# Some standard theorems#

Theorem 18 (Egorov’s theorem)

Let $$\mu$$ be a finite measure on a set $$X$$ and $$f_n\colon X \to \mathbb R$$ a sequence of $$\mu$$-measurable functions such that $$f_n \to f$$ pointwise $$\mu$$-a.e. Then for every $$\epsilon>0$$ there exists a measurable $$G\subset X$$ with $$\mu(X\setminus G)<\epsilon$$ and $$f_n\to f$$ uniformly on $$G$$.

Proof. Fix $$\epsilon >0$$, $$k\in\mathbb N$$ and for each $$n\in\mathbb N$$ let

$B_{n,k}=\{x\in X : |f_n(x)-f(x)|>1/k \text{ for some } m\geq n\}.$

By assumption, the $$B_{n,k}$$ are measurable sets that monotonically decrease to a $$\mu$$-null set as $$n\to \infty$$. Therefore, there exists $$n\in\mathbb N$$ such that $$\mu(B_{n,k})<\epsilon 2^{-k}$$. Let

$G_k = X\setminus B_{n,k} \quad \text{and} \quad G=\bigcap_{k\in\mathbb N}G_k.$

Then $$\mu(X\setminus G)<\epsilon$$ and, for each $$x\in G$$ and each $$k\in\mathbb N$$, $$G\subset G_k$$, so there exists $$n\in\mathbb N$$ such that

$|f(x)-f_m(x)| < 1/k$

for all $$m\geq n$$. That is, $$f_m\to f$$ uniformly on $$G$$, as required.

Theorem 19 (Lusin’s theorem)

Let $$\mu$$ be a finite Borel measure on a metric space $$X$$ and let $$f\colon X\to\mathbb R$$ be $$\mu$$-measurable. Then for every $$\epsilon>0$$ there exists a closed $$C\subset X$$ with $$\mu(X\setminus C)<\epsilon$$ such that $$f|_C$$ is continuous.

Proof. Fix $$\epsilon>0$$ and for each $$i\in\mathbb Z$$ let

$X_i = f^{-1}([i\epsilon, (i+1)\epsilon)),$

a collection of disjoint Borel sets which cover $$X$$. Since $$\mu(X)<\infty$$, there exists $$n\in\mathbb N$$ such that

$\mu\left(X\setminus \bigcup_{i=1}^n X_i\right)<\epsilon.$

Since $$\mu$$ is Borel, for each $$1\leq i \leq n$$ there exists a closed $$C_i\subset X_i$$ with $$\mu(X_i\setminus C_i)<\epsilon/n$$.

For a moment fix $$1\leq i \leq n$$ and let

$D=\bigcup_{1\leq j\neq i \leq n}C_j.$

Since $$D$$ is closed, the sets $$B(D,\delta)$$ monotonically decrease to $$D$$ as $$\delta \to 0$$, which is disjoint from $$C_i$$. Therefore there exists $$\delta_i>0$$ such that

$C'_i := C_i \setminus B(D,\delta_i)$

satisfies $$\mu(C_i \setminus C'_i)<\epsilon/n$$. Note that $$C'_i$$ is closed.

Let $$\delta :=\min_{1\leq i \leq n}\delta_i>0$$ and $$C_\epsilon= \bigcup_{i=1}^n C'_i$$, a closed set. For any $$1\leq i\neq j\leq n$$ and $$x\in C'_i$$ and $$y\in C'_j$$, $$d(x,y)\geq \delta$$. Therefore, if $$x,y\in C_\epsilon$$ with $$d(x,y)<\delta$$, $$|f(x)-f(y)|<\epsilon$$. Repeat this for each $$k\in\mathbb N$$ with $$\epsilon_k = 2^{-k}\epsilon/3$$ and let $$C=\bigcap_{k\in\mathbb N}C_{\epsilon_k}$$, so that $$\mu(X\setminus C)<\epsilon$$. Then $$f$$ is continuous on $$C$$.

Definition 26 (Absolutely continuous and singular measures)

Let $$\mu,\nu$$ be measures on a set $$X$$. We say $$\nu$$ is absolutely continuous with respect to $$\mu$$, written $$\nu\ll \mu$$, if for every $$S\subset X$$, $$\mu(S)=0 \implies \nu(S)=0$$. We say that $$\nu$$ is singular with respect to $$\mu$$, written $$\nu \perp \mu$$ if there exists $$A\subset X$$ with $$\mu(X\setminus A)=0=\nu(A)$$.

