Riesz's subsequence theorem

Remarks

Choose $n_1 < n_2 < \cdots$ with $\mu(E_k) < 2^{-k}$, where $E_k = \set{x : \abs{f_{n_k}(x) - f(x)} \ge 2^{-k}}$. Let $E = \bigcap_{j} \bigcup_{k \ge j} E_k$; then $\mu(E) \le \sum_{k \ge j} 2^{-k} = 2^{1-j}$ for every $j$, so $\mu(E) = 0$. For $x \notin E$, one has $\abs{f_{n_k}(x) - f(x)} < 2^{-k}$ for all large $k$, so $f_{n_k}(x) \to f(x)$. Example: the sliding indicator functions $f_{2^k + j} = \mathbf{1}_{[j 2^{-k}, (j+1) 2^{-k}]}$ on $[0, 1]$ converge to $0$ in measure but at no point; the subsequence $f_{2^k}$, the indicators of $[0, 2^{-k}]$, converges to $0$ at every $x \ne 0$. The passage to a subsequence is necessary, as this example shows.

Going deeper

For a student

The result is a standard tool to pass from integral estimates to pointwise statements. For example, if $f_n \to f$ in $L^p$, then $f_n \to f$ in measure, so a subsequence converges to $f$ almost everywhere; this is a step in the proof that $L^p$ is complete. On a space of finite measure, it gives a characterisation of convergence in measure. There, $f_n \to f$ in measure if and only if every subsequence has a further subsequence converging to $f$ almost everywhere.