$\lim_{n\to\infty}\int_{\Omega}|f_n - f|dx = 0$, if $\forall \epsilon > 0, \exists N | \forall n,m > N, \int_{\Omega}|f_n - f_m|dx < \epsilon$.

Question

I have to show that if $\Omega \subseteq \mathbb{R}$ measurable, and $\{f_n\}$ is a sequence of Lebesgue integrable functions on $\Omega$ then exists an integrable function $f$ on $\Omega$ such that

$$\lim_{n\to\infty}\int_{\Omega}|f_n - f|dx = 0$$

if $$\forall \epsilon > 0, \exists N | \forall n,m > N, \int_{\Omega}|f_n - f_m|dx < \epsilon$$.

This is equivalent to a Cauchy sequence in Measure.

I was trying to use the fact that each $f_n$ is Lebesgue integrable, and claim that for each $f_n$, exists a sequence of simple function $\phi_k$ such that $\phi_k \to f_n$ as $k\to \infty$, and $|\phi_k| \leq |f_n|$ and then use the dominated convergence theorem, since each $f_n$ is integrable and dominate $\phi_k$ then

$$\lim_{k\to \infty}\int_{\Omega}\phi_k = \int_{\Omega}f_n$$

I want to show that $\phi_k$ is the limit of $f_n$ and the integral of that integrable function is the limit of the Cauchy sequence in measure. But I get stuck here. Could anyone helps me finish the proof or I'm not going in the right direction?

Answer 1

HINT

I suggest to take a look at the proof of this theorem:

A normed space $X$ is complete iff every absolutely convergent series converges in $X$

You can use the technique of the proof of this theorem to construct a convergence subsequence of your Cauchy sequence.

Then use the fact that if a Cauchy sequence has a convergent subsequence to some limit $x$ than it converges to the same limit $x$

After you find the limit prove that is integrable.