How can I prove that if a sequence of functions $\{f_n\}$ that converges to $f$ in measure on a space of finite measure, then there exists a subsequence of $\{f_n\}$ that converges to $f$ almost everywhere?


Let $(X,\mathcal{A},\mu)$ be a measure space and $(f_n)_{n \in \mathbb{N}}$ such that $f_n \to f$ in measure, i.e.

$$\mu(|f_n-f|>\varepsilon) \stackrel{n \to \infty}{\to} 0$$

for any $\varepsilon >0$. Setting $\varepsilon=2^{-k}$, $k \in \mathbb{N}$, we can choose $n_k$ such that

$$\mu(|f_n-f|> 2^{-k}) \leq 2^{-k}$$

for all $n \geq n_k$. Without loss of generality, $n_{k+1} \geq n_k$ for all $k \in \mathbb{N}$. Set

$$A_k := \{x \in X; |f_{n_k}(x)-f(x)| > 2^{-k}\}.$$

As $$\sum_{k \geq 1} \mu(A_k) \leq \sum_{k=1}^{\infty} 2^{-k} < \infty,$$ the Borel-Cantelli lemma yields

$$\mu \left( \limsup_{k \to \infty} A_k \right) =0.$$

It is not difficult to see that this implies

$$\lim_{k \to \infty} f_{n_k}(x) =f(x)$$

$\mu$-almost everywhere.

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    $\begingroup$ where did you use $(X,\mathcal{A},\mu)$ was a finite measure space? $\endgroup$ – JonSK Jan 17 '16 at 1:07
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    $\begingroup$ @JonSK I used the Borel Cantelli lemma. $\endgroup$ – saz Jan 17 '16 at 7:33
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    $\begingroup$ Are you sure that Borel-Cantelli requires a finite measure space? Wikipedia says it doesn't, and the proof there makes sense to me. $\endgroup$ – Matthew Kvalheim Aug 11 '18 at 2:45
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    $\begingroup$ @MatthewKvalheim You are right; it should work without this assumption. I thought the finiteness would be necessary to use the continuity of the measure, but it isn't. $\endgroup$ – saz Aug 12 '18 at 14:38
  • $\begingroup$ It won't let me comment, but the answer by @saz is incorrect. Consider the case where $|f_{n_k}(x) - f(x)| = 2^{-(k-1)}$ for all $k$. Clearly, $\lim_{k \to \infty}f_{n_k}(x) = f(x)$, but also $|f_{n_k}(x) - f(x)| > 2^{-k}$ for all $k$, so in particular $x \in \limsup_{k \to \infty} A_k$. So they are not equivalent. However, $x \not\in \limsup_{k \to \infty} A_k$ does imply that $\lim_{k \to \infty}f_{n_k}(x) = f(x)$, and this occurs a.s. $\endgroup$ – blair Jan 5 '20 at 0:32

We can enhance the result to almost uniform convergence by going to subsequences. Fix $e$, let $E_{n,k} = {x: |f_n - f| > 1/k}$, then $\lim \mu E_{n,k} = 0$, we can pick $n_j$, such that $\mu E_{n_j,k} < e/2^{k+j}$, then we have $\mu \bigcup_j E_{n_j,k} < e/2^k$ and $\mu \bigcup_{k} \bigcup_j E_{n_j,k} < e$, let the last union of sets be $T$, then on $T^c$ we have uniform convergence of $f_n$. cf. Egoroff's theorem.


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