If $f_n\to f$ in $L^1(\Bbb R)$ and $f_n\to g$ in $L^2(\Bbb R)$,deduce that $f=g$ a.e. 
Let $f_n$ be a sequence in $L^1(\Bbb R)\cap L^2(\Bbb R)$ ;$f\in  L^{1}(\Bbb R)$ and $g\in L^2(\Bbb R)$.
If $f_n\xrightarrow {L_1(\Bbb R)} f, f_n\xrightarrow {L_2(\Bbb R)} g$,then show that $f=g$(almost everywhere).

Attempt:
We have  $\int_\Bbb R |f_n-f|<\infty $  and $\int_\Bbb R |f_n-g|^2<\infty .$
Now since $f_n\xrightarrow {L_1(\Bbb R)}f\implies $ there exists a subsequence $f_{n_k}$ of $f_n$ converging pointwise to $f$ a.e.  $\implies ||f_{n_k}-f||_{L_1(\Bbb R)}\to 0\implies \int _\Bbb R|f_{n_k}-f|\to 0$.
Similarly there exists a subsequence $f_{n_k}$ of $f_n$ converging pointwise to $g$ a.e. $\implies ||f_{n_k}-g||_{L_2(\Bbb R)}\to 0\implies \int _\Bbb R|f_{n_k}-g|^2\to 0$.
Two questions:

*

*Is it true that $\int _\Bbb R|f_{n_k}-g|^2\to 0\implies \int _\Bbb R|f_{n_k}-g|\to 0$.


*If I can show that $\int |f-g|=0$ then that will imply that $f=g $ a.e.
But I am stuck on these questions.Please help.
 A: Because $f_n\to f$ in $L^1$, there is a subsequence converging to $f$ pointwise a.e.. Then that subsequence still converges in $L^2$ to $g$, so it has a further subsequence converging pointwise a.e. to $g.$  This subsubsequence now converges pointwise to both $f$ and to $g$ a.e. implying that $f=g$ a.e. 
This is likely the method Errol. Y hinted at in a comment. 

Regarding your attempts and related questions: 


*

*Taking two different subsequences with no relation between them would make it harder to show they have the same pointwise limit.  

*No you cannot conclude from $\int|f_n-g|^2\to 0$ that $\int|f_n-g|\to 0$, and you cannot even conclude that $\int|f_n-g|$ is ever finite, because you do not no a priori that $g$ is in $L^1$.  (Of course we know in this problem that it ends up being true because $g=f$.)

*If you could show that $\int|f-g|=0$, then it would follow that $f=g$ a.e., but I don't see a way to go this way, again especially because you do not know at first even that $g$ is in $L^1$.  An $L^2$ limit of $L^1$ functions need not be in $L^1$.  The conclusion of this problem does show that $g$ is in $L^1$, by showing that it equals $f$.  

