Measure theory class notes - 15 October 2010, class 21 1

Absolute continuity and singularity

Suppose Ω is a nonempty set, and A is a σ-field on it. Let µ and ν be measures on (Ω, A ). Definition. ν is said to be absolutely continuous with respect to µ, denoted as ν  µ, if for all A ∈ A , µ(A) = 0 =⇒ ν(A) = 0.

This was the situation we encountered in the Radon-Nikodym theorem. At the other end of the spectrum, we have the following:

Definition. ν is said to be singular with respect to µ, denoted as µ ⊥ ν, if there exists A ∈ A such that ν(A) = 0 = µ(Ω \ A).

This says that µ and ν are supported on disjoint sets.

Now suppose µ and ν are both σ-finite. Then we can write ν as ν1 + ν2, where ν1 is absolutely continuous with respect to µ, and ν2 is singular with respect to µ. We will not prove this now.

Function spaces

We now do some analysis, using the theory we have developed. Studying limits is important, since that is often the only way to obtain certain objects. For example, irrational real numbers can often be obtained only via a limiting operation. For a space (Ω, A , µ), for p ∈ R, p ≥ 1, define  Z  Lp(Ω, A , µ) = f : f real measurable , |f|p dµ < ∞

Lp(Ω, A , µ) will be denoted simply by Lp if there is no ambiguity. For f ∈ Lp, define

1 Z  p p kfkp = |f| dµ

Suppose we take Ω = {1, . . . , n}, A = 2Ω, and µ(A) = |A|, the counting measure. Then the set of all real functions on Ω is just Rn. Taking p = 2, we get the usual Euclidean norm on Rn:

1 v n Z  2 u 2 uX 2 kfk2 = f dµ = t (f(n)) i=1

∞ We now look at some modes of convergence of functions: suppose {fn}n=1 is a sequence of mea- surable functions and f is a measurable .

1. fn → f as n → ∞ if µ({ω : fn(ω) 6→ f(ω)}) = 0.

2. fn → f as n → ∞ in measure if for all  > 0,

µ({ω : |fn(ω) − f(ω)| > }) → 0 as n → ∞ Measure theory class notes - 15 October 2010, class 21 2

p 1 p 3. fn → f as n → ∞ in L (also denoted as fn −→ f) if each fn − f, fn, and f are in L and p kfn − fkp → 0.

Here are some examples. Consider our usual (R, B, λ).

1. Let fn = 1[n,∞), f = 0. Then fn → f almost everywhere (in fact, everywhere), however, for all n ∈ N,  1  λ ω : |fn(ω) − f(ω)| > 2 = λ([n, ∞)) = ∞ ∞ So {fn}n=1 does not converge to f in measure.

∞ 2. Define a sequence of indicator functions {fn}n=1 as follows: the first element is 1[0,1). The next two elements are 1 1 and 1 1 . The next four elements are 1 1 , 1 1 2 , 1 2 3 , 1 3 ; [0, 2 ) [ 2 ,1) [0, 4 ) [ 4 , 4 ) [ 4 , 4 ) [ 4 ,1) and so on. In general, the elements from positions 2n to 2n+1 − 1 are indicator functions of sets obtained by dividing [0, 1) into 2n pieces in the obvious way. As before, let f be the zero function.  ∞ Now consider the sequence λ({ω : fn(ω) − f(ω) > }) n=1. For  ≥ 1 this is the zero sequence; for 0 <  < 1, this sequence consists of a sequence of blocks, the nth block having n−1 1 2 numbers, each each number being 2n−1 . So this sequence converges to 0, and fn → f in measure as n → ∞. ∞ However, fn 6→ f almost everywhere. In fact, for every ω ∈ [0, 1), {fn(ω)}n=1 does not converge to f(ω) = 0. This is because, for a fixed ω, in the block of indices from 2n to n+1 ∞ 2 −1, there is some index k such that fk(ω) = 1. {fn(ω)}n=1 has a subsequence consisting of all 1s, and so does not converge to 0.

Here the set where fn differs from f becomes small, but keeps “moving around” all over [0, 1), preventing almost everywhere convergence.

1We will soon see that Lp is a vector space under the usual operations.