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If $f_n\to f$ in measure, is there a subsequence s.t. $f_{n_k}\to f$ a.e. ? The convergence in measure is a little be abstract to me, I don't really see what it means (even if I know the definition). So I have difficulties to answer this question.

user301068
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1 Answers1

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If $f_n\to f$ in measure then $|f_n-f|\to 0$ in measure. So let prove that if $g_n\to 0$ in measure, then there is a subsequence $g_{n_k}\to 0$.

It's in fact a consequence of Borel-cantelli lemma. You have by definition that $$m\left(|g_n|>\frac{1}{p}\right)\underset{n\to \infty }{\longrightarrow }0$$ for all $p\in\mathbb N$. So fix $p\in\mathbb N$. There is $k_p$ s.t. $$m\left(|g_{k_p}|>\frac{1}{p}\right)<\frac{1}{2^p}.$$

Therefore $$\sum_{p\in\mathbb N^*}m\left(|g_{k_p}|>\frac{1}{p}\right)<\infty $$ and thus by Borel-Cantelli, $$m\left(\limsup_{p\to\infty }\left\{|g_{k_p}|>\frac{1}{p}\right\}\right)=0,$$

what prove the claim.

Convergence in measure

You have that $f_n\to f$ if $$\forall \varepsilon>0, \lim_{n\to \infty }m\{x\in\mathbb R\mid |f_n(x)-f(x)|>\varepsilon \}=0$$ so the convergence in measure is a weakness convergence.

Surb
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  • very nice answer. Maybe though choose a different symbol for $m \in \mathbb N$, since $m$ already is our measure? :) – Ant Jan 29 '16 at 18:20
  • You are right, I corrected it. Thank you. – Surb Jan 29 '16 at 18:22
  • Thank you :-) So does it mean that $f_n\to f$ in measure if $m{x\mid \lim_{n\to \infty }f_n(x)=f(x)}\neq 0$ ? – user301068 Jan 29 '16 at 18:38
  • The last paragraph is simply wrong? Whatever this is it has noting to do with convergence in measure. – T'x Jul 04 '18 at 15:38
  • @T'x: Indeed, I corrected it... (I don't know why I wrote that 2 years ago). Thank you for pointing this mistake. – Surb Jul 04 '18 at 16:06