I'm studying Capinski - Copp Measure - Integral Probability. Specifically, at the proof of Thm 5.1 p.130 ($L^1(E)$ is Complete) they write:
Firstly, they consider a sequence ${f_n}$ in $L^1(E)$, after some work they produce a subsequence of ${f_n}$ that converges to some $f(x)$ for every $x\in E$. Then, they say 'Since the sequence of real numbers ${f_n(x)}$ is a Cauchy' we have that the ${f_n}$ also converge to $f(x)$.
I missing the argument that ${f_n(x)}$ is Cauchy on $\mathbb{R}$. It follows from the fact that ${f_n}$ is Cauchy on $L^1(E)$ ? And if this is the case how can I prove it?
I do not know the full detail of the proof in your book so I will explain based on the proof I know. (The last page of following lecture note introduces the proof I know. It proves the completeness of $L^2$, but the argument works also for $L^1$.)
In fact, the sequence $\langle f_n\rangle$ need not be Cauchy. For example, consider the following sequence over $E=[0,1]$: $$\chi_{[0,1]},\, \chi_{[0,1/2]},\, \chi_{[1/2,1]}, \, \chi_{[0,1/3]},\, \chi_{[1/3,2/3]},\,\cdots$$ Here $\xi_A$ is the characteristic function of $A$. Then the sequence converges to 0 in $L^1$-sense (so it is Cauchy in $L_1$-sense), but it does not have a pointwise limit. Moreover, you can examine that the sequence is not Cauchy over any point (that is, $\langle f_n(x)\rangle$ is not Cauchy for any $x\in[0,1]$.)