 2 R and Z x F (x) := f (s)ds: a Prove that F :[a; 1) ! R is continuous. 2.