TITLE: Non-Borel a.e limit of Borel functions
QUESTION [5 upvotes]: As a homework assignment I'm supposed to prove or disprove Borel measurability is closed under a.e convergence. I think this is not true because the Borel $\sigma$-field is not complete. However, I'm not sure how to construct or describe a counterexample.
The proof in case of a complete domain $X$ goes by observing that measurability is preserved under a.e equality and limits, so also by a.e convergence. I just don't see how to get out of this a counterexample to the claim.

REPLY [6 votes]: As the Borel-$\sigma$-algebra is not complete, there is a non-measurable set $A\subseteq \mathbf R$ of outer measure zero. Let $f = \chi_A$ (the characteristic function of $A$) and $f_n = 0$. Then $f_n \to f$ almost everywhere (namely outside of $A$), then $f_n$ are Borel-measurable, but $f$ is not.