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How can I prove that if a sequence of functions $\{f_n\}$ that converges to $f$ in measure on a space of finite measure, then there exists a subsequence of $\{f_n\}$ that converges to $f$ almost everywhere?

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2 Answers 2

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Let $(X,\mathcal{A},\mu)$ be a measure space and $(f_n)_{n \in \mathbb{N}}$ such that $f_n \to f$ in measure, i.e.

$$\mu(|f_n-f|>\varepsilon) \stackrel{n \to \infty}{\to} 0$$

for any $\varepsilon >0$. Setting $\varepsilon=2^{-k}$, $k \in \mathbb{N}$, we can choose $n_k$ such that

$$\mu(|f_n-f|> 2^{-k}) \leq 2^{-k}$$

for all $n \geq n_k$. Without loss of generality, $n_{k+1} \geq n_k$ for all $k \in \mathbb{N}$. Set

$$A_k := \{x \in X; |f_{n_k}(x)-f(x)| > 2^{-k}\}.$$

As $$\sum_{k \geq 1} \mu(A_k) \leq \sum_{k=1}^{\infty} 2^{-k} < \infty,$$ the Borel-Cantelli lemma yields

$$\mu \left( \limsup_{k \to \infty} A_k \right) =0.$$

It is not difficult to see that this implies

$$\lim_{k \to \infty} f_{n_k}(x) =f(x)$$

$\mu$-almost everywhere.

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    $\begingroup$ I just came across this answer and I'm struggling to understand the last step. Why is $\limsup\limits_{k \rightarrow \infty} A_k$ the opposite of $\{ f_{n_k}(x) \rightarrow f(x), k \rightarrow \infty \}$? $\endgroup$
    – Lime91
    Jun 6, 2023 at 19:19
  • $\begingroup$ @Lime91 $\limsup A_k$ is the set of points which are in infinitely many $A_k$'s. So $x\notin\limsup A_k$ means $x$ is in at most finitely many $A_k$'s, so there is an $N_x$ with $k\geq N_x\implies x\notin A_k \Longleftrightarrow |f_{n_k}(x)-f(x)|\leq 2^{-k}$. This implies pointwise convergence at $x$. $\endgroup$ Dec 4, 2023 at 14:03
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We can enhance the result to almost uniform convergence by going to subsequences. Fix $e$, let $E_{n,k} = {x: |f_n - f| > 1/k}$, then $\lim \mu E_{n,k} = 0$, we can pick $n_j$, such that $\mu E_{n_j,k} < e/2^{k+j}$, then we have $\mu \bigcup_j E_{n_j,k} < e/2^k$ and $\mu \bigcup_{k} \bigcup_j E_{n_j,k} < e$, let the last union of sets be $T$, then on $T^c$ we have uniform convergence of $f_n$. cf. Egoroff's theorem.

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