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I have seen a similar exercise in Royden (if I recall correctly), but the statement included Cauchy convergence in measure.

In this case, there are measurable functions $f_n:X\rightarrow\mathbb{R}$ such that $\{f_n\}$ is Cauchy a.e.

I need to prove that there exists a measurable function $f$ for which $f_n\to f$ a.e.

Every Cauchy sequence has a finite limit then, intuitively, it seems that we can find a function $f$ which is finite-valued so that $f_n \to f$ a.e.

I can't, however, justify this in a rigorous way. Can you provide a hint, please?

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    $\begingroup$ For any point $x$ outside a set of measure zero $N,$ the sequence of real numbers $f_n(x)$ is fundamental, hence converging to some $f(x)$ and on $N$ we define $f = 0.$ Q.E.D. $\endgroup$
    – William M.
    Feb 4 '19 at 21:37
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Let $\mu$ be the measure on $X,$ and let $A=\{x\in X: f_n(x) \text { is Cauchy }\}.$ We are given $X\setminus A$ is measurable, with $\mu(X\setminus A)=0.$ Hence $A$ is measurable. For $x\in A,$ we have $f_n(x)$ Cauchy, hence convergent, hence $f_n(x)\to \liminf f_n(x).$ But $\liminf f_n(x)$ is a measurable function on $A.$ Hence the function

$$f(x) = \begin{cases} \liminf f_n(x),&x\in A\\ 0,&x\in X\setminus A\end{cases}$$

is measurable on $X,$ and $f_n(x)\to f(x)$ $\mu$-a.e.

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  • $\begingroup$ Downvoter: Why? $\endgroup$
    – zhw.
    Aug 15 '19 at 20:49
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The set, on which $(f_n)_n$ is a Cauchy-sequence, can be written as $$\Omega = \bigcap_{k=1}^\infty \bigcup_{N=1}^\infty \bigcap_{n,m \ge N} \{ |f_n - f_m| < 1/k\}.$$ Since $\mathbb{R}$ is complete, for any $\omega \in \Omega$ the limes $\lim_{n \rightarrow \infty} f_n(\omega)$ exists in $\mathbb{R}$, Your assumptations says also that $\mu(\Omega^c) =0$, i.e. $\Omega^c$ is a nullset. Next define $$f:= \lim_{n \rightarrow \infty} f_n(\omega) 1_\Omega(\omega).$$ Then $f_n \rightarrow f$ a.e. and, since $f_n 1_{\Omega}$ is measurable for any $n \in \mathbb{N}$, also $f$ is measurable (as a pointwise limes).

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  • $\begingroup$ Can you please add more details on the $\Omega$ set? What does it stand for? $\endgroup$
    – Don Draper
    Feb 2 '19 at 17:23

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