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.

Theorem 20 (Radon–Nikodym theorem)

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

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

contradicting (34).

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.