If $(f_n)_n$ converges in measure m to both $f$ and $g$, then $f = g$ almost everywhere (alternative proof). Let $(X, M, m)$ be a finite measurable space and $f, (f_n)_n, g$ measurable functions. Then the following hold: 

Proposition 1: 
  If $(f_n)_n$ converges in measure m to both $f$ and $g$, then $f = g$ 
   almost everywhere. 


Proposition 2:
  $(f_n)_n$ converges in measure to f iff for every subsequence $(f_{k_n})_n$ of $(f_n)_n$, there exists a further subsequence $(f_{k_{l_n}})_n$ such that $(f_{k_{l_n}})_n$ converges to $f$ m-almost everywhere.

I am interested in the proof of proposition 1 as a direct collorary of proposition 2. It was stated in class that the proof should follow naturally but I'm struggling to see this. 
Please note that I am aware of the 'common' proof  of proposition 1 as found e.g. here. 
 A: The sequence $(f_n)_{n \in \mathbb{N}}$ is trivially a subsequence of $(f_n)_{n \in \mathbb{N}}$, and therefore it follows from Proposition 2 that there exists a subsequence $(f_{n_k})_{k}$ such that $f_{n_k} \to f$ $m$-almost everywhere. Applying Proposition 2 once more we find that $f_n \to g$ in measure implies that there exists a subsequence $(f_{n_{k_{\ell}}})_{\ell}$ of $(f_{n_k})_{k}$ which converges $m$-almost everyhwere to $g$. However, $(f_{n_{k_{\ell}}})_{\ell}$ is a subsequence of $(f_{n_k})_k$, and therefore we also have $f_{n_{k_{\ell}}} \to f$ $m$-almost everywhere. Hence, by the uniqueness of pointwise limits, we conclude $f=g$ almost everywhere.
A: It is not hard to see that, if $(f_{n})$ converges to $f$ (or $g$) in measure, then all the subsequences $(f_{n_{k}})$ converges to $f$ (or $g$) in measure as well.
Now given $(f_{n})$ converges to $f$ in measure, then by Proposition 2 there exists some subsequence $(f_{n_{k}})$ converges to $f$ $m$-almost everywhere, say, $f_{n_{k}}(x)\rightarrow f(x)$ for all $x\in X-N_{1}$, $m(N_{1})=0$. 
Now for this subsequence $(f_{n_{k}})$, it also converges to $g$ in measure, again by Proposition 2, then there exists some subsequence $(f_{n_{k_{l}}})$ such that $f_{n_{k_{l}}}(x)\rightarrow g(x)$ for all $x\in X-N_{2}$, $m(N_{2})=0$.
Now $m(N_{1}\cup N_{2})=0$ and for all $x\in X-(N_{1}\cup N_{2})$, we have both $f_{n_{k_{l}}}(x)\rightarrow f(x)$ and $f_{n_{k_{l}}}(x)\rightarrow g(x)$, so $f(x)=g(x)$ for all such $x$, hence $f=g$ $m$-almost everywhere.
