#### [SOLVED] Convergence in measure implies convergence almost everywhere of a subsequence

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?

#### 1 comments #### @saz 2014-11-04 17:10:55

Let $(X,\mathcal{A},\mu)$ be a finite 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 is equivalent to

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

$\mu$-almost everywhere. #### @JonSK 2016-01-17 01:07:39

where did you use $(X,\mathcal{A},\mu)$ was a finite measure space? #### @saz 2016-01-17 07:33:31

@JonSK I used the Borel Cantelli lemma. #### @Matthew Kvalheim 2018-08-11 02:45:58

Are you sure that Borel-Cantelli requires a finite measure space? Wikipedia says it doesn't, and the proof there makes sense to me. #### @saz 2018-08-12 14:38:37

@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.

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