Suppose that a succession of measurable functions {$f_{n}(x)$} converges in almost all points towards a limit function $f(x)$. Prove that {$f_{n}(x)$} converges in almost all points towards a limit function $g(x)$ iff $g(x)$ is equivalent to $f(x)$.
My attempt: [Necessity] Let $A_{f}$={$x \in X: \lim_{n\rightarrow\infty}{f_{n}(x)=f(x)}$} and $A_{g}$= {$x \in X: \lim_{n \rightarrow \infty}{f_{n}(x)=g(x)}$}. Then $X-A_{f}$ and $X-A_{g}$ are measurables because are the sets where points no converges then $\mu (X-A_{f})=\mu (X-A_{g})=0$. Then {$x \in X : f(x) \neq g(x)$}=$(X-A_{f}) \cap (X-A_{g})$ then the measure is equal to zero.
Could you check this proof? and how could I do the sufficience?
It looks good.
For the other implication, consider the set $\{f_n\to f\}\cap \{f=g\}$, which has full measure by similar arguments to what you wrote.