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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.

z3wood
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1 Answers1

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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 $f$.

z3wood
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