Chapter 4. Measurable Functions. 4.1 Measurable Functions
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1 Chapter 4 Measurable Functions If X is a set and A P(X) is a σ-field, then (X, A) is called a measurable space. If µ is a countably additive measure defined on A then (X, A, µ) is called a measure space. In this chapter we will introduce the family of measurable functions for which we will seek to define the Lebesgue integral. We will prove the very important fact that pointwise limits of measurable functions must be measurable. This is encouraging because pointwise limits of Riemann integrable functions need not be Riemann integrable Measurable Functions Definition Let (X, A, µ) be a measure space. i. If f : X R we say that f is A-measurable provided that for all a R. 2 f 1 (, a) = {x X f(x) < a} A ii. If f : X C, the complex numbers, we write f(x) = u(x) + iv(x) for real-valued functions u = Rf and v = If. We say f is A-measurable provided that both u and v are A-measurable. iii. If f : X S, where S is a topological space, we say that f is A- measurable provided that f 1 (G) A for every open set G S. 1 See Example This definition is motivated by Section
2 It is necessary to prove that the three parts of the definition of measurability of a function are consistent. Exercise Show that if f : X R is A-measurable, then f 1 (G) A for every open set G R. Show that the concepts of measurability in Definition for both real and complex valued functions are consistent with the concept of measurability for a topological space valued function. For example, show that f : X R satisfies the definition of measurability if and only if f 1 (G) is measurable for each open set G R. The reader may note correctly that the concept of measurability has a formal similarity to the definition of continuity. A function between topological spaces is called continuous provided that the inverse image of each open set is open. Of course for measurability the inverse image of an open set must be measurable, and measurability is by no means synonymous with open. In R n, every open set is Lebesgue measurable, but the converse is clearly false. The following exercise gives a very useful equivalent form of measurability for functions. Exercise Let f : X R n, which can be regarded as an example of a topological space. Prove that f is measurable if and only if f 1 (B) L for each Borel set B. (Hint: Show that the family is a σ-field.) S = {A P(R n ) f 1 (A) L} It will be important to know that many combinations of measurable functions and many functions of measurable functions are again measurable. To investigate this we need the following definition. Definition Let (X, A, µ) be any measure space arising from the Hopf extension theorem, so that we have a concept of the field of Borel sets generated by a field of elementary sets. If f : X S, where S is a topological space, we say that f is a Baire function provided that f 1 (G) is a Borel set for each open set G S. Clearly, every Baire function is measurable and every continuous function (from R n to S) is a Baire function. The indicator function of a measurable set that is not a Borel set would be an example of a measurable function that is not a Baire function. 46
3 Theorem Suppose each of the functions f 1, f 2,..., f n is an A-measurable real-valued function defined on X. Let Φ : R n R be a Baire function. Then F = Φ(f 1, f 2,...,f n ) is an A-measurable function defined on X. Proof. We need to show that for each open set G R we have F 1 (G) A. Denote f = (f 1, f 2,...,f n ) : X R n. We claim that f is measurable. In fact, each open set G R n can be written as a disjoint union of countably many open blocks of the form B = n (a i, b i ) i=1 which is a Cartesian product of n open intervals. Thus Then f 1 (B) = n i=1 f 1 i (a i, b i ) A. F 1 (G) = (Φ f) 1 (G) = f 1 ( Φ 1 (G) ). Since Φ 1 (G) is a Borel set, the theorem is true by the result of Exercise We remark that in Theorem it would have sufficed to have Φ defined on a set D R n provided (f 1,..., f n ) : X D. Remark It follows from Theorem that such combinations of measurable functions as the following must be measurable. c 1 f 1 + c 2 f 2 where c 1 and c 2 are constants f 1 f 2, the product of two measurable functions f 1 f 2 where f 2 is nowhere zero In particular, f 1 is also measurable. Thus f is measurable if and only if {x f(x) > α} A for every α R. Corollary If f 1 and f 2 are A-measurable real-valued functions, then each of the following functions is A-measurable. i. max(f 1, f 2 ) = f 1 f 2, where f 1 f 2 (x) = max(f 1 (x), f 2 (x)) for each x X. 47
