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:
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.