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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!


marked as duplicate by Alex M., YuiTo Cheng, metamorphy, Mars Plastic, The Count Aug 5 at 0:05

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  • $\begingroup$ What goes wrong when the measure space is not finite? Why does the finiteness help against this problem? $\endgroup$ – Listing Nov 25 '13 at 8:28
  • $\begingroup$ 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 $p$ $\endgroup$ – seriously divergent Nov 25 '13 at 9:02
  • $\begingroup$ @AntonioMontana According to Proof Wiki, finiteness is not needed $\endgroup$ – AlexR Nov 25 '13 at 9:45
  • 3
    $\begingroup$ @ 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 $\endgroup$ – seriously divergent Nov 25 '13 at 9:53
  • $\begingroup$ In Cohn's Measure Theory, Proposition 3.1.2, the converse is proven without assuming the measure is finite. $\endgroup$ – MaudPieTheRocktorate Jan 28 '17 at 2:06

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, $$\mu\left(\limsup_{k\to \infty}A_k\right)=\mu\left(\bigcap_{j=1}^\infty\bigcup_{k\geqslant j}A_k\right)\geqslant \delta.$$


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?


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