Let $$\mu$$ be a measure on a set $$X$$ and let $$f\colon X\to \mathbb R^+$$. The set valued function

$\nu(S) = \int_S f\, \mathrm{d}\mu$

defines a measure on $$X$$. Note also that $$\nu \ll \mu$$. The Radon-Nikodym theorem provides the converse to this statement.

Let $$\mu,\nu$$ be finite measures on a set $$X$$ such that $$\nu\ll \mu$$. There exists a $$\mu$$ and $$\nu$$-measurable $$f\colon X \to \mathbb R^+$$ such that

$\nu(S) = \int_S f\, \mathrm{d}\mu$

for all $$\mu$$ and $$\nu$$-measurable $$S\subset X$$. The function $$f$$ is called the Radon-Nikodym derivative of $$\nu$$ with respect to $$\mu$$.

Proof. Let $$\mathcal F$$ be the set of all $$\mu$$ and $$\nu$$-measurable $$f\colon X\to \mathbb R^+$$ such that

$\int_S f \, \mathrm{d}\mu \leq \nu(S)$

for all $$\mu$$ and $$\nu$$-measurable $$S\subset X$$. Note that $$0\in\mathcal F$$ and

(33)#$f,g\in \mathcal F \Rightarrow \max\{f,g\}\in\mathcal F.$

Let

$M=\sup\left\{\int f\, \mathrm{d}\mu: f\in\mathcal F\right\},$

so that $$0\leq M\leq \nu(X)<\infty$$, and let $$f_i\in\mathcal F$$ be such that

$\int f_i\, \mathrm{d}\mu \to M.$

(33) implies that we may suppose the $$f_i$$ monotonically increase. Let $$f\colon X\to \mathbb R^+$$ be the pointwise limit of the $$f_i$$. Then $$f$$ is $$\mu$$ and $$\nu$$-measurable and, by the monotone convergence theorem, $$f\in\mathcal F$$ and $$\int f\, \mathrm{d}\mu \geq M$$. Thus

$\label{f_equality}\int f\, \mathrm{d}\mu = M.$

We claim that $$f$$ satisfies the conclusion of the proposition. Indeed, suppose that $$B\subset X$$ is $$\mu$$-measurable but

$\nu(B) > \int_B f\, \mathrm{d}\mu$

and let $$\epsilon>0$$ be such that

(34)#$\nu(B) > \int_B f +\epsilon \, \mathrm{d}\mu.$

Let $$\mathcal S$$ be the collection of all $$\mu$$ and $$\nu$$-measurable $$S\subset B$$ such that

$\nu(S) \leq \int_S f+\epsilon \, \mathrm{d}\mu.$

We claim that there exists a $$\mu$$ and $$\nu$$-measurable $$G\subset B$$ of positive $$\mu$$-measure such that each $$\mu$$ and $$\nu$$-measurable $$G'\subset G$$ of positive $$\mu$$-measure is not contained in $$\mathcal S$$. Indeed, if not, then each $$G\subset B$$ of positive $$\mu$$-measure contains an element of $$\mathcal S$$ of positive $$\mu$$-measure. Thus Lemma 15 gives a countable disjoint decomposition of $$\mu$$-almost all of $$B$$ into elements of $$\mathcal S$$. Since $$\nu\ll \mu$$ this implies

$\nu(B) \leq \int_B f+\epsilon \, \mathrm{d}\mu,$

Note that $$f+\epsilon G \in\mathcal F$$. Indeed, if $$S\subset X$$ is $$\mu$$-measurable,

$\begin{split}\int_S f+\epsilon G\, \mathrm{d}\mu &= \int_{S\setminus G} f + \int_{S\cap G} f+\epsilon \, \mathrm{d}\mu\\ &\leq \nu(S\setminus G) + \nu(S\cap G)\\ &=\nu(S).\end{split}$

On the other hand, since $$\mu(G)>0$$,

$\int f+\epsilon G \, \mathrm{d}\mu = M+\epsilon \mu(G)>M,$

contradicting the definition of $$M$$.

Theorem 21 (Lebesgue decomposition theorem)

Let $$\mu,\nu$$ be finite measures on a set $$X$$. There exists a $$\nu$$-measurable $$A\subset X$$ with $$\mu(X\setminus A)=0$$ such that, for all $$S\subset A$$, $$\mu(S)=0 \Rightarrow \nu(S)=0$$. That is, $$\nu=\nu_{ac}+\nu_{\perp}$$ with $$\nu_{ac}\ll \mu$$ and $$\nu_{\perp}\perp \mu$$.

