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Let $ ( \Omega , \Sigma , \mu ) $ be a measure space, such that $ \mu $ is $ \sigma $ - finite.

  • A sequence $ (f_n )_{ n \geq 0 } $ of measurable functions is said to be a locally Cauchy sequence in measure if :

$ \forall A \in \Sigma \ $ : $ \ \mu (A) < + \infty $

$ \forall \epsilon > 0 \ $ : $ \ \displaystyle \lim_{ p,q \to + \infty } \mu ( \{ \omega \in A \ : \ | f_p ( \omega ) - f_q ( \omega ) | \geq \epsilon \} ) = 0 $

  • A sequence $ (f_n )_{ n \geq 0 } $ of measurable functions is said to be a Cauchy $ \mu $ - almost everywhere if :

$ \exists N \in \Sigma \ $ : $ \ \mu (N) = 0 $

$ \forall \omega \in \Omega \backslash N \ \ \forall \epsilon > 0 \ \ \exists n_{ \epsilon , \omega } \in \mathbb{N} \ \ \forall p,q \in \mathbb{N} \ : \ p,q \geq n_{ \epsilon , \omega } \ \ \Longrightarrow \ \ | f_p ( \omega ) - f_q ( \omega ) | < \epsilon $

Question :

How to show that, if $ (f_n )_{ n \geq 0 } $ is a locally Cauchy sequence in measure, then there exists a subsequence $ (f_{n_{k}})_{ k \geq 0 } $ which is Cauchy $ \mu $ - almost everywhere.

My try :

I don't know if we have to pass to a subsequence $ (f_{n_{k}} )_{ k \geq 0 }$ which is chosen such that if $ E_j = \{ x \in A : | f_{n_{j}} (x) - f_{n_{j+1}}(x)| \geq 2^{-j}\}$ with : $ \mu (A) < + \infty $. then $\mu(E_j) \leq 2^{-j}$. Let $F_k = \cup_{k}^\infty E_j$, so $\mu(F_k) \leq 2^{1-k}$ by subadditivity of $\mu$. For $x \notin F_k$ and $i \geq j \geq k$, we can show that $$|f_{n_{j}} (x) - f_{n_{i}} (x)| \leq \sum_{l=j}^{i-1} |f_{n_{l+1}}(x) - f_{n_{l}} (x)| \leq 2^{1-j}$$ by definition of the subsequence. So, $ (f_{n_{k}} )_{ k \geq 0 } $ is locally pointwise Cauchy on $\complement F_k $. Let $F = \cap_1^\infty F_k$, which has $\mu(F) = 0$. But, I am still unclear as to where to pursuit my reasoning. .

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You are on the right track and are almost done. What you have showed, by taking $j$ sufficiently large in your inequality, is that $\forall \omega \in A\setminus F$ and $\epsilon > 0$, $\exists n_{\epsilon,\omega}$ such that $p,q \ge n_{\epsilon,\omega} \implies |f_p(\omega)-f_q(\omega)| < \epsilon$.

Using $\sigma$-finiteness, say $\Omega = \sqcup_n A_n$ where $\mu(A_n) < \infty$. Then do what you did above, calling $A$, $A_n$ and calling $F$, $F_n$. Then, $\mu(\cup_n F_n) = 0$, and you are done.

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