convergence divergence - Cauchy in measure implies uniqueness of almost everywhere limits for subsequences

I tried this exercise all the past week, but a I can´t do it.

We know that if $\{f_n\}$ is Cauchy in measure, then, there is a measurable function $h$ such that $\{f_n\}$ converges in measure to $h$. Moreover, there is a subsequence that converges to $h$ almost uniformly and almost everywhere to $h$, but...

If $\{f_n\}$ is Cauchy in measure and there are subsequences $\{f_{n_k}\}$ and $\{f_{m_k}\}$ and measurable functions $f,g$ such that $\{f_{n_k}\}$ and $\{f_{m_k}\}$ converge to $f$ and $g$ almost everywhere, respectively, then $f=g$ almost everywhere.

I tried to show that $f=h=g$ a.e., but I can't solve it. Any help (or hint) is welcome.