4 ii. min(f 1, f 2 ) = f 1 f 2, where f 1 f 2 (x) = min(f 1 (x), f 2 (x)) for each x X. iii. f + = f 0, known as the positive part of f. iv. f = ( f) 0, known as the negative part of f. (The reader should note that the negative part of a real-valued function is positive.) v. f = f + + f Proof. It suffices to observe that max(x 1, x 2 ) and min(x 1, x 2 ) are both continuous functions from R 2 to R, making each of these a Baire function. For the last part we use the fact that Φ(x 1, x 2 ) = x 1 + x 2 is continuous and thus a Baire function. 4.2 Limits of Measurable Functions In the study of point-wise limits of measurable functions and integrable functions, we will consider sequences of functions for which f n (x) diverges to ± for some values of x. Thus it is helpful to extend the concept of real numbers to the set R = R {± }. Measurability for an extended real valued function means that for each α R the set f 1 [α, ] = {x f(x) α} A. Theorem Let (X, A, µ) be a measure space and let {f n n N} be any sequence of measurable functions from X to R. Then each of the five functions defined as follows is A-measurable. i. f (x) = inf{f n (x) n N} for all x X ii. f (x) = sup{f n (x) n N} for all x X iii. f(x) = lim inf f n (x) for all x X iv. f(x) = lim sup fn (x) for all x X v. f(x) = lim n f n (x) provided the limit exists for all x X. 48
5 Proof. For the first part, we observe that each a R, f 1 [α, ] = {x inf f n(x) α} = n n N Thus f is A-measurable. Since f (x) inf n ( f n(x)) fn 1 [α, ] A. it follows that f is A-measurable as well. Note next that since i n = inf{f k (x) k n} is an increasing sequence of extended real numbers i n, we have lim inf f n (x) = f(x) = sup{inf{f k (x) k n} n N} so that f is A-measurable being the supremum of a sequence of measurable functions given as infima. Also, lim sup f n (x) = f(x) = inf{sup{f k (x) k n} n N} with the result that f is A-measurable. Finally, we note that f(x) = lim n f n (x) exists if and only if f(x) = f(x). Thus f is A-measurable provided the point-wise limit exists on X. Exercise Suppose f n : X R is a measurable function for each n N, where (X, A, µ) is a measure space. Prove that the set is a measurable set. S = {x lim n f n (x) exists} The reader should be able to give examples of point-wise convergent sequences of continuous functions for which the limit is not continuous, and examples of point-wise convergent sequences of Riemann integrable functions for which the limit is not Riemann integrable. We see already that measurability must be a valuable concept for point-wise convergence since point-wise convergence does preserve measurability. 49
6 Definition Let (X, A, µ) be a measure space and let N be the set of all null-sets. We say that x has some property P A µ almost everywhere if and only if there is a set N N such that x has the property P for all x X \ N. This is commonly expressed as µ-almost everywhere (µ-a.e.) or as almost everywhere (a.e.) provided that there will be no confusion as to which measure µ or σ-algebra A is in use. Corollary Let f n be a sequence of measurable functions on the complete measure space (X, A, µ). Suppose f n (x) f(x) almost everywhere on X. Then the function defined by f(x) = lim n f n (x) is defined almost everywhere and is measurable. Proof. This follows from Theorem It is commonly understood in this context that although the function f is defined everywhere by the given limit except on a null-set N, we may assign f arbitrary values on N itself. Then the completeness of the measure space tells us that the resulting function f remains A µ measurable regardless of how values are assigned to f within the null-set N. Exercise Suppose f : X R is a measurable function that has finite values almost everywhere on the measure space (X, A, µ) where µ(x) > 0. Prove that there is a set of positive measure on which f is bounded. Exercise Suppose the measurable function f : R n R has the special property that for each fixed vector c R n, the translation of f given by f c (x) = f(x + c) is equal almost everywhere to f(x) itself. Prove that f(x) is equal almost everywhere to a constant function. (Hints: Consider both the sum and the terms of n Z l(f 1 [n, n + 1)). Apply Exercise to select the special value of n. Divide and conquer!) 4.3 Simple Functions & Egoroff s Theorem Definition A function f : X R is called A-simple (or simple if there will be no confusion regarding the σ-field A that is under consideration) if and only if f is A-measurable and {f(x) x X} is a finite set. The class of simple functions is denoted by S. 50
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