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Measurable Functions and Simple Functions Measure theory class notes - 8 September 2010, class 9 1 Measurable functions and simple functions The class of all real measurable functions on (Ω, A ) is too vast to study directly. We identify ways to study them via simpler functions or collections of functions. Recall that L is the set of all measurable functions from (Ω, A ) to R and E⊆L are all the simple functions. ∞ Theorem. Suppose f ∈ L is bounded. Then there exists an increasing sequence {fn}n=1 in E whose uniform limit is f. Proof. First assume that f takes values in [0, 1). We divide [0, 1) into 2n intervals and use this to construct fn: n− 2 1 k − k k+1 fn = 1f 1 n , n 2n ([ 2 2 )) Xk=0 k k+1 k Whenever f takes a value in 2n , 2n , fn takes the value 2n . We have • For all n, fn ≤ f. 1 • For all n, x, |fn(x) − f(x)|≤ 2n . This is clear from the construction. • fn ∈E, since fn is a finite linear combination of indicator functions of sets in A . k 2k 2k+1 • fn ≤ fn+1: For any x, if fn(x)= 2n , then fn+1(x) ∈ 2n+1 , 2n+1 . ∞ So {fn}n=1 is an increasing sequence in E converging uniformly to f. Now for a general f, if the image of f lies in [a,b), then let f − a1 g = Ω b − a (note that a1Ω is the constant function a) ∞ Im g ⊆ [0, 1), so by the above we have {gn}n=1 from E increasing uniformly to g. Let fn = a1Ω +(b − a)gn ∞ Then {fn}n=1 is a sequence from E increasing uniformly to f. Theorem. Suppose f ∈ L is nonnegative (that is, f(x) ≥ 0 for all x ∈ Ω). Then there exists an ∞ increasing sequence {fn}n=1 in E whose pointwise limit is f. 1 Proof. As in the previous proof, we divide the range of f into intervals of length 2n . However, f could be unbounded and we may need infinitely many of such intervals, which will not help in defining a simple function. So at the nth stage, we divide only [0,n), and not the entire [0, ∞). More precisely, let n− n2 1 k − k k+1 fn = 1f 1 n , n 2n ([ 2 2 )) Xk=0 • For all n, fn ≤ f. This is clear from the construction of fn (note that f ≥ 0). 1 • For all n, x, if f(x) <n, then |fn(x) − f(x)| < 2n . Measure theory class notes - 8 September 2010, class 9 2 • fn ∈E, since fn is a finite linear combination of indicator functions of sets in A . • fn ≤ fn+1: if fn(x) = 0 then clearly fn(x) ≤ fn+1(x), otherwise f(x) <n<n +1 and from the construction it is easy to see that fn(x) ≤ fn+1(x). 1 Now for any x ∈ Ω, there is some integer m such that m>f(x). For all n ≥ m, |fn(x)−f(x)| < 2n , and so limn→∞ fn(x)= f(x). For f : Ω → R, define f +,f − : Ω → R f +(x) = max{f(x), 0} f −(x)= − min{f(x), 0} Both f + and f − are nonnegative, and f = f + − f −. If f is measurable, so are both f + and f −. Every measurable function is the difference of two nonnegative measurable functions. This is useful because by the above theorem, nonnegative measurable functions are easier to deal with than general measurable functions. Theorem. Suppose Ω is a nonempty set and A is a σ-field on it. F is a field such that σ(F )= A . Suppose F is a collection of real measurable functions on Ω which is a vector space under the natural operations and closed under monotone pointwise limits. Suppose for all A ∈ F , 1A ∈ F . Then F = L, the set of all real measurable functions. Proof. We first show that indicator functions of all sets in A belong to F . Let B = {A ∈ A :1A ∈ F } We are given that F ⊆ B. Since F is closed under monotone pointwise limits, B is a monotone class. By the monotone class theorem, B is a σ-field, and so B = A . So F has: • 1A for all A ∈ A . • By taking finite linear combinations of the above (since F is a vector space), all simple functions. • By taking increasing pointwise limits of simple functions, all nonnegative measurable func- tions. • By taking differences of nonnegative measurable functions, all measurable functions. Sometimes we are concerned only about bounded measurable functions. We have a similar theorem for that case. We need a definition first. R ∞ Definition. A sequence of Ω → functions {fn}n=1 is said to converge bounded pointwise (ab- breviated as bp) to f : Ω → R if ∞ • {fn}n=1 converges to f pointwise. • There is a C ∈ R such that for all n and x, |fn(x)|≤ C. Measure theory class notes - 8 September 2010, class 9 3 In other words, bounded pointwise convergence is just pointwise convergence along with uniform boundedness of the sequence of functions. Here are some examples of [0, 1] → R functions which illustrate various modes of convergence. n fn gn hn 1 1 n 1 1 0n 1 0 1 0n 1 ∞ ∞ {fn}n=1 converges to the zero function pointwise, but not bounded pointwise or uniformly. {gn}n=1 ∞ converges to the zero function uniformly and bounded pointwise. {hn}n=1 converges to the zero function bounded pointwise but not uniformly. Theorem. Suppose Ω is a nonempty set and A is a σ-field on it. F is a field such that σ(F )= A . Suppose F is a collection of bounded real measurable functions on Ω which is a vector space under the natural operations and closed under bounded pointwise limits. Suppose for all A ∈ F ,1A ∈ F . Then F is the set of all bounded real measurable functions. Proof. As before, we first show that indicator functions of all sets in A belong to F . Let B = {A ∈ A :1A ∈ F } We are given that F ⊆ B. Since F is closed under bounded pointwise limits, B is a monotone class (all indicator functions are bounded by 1). By the monotone class theorem, B is a σ-field, and so B = A . So F has: • 1A for all A ∈ A . • By taking finite linear combinations of the above (since F is a vector space), all simple functions. • By taking increasing bounded pointwise limits of simple functions, all nonnegative bounded measurable functions (if |f(x)| ≤ C for all x, we can get an increasing sequence of simple functions converging to it pointwise (even uniformly, in fact), all of which take values between 0 and C). • By taking differences of nonnegative bounded measurable functions, all bounded measurable functions (if f is bounded, so are f + and f −). Measure theory class notes - 8 September 2010, class 9 4 Defining integration We have studied the properties of the class of measurable functions, now we define their integrals. A R Suppose (Ω, ,µ) is a measure space, and f : Ω → is measurable. How can we define Ω fdµ? If we could define the integral for f + and f −, we could define R fdµ = f +dµ − f −dµ ZΩ ZΩ ZΩ because we want integration to be a linear operation. So we have reduced the problem from general measurable functions to nonnegative measurable functions. If f is nonnegative, how can we define Ω fdµ? If s is a nonnegative simple function, say taking values a1,...,ak, then we can define R k −1 sdµ = aiµ f ({ai}) Z Ω Xi=1 (this corresponds to our notion of “area under the curve”) Since we want integration to be monotonic, we should have fdµ ≥ sup sdµ : s simple ,s ≤ f ZΩ ZΩ Since f is nonnegative, we know that nonnegative simple functions s with s ≤ f exist, in fact we can get a sequence of them converging to f. Though it is not immediately clear, it makes sense to define the integral of f by taking the above equation to be an equality. We will see this in the next class..
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