Egorov’s Theorem
Published:
This is week two of re-writing a proof from Axler’s book Measure, Integration & Real Analysis by Sheldon Axler. The proof covered here is Egorov’s Theorem.
Lemma 1: Suppose \((X,\mathcal{S},\mu )\) is a measure space and \(E_1\subseteq E_2\subseteq \dots\) is an increasing sequence of sets in \(\mathcal{S}\). Then \(\mu (\cup_{k=1}^{\infty} E_k)=\lim_{k\to \infty }\mu (E_k)\).
Lemma 2: Suppose \((X,\mathcal{S},\mu )\) is a measure space and \(D,E\in\mathcal{S}\) are such that \(D\subseteq E\). Then \(\mu (E\setminus D)=\mu (E)-\mu (D)\) provided that \(\mu (D)<\infty\).
Def. (Pointwise Convergence): Sequence \(f_1,f_2,\dots\) converges pointwise on \(X\) to \(f\) if \(\lim_{k\to\infty}f_{k}(x)=f(x)\) for each \(x\in X\). In other words, \(f_1,f_2,\dots\) converges pointwise on \(X\) to \(f\) if for each \(x\in X\) and every \(\epsilon >0\), there exists \(n\in\mathbb{Z}_+\) such that \(\vert f_{k}(x)-f(x)\vert < \epsilon\) for all integers \(k\geq n\) and all \(x\in X\).
Def. (Uniform Convergence): Sequence \(f_1,f_2,\dots\) converges uniformly on \(X\) to \(f\) if for every \(\epsilon >0\), there exists \(n\in\mathbb{Z}_{+}\) such that \(\vert f_k (x)-f(x)\vert <\epsilon\) for all integers \(k\geq n\) and all \(x\in X\).
Egorov’s Theorem: suppose \((X,\mathcal{S},\mu )\) is a measure space with \(\mu (X)<\infty\). Suppose \(f_1,f_2,\dots\) is a sequence of \(\mathcal{S}\)-measurable functions from \(X\) to \(\mathbb{R}\) that converge pointwise on \(X\) to a function \(f:X\to\mathbb{R}\). Then for every \(\epsilon > 0\), there exists aa set \(E\in\mathcal{S}\) such that \(\mu (X\setminus E)<\epsilon\) and \(f_1,f_2,\dots\) converges uniformly to \(f\) on \(E\).
Proof: Temporarily fix \(n\in\mathbb{Z}_{+}\). For \(m\in\mathbb{Z}_{+}\), let \(A_{m,n}=\cap_{i=1}^{\infty}\{\ x\in X : |f_k(x)-f(x)|< \frac{1}{n} \}\) and note that \(\cup_{m=1}^{\infty}A_{m,n}=X\) by the Def. of pointwise convergence. Note that each \(A_{m,n}\in\mathcal{S}\) since \(f_k-f\) is \(\mathcal{S}\)-measurable for all \(k\in\mathbb{Z}_{+}\) and that \(A_{1,n}\subseteq A_{2,n}\subseteq \dots\) is an increasing sequence of sets. Thus, Lemma 1 holds and we have that \(\lim_{m\to\infty}\mu (A_{m,n})=\mu (\cup^{\infty}_{m=1}A_{m,n})=\mu (X)~.\) By Def. of limit, there exists \(m_n\in\mathbb{Z}^{+}\) such that \(\mu (X)-\mu (A_{m_n,n})<\frac{\epsilon}{2^n}\) assuming \(\epsilon >0\). We claim \(E:= \cap_{n=1}^{\infty}A_{m_n,n}\) will satisfy our core claim of the proof. Observe the following,
\[\begin{equation} \begin{split} \mu (X\setminus E) & = \mu (X\setminus \cap_{n=1}^{\infty} A_{m_n,n} ) & \qquad \text{//Def. E}\\ & = \mu (\cup_{n=1}^{\infty}X\setminus A_{m_n,n} ) & \qquad \text{//De Morgen's Law of set diff.} \\ & \leq \sum_{n=1}^{\infty} \mu (X\setminus A_{m_n,n}) & \qquad \text{//countable subadditivity} \\ & = \sum_{n=1}^{\infty}\mu (X)-\mu (A_{m_n,n}) & \qquad \text{//Lemma 2}, \mu (X)<\infty , \; \& \; A_{m_n,n}\subseteq X \\ & < \epsilon & \quad \text{//}\sum_{k=1}^{\infty}\frac{1}{2^k}=1 \end{split} \end{equation}\]To complete the proof, we must verfy that \(f_1,f_2,\dots\) converge uniformly to \(f\) on \(E\). Suppose \(\epsilon ' >0\). Let \(n'\in\mathbb{Z}_{+}\) be such that \(\frac{1}{n'}<\epsilon '\). Then \(E\subseteq A_{m_n,n}\) which implies that \(|f_k(x)-f(x)|<\frac{1}{n'} <\epsilon '\) for all \(k>m_n\) and all \(x\in E\). Hence, \(f_1,f_2,\dots\) does indeed converge uniformly to \(f\) on \(E\). \(\square\)