Here is an edited fragment from an exercise:
Let $(X, \mathcal A, \mu)$ be a measure space, $(f_n)$ be a sequence of such and such functions. If $f(x)= \lim f_n(x)$ exists for almost every $x\in X$ then $f$ has such and such properties (in particular $f$ is measurable).
I'm confused about the role of $f$. Is it already given in the statement? E.g. if in the proof, one would set $f$ to zero on some convenient set of measure zero, it wouldn't be OK? (A proof which I encountered does exactly that.)
The question probably has little to do with measure theory but that a general "if something shows up in the assumption, must it be fixed from there on?" but I've added the context just in case.
If the measure space is not complete, then following statement is, in general, not correct.
Just pick a non-measurable set $N$ such that there exists $M \supseteq N$ measurable with $\mu(M)=0$ (such a set $N$ exists if the measure space is not complete). Then $f_n(x) := 0$ converges for almost every $x \in X$ to $f(x) := 1_N(x)$, but $f$ is not measurable.
However, the following statement holds true:
Note that the function $\tilde{f}$ satisfies in particular $$\tilde{f}(x) = \lim_{n \to \infty} f_n(x)$$ for $\mu$-almost all $x \in X$.