Proof of Almost uniform convergence implies Convergence almost everywhere I read the same proof that almost uniform convergence implies convergence almost everywhere on several sources (Friedman's Foundations of Modern Analysis and online sources), and they all seem to use the same proof:
Proof taken from Proofwiki, https://proofwiki.org/wiki/Convergence_a.u._Implies_Convergence_a.e.
However, I have a problem with this proof. While the set on which $f_n$ does not converge to $f$ is definitely a subset of $B$, I do not see why the reverse inclusion (that $f_n$ does not converge to $f$ for every element of $B$) is true. Hence, might it not be possible that $f_n$ converges to $f$only on a proper subset of $B$ that just happens to be non-measurable? (Which is possible since the measure space is not specified to be complete.) Then the proof would be false.
I would really appreciate help in understanding why my critique of the proof does not hold. Thanks in advance.
 A: Actually if $(f_n)_{n \in \mathbf{N}}$ is a sequence of measurable functions to a complete space, then the set of convergence points of $(f_n)$ is always measurable. In your example, $f_n : D \to \mathbf{R}$ and $\mathbf{R}$ is a complete space.
$$\{x \in D \; \vert \; (f_n(x)) \text{ converges}\}=\bigcap_{\epsilon \in \mathbf{Q_+^*}}\bigcup_{N \in \mathbf{N}}\bigcap_{n,m\ge N}\{x \in D \; \vert \; |f_n(x)-f_m(x)| \le \epsilon\}$$
And $\{x \in D \; \vert \; |f_n(x)-f_m(x)| \le \epsilon\}=|f_n-f_m|^{-1}([0,\epsilon])$ is measurable because $|f_n-f_m|$ is still measurable.
A: This answer seeks to address your confusion over the definition of "almost everywhere" (a.e.). According to Rudin - Real and Complex Analysis, something occurs a.e. if the set where it doesn't occur is a subset of a set of measure 0. If the measure isn't complete the actual set on which it doesn't occur could be non-measurable. This fact is used, for example, in Rudin's book in Chapter 8 when proving that (under certain conditions) if the projection of a subset of $U \subset X \times Y$ onto $X$ has measure 0 for almost all $y \in Y$, $U$ has measure 0.
