Egorov's Theorem

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $D \in \Sigma$ be such that $\mu \left({D}\right) < +\infty$.

Let $\left({f_n}\right)_{n \in \N}, f_n: D \to \R$ be a sequence of $\Sigma$-measurable functions.

Suppose that $f_n$ converges a.e. to $f$, for some $\Sigma$-measurable function $f: D \to \R$.


Then $f_n$ converges a.u. to $f$.


Proof

Let $\epsilon > 0$ be arbitrary.

By definition of convergence a.e., there is a set $E \in \Sigma$ such that $E \subseteq D$, $\mu \left({E}\right) = 0$ and:

$\displaystyle \lim_{n \to \infty} f_n \left({x}\right) = f \left({x}\right)$

for each $x \in A$ where $A = D \setminus E$ is the set difference of $D$ with $E$.


For all $n, k \in \N$, define $A_{n, k}$ by:

$A_{n, k} = \left\{ {x \in A : \left|{ f_n \left({x}\right) - f \left({x}\right) }\right| \ge \dfrac 1 k}\right\}$

Also, define $B_{n, k}$ by:

$B_{n, k} = \displaystyle \bigcup_{i \mathop = n}^\infty A_{i, k}$


Since $f_n$ converges pointwise to $f$ on $A$, it follows by definition of convergence that for each $x \in A$ and $k \in \N$:

$\left|{ f_i \left({x}\right) - f \left({x}\right) }\right| < \dfrac 1 k$

for all $i \in \N$ sufficiently large.

Thus, when $k$ is fixed, no element of $A$ belongs to $A_{n, k}$ for infinitely many $n$.

Hence, by Definition 2 of Limit Superior of Sequence of Sets:

$\displaystyle \limsup_{n \to \infty} A_{n, k} = \varnothing$


So we have:

\(\displaystyle 0\) \(=\) \(\displaystyle \mu \left({\varnothing}\right)\) Measure of Empty Set is Zero
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\limsup_{n \to \infty} A_{n, k} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\bigcap_{n \mathop = 0}^\infty B_{n, k} }\right)\) Definition of Limit Superior of Sequence of Sets
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\limsup_{n \to \infty} B_{n, k} }\right)\) by Definition 2 of Limit Superior of Sequence of Sets
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\lim_{n \to \infty} B_{n, k} }\right)\) since $B_{n, k}$ is a decreasing sequence of sets
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \to \infty} \mu \left({B_{n, k} }\right)\) since $B_{n, k} \subseteq D$, $\mu \left({D}\right) < \infty$ and by Continuity of Measure

Thus, since $k$ was arbitrary, we can associate some $n_k \in N$ with each $k$ such that $\mu \left({B_{n_k, k}}\right) < \dfrac {\epsilon} {2^{k+1}}$.

Setting $B = \displaystyle \bigcup_{k \mathop \in \N} B_{n_k, k}$, we have:

$\displaystyle \mu \left({B}\right) \le \sum_{k \mathop \in \N} \mu \left({ B_{n_k, k} }\right) < \sum_{k \mathop \in \N} \frac {\epsilon} {2^{k+1}} = \epsilon$

by the countable additivity of $\mu$ and by Sum of Infinite Geometric Progression.


Also, given any $k$, we have that $x \in A \setminus B$ implies $x \notin B_{n_k, k}$, which means:

$\left|{ f_i \left({x}\right) - f \left({x}\right) }\right| < \frac 1 k$

for all $i \ge n_k$.

Since this is true for all $x \in A \setminus B$, it follows that $f_n$ converges to $f$ uniformly on $A \setminus B$.


Finally, note that $A \setminus B = D \setminus \left({E \cup B}\right)$, and:

$\mu \left({E \cup B}\right) \le \mu \left({B}\right) + \mu \left({E}\right) = \mu \left({B}\right) + 0 < \epsilon$.

Since $\epsilon$ was arbitrary, it follows that $\mu \left({E \cup B}\right) = 0$.

Hence $f_n$ converges to $f$ almost uniformly.

$\blacksquare$


Also see


Source of Name

This entry was named for Dmitri Fyodorovich Egorov.