$f_n\rightarrow g$ in $L_1$ and $f_n\rightarrow h$ in $L_2$ .Then $g=h $almost everywhere $f_n\rightarrow g$ converges in $L_1$  and $f_n\rightarrow h$ converges in $L_2$  
how to show: $g=h$ almost everywhere 
Attempt:
convergent in $L_1$ implies convergent in $L_2$. then by triangle inequality g-h converges to zero in $L_2$. convergent in $L_2$ implies convergent in measure. so $\mu\lbrace x: |g-h|>\alpha\rbrace$ goes to zero. does this imply our conclusion?
 A: We need two ingredients:


*

*If a sequence $(f_n)_n$ converges in some $L^p$ then there exists a subsequence $(f_{n_k})_k$  that converges almost everywhere.

*A subsequence of a convergent sequence converges to the same limit.


Now, since $(f_n)_n$ converges to $g$ in $L^1$ then, (by 1.), there there exists a subsequence $(f_{n_k})_k$  that is simply convergent to $g$ on $\Bbb{R}\setminus {\cal N}$, where ${\cal N}$ is $\mu$-negligable.
Since $(f_n)_n$ converges to $h$ in $L^2$, the subsequence $(f_{n_k})_k$ converges also, (by 2.) to $h$. Thus, using (1.) again, there is a subsequence $(f_{n_{k_r}})_r$ that is 
simply convergent to $h$ on $\Bbb{R}\setminus {\cal N}'$ where ${\cal N}'$ is $\mu$-negligable.
Now, for every $x$ in $\Bbb{R}\setminus ({\cal N}\cup{\cal N}')$, the sequence
$(f_{n_{k_r}}(x))_r$ converges to $h(x)$ and to $g(x)$ (as subsequence of $(f_{n_k}(x))_k$).
This proves that $g(x)=h(x)$ for every $x$ in $\Bbb{R}\setminus ({\cal N}\cup{\cal N}')$,
which is the desired conclusion.
