Egorov's Theorem

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $D \in \Sigma$ be such that $\map \mu D < +\infty$.

Let $\sequence {f_n}_{n \mathop \in \N}, f_n: D \to \R$ 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 $E \subseteq D$, $\map \mu E = 0$ and:

$\displaystyle \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$

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:

$A_{n, k} = \set {x \in A: \size {\map {f_n} x - \map f x} \ge \dfrac 1 k}$

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 $k \in \N$:

$\size {\map {f_i} x - \map f x} < \dfrac 1 k$

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:

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


So we have:

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

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

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

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

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


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

$\size {\map {f_i} x - \map f x} < \frac 1 k$

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 $A \setminus B = D \setminus \paren {E \cup B}$, and:

$\map \mu {E \cup B} \le \map \mu B + \map \mu E = \map \mu B + 0 < \epsilon$

Since $\epsilon$ was arbitrary, it follows that:

$\map \mu {E \cup B} = 0$

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

$\blacksquare$


Also see


Source of Name

This entry was named for Dmitri Fyodorovich Egorov.