if $f_n\geq 0$ a.e. and $f_n\longrightarrow f$ in measure then $f\geq 0$ a.e.

Let $(\Omega, \Sigma, \mu)$ be a measurable space and $(f_n)_{n=1}^\infty$,$f$ measurable functions. Prove that if $f_n\geq 0$ a.e. and $f_n\longrightarrow f$ in measure then $f\geq 0$ a.e.

Since $f_n\longrightarrow f$ in measure there exists a subsequence $f_{n_k}$ such that $f_{n_k}\longrightarrow f$ a.e. Now I don't know how to proceed. Is enough to notice that since we know that the statement is true for pointwise a.e convergence then is true for $f_{n_k}\longrightarrow f$ and by limit uniqueness must be $f\geq 0$ a.e. ? Or, on the contrary, I need to construct a subsequence of $f_{n_{k_i}}$ of $f_{n_k}$?

$\{x: f(x) <0\}\subseteq \bigcup_k \{x: f_{n_k} (x) <0 \} \cup A$ where $A$ is the set of points where $(f_{n_k}(x))$ does not converge to $f(x)$>

$f_{n}\geq 0$ a.e., in other words, for each $n=1,2,...$, there is a set $N_{n}$ with $|N_{n}|=0$ and that $f_{n}(x)\geq 0$ for all $x\in(N_{n})^{c}$.

Now the subsequence $(f_{n_{k}})$ is such that $f_{n_{k}}(x)\rightarrow f(x)$ a.e., so a set $N$ is such that $f_{n_{k}}(x)\rightarrow f(x)$ for all $x\in N^{c}$.

Now $|\bigcup_{n}N_{n}\cup N|=0$ and for all $x\in(\bigcup_{n}N_{n}\cup N)^{c}$ we have $f_{n_{k}}(x)\geq 0$ and $f_{n_{k}}(x)\rightarrow f(x)$, this implies $f(x)\geq 0$ and hence $f\geq 0$ a.e. The later almost everywhere is due to $\bigcup_{n}N_{n}\cup N$.