Proof. Let $$\mathcal S$$ be the set of all $$\nu$$-measurable $$S\subset X$$ with $$\mu(S)=0$$. By Lemma 15, there exists $$S_i\in \mathcal S$$ such that each $$S\in\mathcal S$$ with

$S\subset A:=X\setminus \bigcup_{i\in\mathbb N}S_i$

satisfies $$\nu(S)=0$$. Since $$\mu(X\setminus A)=0$$, this is the required decomposition.

## Exercises#

Example 55

Let $$\mu$$ be a Borel measure on a metric space $$X$$, $$f\colon X \to \mathbb R$$ $$\mu$$-integrable and $$\epsilon>0$$.

1. Show that there exists a simple function $$s$$ such that

$\int_X |f-s|\, \mathrm{d}\mu <\epsilon.$
2. If $$f$$ is positive show that we may require $$0\leq s \leq f$$ in the previous point.

3. Show that if $$\mu$$ is finite, there exists $$g\in C(X)$$ with

$\int_X |f-g| \, \mathrm{d}\mu <\epsilon.$
4. Show that the previous point may fail if $$\mu$$ is only $$\sigma$$-finite.

Example 56

Prove the following variant of Lusin’s theorem for the case that $$\mu$$ is not finite but $$f$$ is $$\mu$$-integrable: for every $$\epsilon>0$$ there exists a closed $$C\subset X$$ with

$\int_{X\setminus C} |f|\, \mathrm{d}\mu<\epsilon$

such that $$f|_C$$ is continuous.

Example 57

Let $$\mu,\nu$$ be measures on a set $$X$$. Show that if $$A\subset X$$ is $$\mu+\nu$$ measurable then it is also $$\mu$$-measurable.

Example 58

Prove the Radon-Nikodym and Lebesgue decomposition theorems for $$\sigma$$-finite measures.

Example 59

Let $$\mu,\nu$$ be finite measures on a set $$X$$ and suppose $$\nu\ll\mu$$. For any $$\mu,\nu$$-measurable $$g\colon X\to \mathbb R^+$$ show that

$\int g\, \mathrm{d}\nu = \int gf\, \mathrm{d}\nu,$

where $$f$$ is the Radon-Nikodym derivative of $$\nu$$ with respect to $$\mu$$.

Example 60

For a measure $$\mu$$ on a set $$X$$, we say a function $$f\colon X\to \mathbb R^n$$ is $$\mu$$-measurable if each component of $$f$$ is $$\mu$$-measurable and define the integral of $$f$$ component by component.

An $$n$$-dimensional vector valued measure is a function

$\nu\colon \{A: A \subset X\} \to \mathbb R^n$

for which there exists a measure $$\mu$$ on $$X$$ and a $$\mu$$-measurable $$f\colon X \to \mathbb S^{n-1}$$ such that

$\nu(A) = \int_A f \, \mathrm{d}\mu$

for each $$A\subset X$$.

1. Show that if

$\nu = \int f \, \mathrm{d}\mu = \int f'\, \mathrm{d}\mu'$

are two representations of a vector valued measure then $$\mu = \mu'$$ (when restricted to the set of $$\mu$$-measurable sets), and hence $$f=f'$$ $$\mu$$-a.e. We denote this unique measure by $$|\nu|$$. It is called the total variation of $$\nu$$.

2. Show that the set of all vector valued measures form a normed vector space when equipped with

$\|\nu\| = |\nu|(X).$
3. Show that this space is complete.

A signed measure is a 1-dimensional vector valued measure.

Example 61

Let $$\Sigma$$ be a $$\sigma$$-algebra on $$X$$. The standard definition of a signed measure on $$\Sigma$$ is a countably additive function

$\mu \colon \Sigma \to \mathbb R.$

The Hahn decomposition theorem states that there exist disjoint $$P,N\in \Sigma$$ with $$X=P\cup N$$ such that:

• For every $$S\in \Sigma$$ with $$S\subset P$$, $$\mu(P)\geq 0$$ and

• For every $$S\in \Sigma$$ with $$S\subset N$$, $$\mu(P)\leq 0$$.

That is, $$\mu|_{P}$$ and $$-\mu|_{N}$$ are (positive) measures.

Use the Lebesgue decomposition and Radon–Nikodym theorems to show that the two definitions of a signed measure agree.