Integrability: Cauchy Sequence

Question

This thread is related to: Spectral Measure: Dominated Convergence

Given a measure space $\Omega$.

Consider a sequence of square integrables: $\int|f_n|^2\mathrm{d}\mu<\infty$

Suppose pointwise convergence: $f_n\to f$.

Does the following hold: $$\int|f_m-f_n|^2\mathrm{d}\mu\to0\implies\int|f-f_n|^2\mathrm{d}\mu\to0$$

The problem is that they may have no dominant at all: $$s_n:=\frac{1}{\sqrt{n}}\chi_{(n,n+1]}\to0:\quad\int|s_m-s_n|^2\mathrm{d}\mu\stackrel{m\neq n}{=}\frac{1}{m}+\frac{1}{n}\to0\quad\int\sup_n|s_n|\mathrm{d}\mu=\sum_{n=1}^\infty\frac{1}{n}=\infty$$

Answer 1

Ok, I got it now: Fatou! :D

Extracting the step within the completeness proofs: $$\int\|F-F_n\|^2\mathrm{d}\mu\leq\liminf\int\|F_m-F_n\|^2\mathrm{d}\mu\to0$$ (Note that this works perfectly for Banach spaces!)

Answer 2

See for a related question Does $L^p$ convergence imply pointwise convergence.

Since you have $(f_n)$ converging to some function $y$ in $L^2$, there exists a subsequence $(f_{n_k})$ converging to $y$ a.e., and therefore since this subsequence also converges pointwise to $f$ we must have $f = y$ a.e.. Since $y \in L^2$ we conclude that $f\in L^2$ as well.