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?
Best Answer
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.