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Fall 2011 - Problem 1

Egorov's theorem

Prove Egorov's theorem, that is:

Consider a sequence of measurable functions $\func{f_n}{\br{0,1}}{\R}$ that converge (Lebesgue) almost everywhere to a measurable function $\func{f}{\br{0,1}}{\R}$ . Then for any $\epsilon > 0$ , there exists a measurable set $E \subseteq \br{0, 1}$ with measure $\abs{E} < \epsilon$ such that $f_n$ converge uniformly on $\br{0, 1} \setminus E$ .

Solution.

Fix $\epsilon > 0$ .

Let $E_n\p{k} = \bigcup_{m \geq n} \set{x \mid \abs{f_m\p{x} - f\p{x}} \geq \frac{1}{k}}$ . Observe that $x \in \bigcap_{n\geq1} E_n\p{k}$ if and only if for all $n \geq 1$ , there exists $m \geq n$ such that $\abs{f_m\p{x} - f\p{x}} \geq \frac{1}{k}$ , i.e., $x$ is not a point of convergence for $\set{f_n}_n$ . Since $f_n \to f$ Lebesgue almost everywhere, $E_{n+1}\p{k} \subseteq E_n\p{k}$ , and $m\p{\br{0, 1}} < \infty$ ( $m$ denotes the Lebesgue measure), we have

\lim_{n\to\infty} m\p{E_n\p{k}} = m\p{\bigcap_{n\geq1} E_n\p{k}} = 0.

Thus, setting $k = 1$ , there exists $n_1 \geq 0$ such that $m\p{E_{n_1}\p{1}} < \frac{\epsilon}{2}$ . Suppose we have chosen $n_1 < \cdots < n_k$ . Then there exists $n_{k+1} > n_k$ such that $m\p{E_{n_{k+1}}} < \frac{\epsilon}{2^{k+1}}$ , since $m\p{E_n\p{k+1}} \to 0$ as $n \to \infty$ . By induction, there exists a strictly increasing sequence $\set{n_k}_k$ such that $m\p{E_{n_k}\p{k}} < \frac{\epsilon}{2^k}$ . Thus, if we set $E = \bigcup_{k\geq1} E_{n_k}\p{k}$ , then

m\p{E} \leq \sum_{k\geq1} m\p{E_{n_k}\p{k}} < \sum_{k\geq1} \frac{\epsilon}{2^k} = \epsilon.

Moreover,

x \in E^c = \bigcap_{k\geq1} \bigcap_{m \geq n_k} \set{x \st \abs{f_m\p{x} - f\p{x}} < \frac{1}{k}} \iff \forall k \geq 1, \forall m \geq n_k : \abs{f_m\p{x} - f\p{x}} < \frac{1}{k}.

$n_k$ only depends on $k$ , so in other words, $f_n$ converges uniformly on $E^\comp$ , as required.