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)

• 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)

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 functions.

• 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.

fn gn hn

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.