Theorem. Let $(X,\mathcal{A},\mu)$ be a complete measure space. Let $f,g\colon X\to [0,\infty]$ be a functions such that $f=g$ a.e, then if $f$ is measurable results that also $g$ is measurable.
This theorem, in general, if the measure space is not complete, is not true.
The completion hypothesis can fall if $f$ and $g$ are functions defined almost everywhere.
Therefore if $f$ is a function a.e. defined, let $N$ the measurable set in which it is not defined, $\mu(N)=0$, then $f$ is measurable as function in $\complement N$.
So I should show the validity of this theorem:
Theorem.2 Let $(X,\mathcal{A},\mu)$ be a measure space. Let $f,g\colon X\to [0,\infty]$ be a functions defined a.e. such that $f=g$ a.e, then if $f$ is measurable results that also $g$ is measurable.
Start of the proof. Let $N_1:=\{x\in\ X\;|\;f\;\text{is not defined}\}$ and $N_2:=\{x\in X\;|\;f(x)\ne g(x)\}$, then by definition $f$ is measurable as a function in $\complement N_1$. Observe that $N_2\subseteq \complement N_1$.
Consider the null set $N:=N_1\cup N_2$ and $\complement N$.
Question. From this point, how can I get to the thesis?
Thanks!