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Suppose $f_n$ are $L^1$ and for every $\epsilon > 0$ the measure $\{x\in \mathbb{R}: |f_n(x) - f(x)| > \epsilon\}$ goes to $0$ as $n \rightarrow\infty$. I want to show that $f_n$ converges to $f$ a.e.

This seems to be trivial and here is my argument: Suppose not, then there exists $\epsilon_0 > 0$ such that $|f_n(x) - f(x)| > \epsilon_0$ on a set of positive measure, which is a contradiction.

I wonder if above argument is true?

Alvis Nordkovich
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  • You have not used $f_n\in L_1$. You would make the claim that convergence in probability implies convergence almost surely. This is not true. – AlvinL Jun 07 '21 at 04:51
  • "Suppose not, then there are some $\delta>0$ and some measurable set $U$ such that $m(U)>\delta$ and such that for all $x\in U$ there is some $\varepsilon>0$ such that for all $N$ there is some $n\ge N$ such that $\lvert f_n(x)-f(x)\rvert>\varepsilon$." How did you continue? –  Jun 07 '21 at 05:07
  • As it is stated, the result is false. Please, see https://math.stackexchange.com/questions/1412091/the-typewriter-sequence/1412099 – Ramiro Jun 07 '21 at 20:22
  • You ca also see https://math.stackexchange.com/questions/3402184/typewriter-sequence-and-a-e-convergence – Ramiro Jun 07 '21 at 20:25

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The proposition is false and it is impossible to prove. If the measure is finite, then the given condition means that $f_n\rightarrow f$ in measure. Remark: It is well-known that $f_n\rightarrow f$ in measure implies that the exists a subsequence $(f_{n_k})$ of $(f_n)$ such that $f_{n_k}\rightarrow f$ a.e. as $k\rightarrow \infty$. If I remember correctly, this result is due to Vitali.

Danny Pak-Keung Chan
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