Cauchy a.e. implies existence of a measurable function to which this sequence converges a.e

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?

• 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. Feb 4 '19 at 21:37

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.

• Downvoter: Why?
– zhw.
Aug 15 '19 at 20:49

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).

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