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 \left({B_\epsilon}\right) < \epsilon\ $ outside of which $f_n\ $ converges uniformly to $f\ $.

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

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

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

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

Hence $\mu \left({B}\right) = 0\ $.

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

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