Convergence a.u. Implies Convergence a.e.

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

Let $$f_n: D \to \R$$ be a sequence of $\Sigma$-measurable functions for $$D \in \Sigma$$.

Then $$f_n\ $$ almost uniformly converges to a function $$f\ $$ on $$D\ $$ only if $$f_n \stackrel{a.e.}{\to} f$$.

Proof
Assume $$f_n\ $$ almost uniformly converges to $$f\ $$ on $$D\ $$.

Then for each $$\epsilon > 0\ $$ there is a $$B_\epsilon \subseteq D$$ with $$\mu(B_\epsilon) < \epsilon\ $$ outside of which $$f_n\ $$ converges uniformly to $$f\ $$.

Thus, $$f_n\ $$ converges pointwise to $$f\ $$ outside of each $$B_\epsilon\ $$.

Next, define $$B \equiv \bigcap_{n\in\N}B_\frac{1}{n}$$.

First, $$B \in \Sigma$$ since $$\Sigma\ $$ is a sigma-algebra, so that $$\mu(B)\ $$ is defined.

Second, note that $$\mu(B) \leq \mu(B_\frac{1}{n}) < \frac{1}{n}$$ for each $$n$$ since the measure is monotonic.

Hence $$\mu(B) = 0\ $$.

But $$x \in D - B$$ only if $$x \notin B_\frac{1}{k}$$ for some $$k \in \N$$, so by the above, $$f_n(x) \to f(x)$$.

It follows by definition that $$f_n\ $$ converges a.e. to $$f\ $$.