Convergence of {$f_n$} to $f$ in $L_1 (\mu)$ are equivalent to each subsequence {$g_n$} in {$f_n$} contains a subsequence {$g_{nk}$} convergent to $f$ in $L_1 (\mu)$
The left side can easily deduce the right side. But how to prove the opposite direction.
Thank in advance.
True. If not, there exists an$\epsilon \gt 0$ , such that for all $k$ , there exists an $n_k \ge k$ satisfying $\int |f_{nk}-f|d\mu \ge \epsilon$. Then we can take it as $f_{nk} $, so the subsequence does not have any subsequence converging to $f$.