$\lim_{n\to\infty}\int_{\Omega}|f_n - f|dx = 0$, if $\forall \epsilon > 0, \exists N | \forall n,m > N, \int_{\Omega}|f_n - f_m|dx < \epsilon$. 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?
 A: 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.
A: For your question, it is equivalent to showing that $L^{1}(\Omega)$
is complete with respect to $||\cdot||_{1}$-norm. Recall that for
an abstract normed space $(X,||\cdot||)$, the norm is complete iff
for any sequence $(x_{n})$ in $X$, $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}x_{k}$
exists whenever $\sum_{k=1}^{\infty}||x_{k}||<\infty$. (The proof
is elementary and is not difficult).
We assume the above proposition without proof (if you really need
it, I can re-produce it for you.). Let $(f_{n})$ be a sequence in
$L^{1}(\Omega)$ such that $\sum_{k=1}^{\infty}||f_{k}||_{1}<\infty$.
For each $n$, define $S_{n}=\sum_{k=1}^{n}|f_{k}|$. Clearly, for
each $x\in\Omega$, $0\leq S_{1}(x)\leq S_{2}(x)\leq\ldots$. Therefore
$S(x)=\lim_{n\rightarrow\infty}S_{n}(x)$ exists and is measurable
(but $S(x)$ may be $+\infty$). By monotone convergence theorem,
$$
\int S(x)dx=\lim_{n\rightarrow\infty}\int S_{n}(x)dx=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}||f_{k}||_{1}<\infty.
$$
 That is, $S$ is integrable. In particular, $S(x)<\infty$ for $x$-a.e.
Denote $A=\{x\in\Omega\mid S(x)<\infty\}$, which is a measurable
set. For each $x\in A$, the series $\sum_{k=1}^{\infty}|f_{k}(x)|$
converges, so $\sum_{k=1}^{\infty}f_{k}(x)$ also converges (which
is a consequence of the completeness of $\mathbb{R}$). Define $f(x)=\sum_{k=1}^{\infty}f_{k}(x)$
if $x\in A$ and $f(x)=0$ if $x\in A^{c}$, then $f$ is measurable.
Lastly, verify that $||\sum_{k=1}^{n}f_{k}-f||_{1}\rightarrow0$ as
$n\rightarrow\infty$.
