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 $D \in \Sigma$.

Let $f_n$ converge a.u. to a function $f$ on $D$.

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

Proof
Assume $f_n$ converge a.u. 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\ $.