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Let $f_n$ be a sequence of measurable functions on a finite measure space. Is it true that

If every subsequence of $\{f_n\}$ has a subsequence which converge to $f$ almost everywhere, then $f_n$ converges to $f$ in measure?

I have proved the converse of this statement, but problem says it is if and only if statement. Thanks in advance for any help!

seriously divergent
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    What goes wrong when the measure space is not finite? Why does the finiteness help against this problem? – Listing Nov 25 '13 at 08:28
  • Actually I used finiteness part for the converse so that i could apply Borel Cantelli(to use continuty from above), but for this statment I am not really sure. As far as I know, $L^p$ domination will be enough to conclude the proof, yet we have only finitness of measure which does not imply that f is in $L^p$ for some $p$ – seriously divergent Nov 25 '13 at 09:02
  • @AntonioMontana According to Proof Wiki, finiteness is not needed – AlexR Nov 25 '13 at 09:45
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    @ AlexR Finiteness is definitely needed, proof-wiki gave false statement: consider $f_n=\chi_{(n,n+1)}$, then $f_n$ goes to zero a.e, but not in measure – seriously divergent Nov 25 '13 at 09:53
  • In Cohn's Measure Theory, Proposition 3.1.2, the converse is proven without assuming the measure is finite. – MaudPieTheRocktorate Jan 28 '17 at 02:06
  • @AlexR I'm viewing that page right now you linked, they didn't wrote a proof, the "prof" section it's empty – Santropedro Apr 16 '17 at 23:39
  • @Santropedro Did you check out the mentioned book? Probably there is a proof in there at the quoted position. – AlexR Apr 17 '17 at 11:40

2 Answers2

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We can show the contrapositive.

Assume that $f_n$ doesn't converge to $f$ in measure. This means that there is $\delta\gt 0$, an $\varepsilon\gt 0$ and a subsequence $(f_{n_k})_k$ such that for each $k$, $$\mu\{|f_{n_k}-f|\gt \varepsilon\}\gt \delta.$$

If $(f_{m_k})_k$ is a subsequence of $(f_{n_k})_k$, we have to show that $f_{m_k}$ doesn't converge almost everywhere to $f$. Define $A_k:=\{|f_{m_k}-f|\gt \varepsilon\}$. Then we have, using finiteness of the measure space, $$\mu\left(\limsup_{k\to \infty}A_k\right)=\mu\left(\bigcap_{j=1}^\infty\bigcup_{k\geqslant j}A_k\right)\geqslant \delta.$$

Davide Giraudo
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This is an easy corollary from Egorov's theorem, which states:

Given some measure space $(X,\Sigma,\mu)$ Let $f_n: E\rightarrow \mathbb{R}$ be a sequence of measurable functions on some $E \in \Sigma, \mu(E)<\infty$. Where $f_n \rightarrow f^*$ pointwise for some $f^*: E \rightarrow \mathbb{R}$. Then for all $\epsilon > 0$ there is a set $F_\epsilon \in \Sigma, F_\epsilon \subset E$ such that $\mu(F_\epsilon) < \epsilon$ and $f_n \rightarrow f^*$ uniformly on $E\backslash F_\epsilon$.

Can you deduce the theorem from Egorov on your own?

Listing
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