# Convergence in measure and almost everywhere

In a finite measure space, let $\{f_{n}\}$ be a sequence of measurable functions. Show that $f_{n} \rightarrow f$ in measure if and only if every subsequence $\{f_{n_{k}}\}$ contains a subsequence $\{f_{n_{k_{j}}}\}$, that converges almost everywhere to f.

I can prove from converge in measure to converge almost everywhere, but I don't know how to write it down for the other direction.

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## 2 Answers

Assume $\{f_n\}$ converges pointwise a.e. to $f$. Fix $\varepsilon > 0$ and define: $$E_N = \{x \in X \mid \exists n > N : |f_n(x) - f(x)| > \varepsilon\}$$

We have $E_1 \supset E_2 \supset \cdots$ and $\mu(E_1) \le \mu(X) < \infty$. Furthermore, since $\{f_n\}$ converges pointwise a.e. to $f$, we have $\mu\left(\bigcap_{N \in \mathbb{N}} E_N\right) = 0$.

Hence, $\lim_{N \to \infty} \mu(E_N) = 0$ and we can find an $N$ for which $\mu(E_N) < \varepsilon$.

It's clear that: $$\{x \in X \mid \forall n > N : |f_n(x) - f(x)| > \varepsilon\} \subset E_N$$

Hence, $\{f_n\}$ converges to $f$ in measure.

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This solution is fine. but I would like ask you a question. Isn't it very obvious that $\mu ({ x\in X | \exists n > N : |f_n(x) - f(x)| > \epsilon}) = 0$, since, $f_n \rightarrow f$ pointwise a.e. ? Where am I making mistakes? –  Topology Aug 20 at 13:23
@Topology If it's very obvious, what's your proof? –  Ayman Hourieh Aug 20 at 16:53

If not, then for some $\varepsilon>0$ and c>0, there exist a subsequence of $f_n(x)$, for example $f_{n_k}(x)$, which satisfies $\forall n_k$,
$m\{x \in X : |f_{n_k}(x) - f(x)| > \varepsilon\}>c$ (*)

Since $f_{n_k}(x)$ has a subsequence $g_n(x)$, which converges almost everywhere to $f(x)$, and that the problem is discussed in a finite measure space, we can say $g_n(x)$ converges to $f(x)$ in measure. However, that contradicts the existence of $f_{n_k}(x)$. $g_n(x)$ is a subsequence of $f_{n_k}(x)$, but it doesn't follow (*).

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