If $f$ is measurable and $f=g$ almost everywhere is $g$ measurable

Question

We were told in lecture, that if two functions $f,g$ are equal almost everywhere. Meaning that for a measure space $(X,A, \mu)$ sucht that the set $\{x \in X | g(x) \neq f(x)\}$ is a measurable nullset, then if $f$ is measurable so is $g$. I understand that this holds for complete measures. But I am stuck showing it for arbitrary measures. Any help would be appreciated.

Answer 1

It's not true.

Suppose $(X, \mathcal A, \mu)$ isn't complete, so that there is some $B \subset C \subset X$ with $\mu (C) = 0$ and $B \notin \mathcal A$. Then, the indicator function $1_B$ is non-measurable, while $1_C$ is measurable.

Now consider the function $g = 1_B - 1_C$. This is non-measurable because $1_B$ is non-measurable. But

$$\{x: 1_C(x) \neq g(x)\} = C.$$