Let $(f_{n})_{n}$ be a sequence of of functions in $L^{1}(\mu)$ ($\mu$ is $\sigma$-finite or finite ) such that it converges to $f$ in measure. Then, if $\lim_n\|f_{n}\|_{L^1}= \|f\|_{L^1}$, then $\lim_n \|f_n-f\|_{L^1}=0$.
Convergence in measure implies that there exists a subsequence $(f_{n_k})$ that $\mu$ almost surely converges to $f$, then $|f_{n_k}|$ converges to $|f|$, $\mu$ almost surely. In this case as $ \lim_k \|f_{n_k}\|_{L^1}= \|f\|_{L^1}$, $\lim_k \|f_{n_k}-f\|_{L^1}=0$. Now i stuck at showing that $\lim_n \|f_n-f\|_{L^1}=0$. Can we say that $\|f_n-f_{n_{k}}\|_{L^1}$ goes to $0$ as the index goes to infinity?
Edit : Not exactly sure if its true, but since $(f_{n_k})$ is also a sequence in $L^{1}(\mu)$, $f$ is in $L^{1}(\mu)$ due to completeness(?). I got two ideas after this point:
1)limit of $(f_{n})$ is finite hence $sup_{n} \int f_{n} <\infty$, so on finite measures $(f_{n})$ is uniformly integrable, which proves the result. Not sure about about the $\sigma$-finite measures.
2)Since $f_n$ converges, it is Cauchy(?) this proves the claim.
Edit 2: It seems that on $\sigma$-finite measure convergence in $L^{1}$ doesn't imply that $f$ is in $L^{1}$, unless the convergence of the sequence is almost uniformly. In finite measure the solution I proposed holds since almost everywhere implies almost uniform convergence by Egoroff's theorem. So the convergence of the subsequence implies that $f$ in $L^{1}$ and the rest is true.
You know already everything, just choose the right sequence :)
Here comes the trick, consider the sequence $(||f_n-f||)$ and choose the subsequence $(n_k)$ that converges to $\limsup_n||f_n-f||$. All the assumptions on $(f_n)$ hold also for $(f_{n_m})$.
Now, you choose a subsubsequence $(n_{m_k})$ of $(n_m)$ as you have suggested in your question. An application of Scheffe's lemma reveals then that $||f_{n_{m_k}}-f||\rightarrow0$.
As $||f_{n_m}-f||\rightarrow\limsup_n||f_n-f||$, we find that $\limsup_n||f_n-f||=0$. Hence $||f_n-f||\rightarrow0$.