Cauchy sequence in mean Suppose $\{f_n\}$ is a sequence of functions in $L_1$. Show that $\{f_n\}$ is a Cauchy sequence in mean if and only if
$\int_E f_n \, d\mu=x_n$ is a Cauchy sequence for real numbers for every measurable set $E,$ and $\{f_n\}$ is a Cauchy sequence in measure.
I am able to prove the first part:
($\Rightarrow$) Since Cauchy sequence in $p$th mean implies Cauchy sequence in measure, we only need to prove that the sequence $\{x_n\}$ is a Cauchy sequence (We can assume that $E=\Omega$ is the entire space), but that is precisely the definition of being a Cauchy sequence in $L_1$, so we are done
I'm having trouble to proving the second part, since I cannot estimate $\int_E |f_n-f_m| \, d\mu$ using $|x_n-x_m|=|\int_E( f_n-f_m ) \, d\mu|$. I was tying to use one Theorem, too:

Theorem: Suppose $\{f_n\}$ is a sequence of functions is $L_p$ and let
$\nu_n(E)$ = $\int_E |f_n|^p \, d\mu$. Then $\{f_n\}$ is a Cauchy sequence in pth mean if an only if $\{f_n\}$ is a Cauchy sequence in measure and the family $\{\nu_n\}$ is equicontinuous at $\varnothing$.

Since $\nu_n(\Omega)<\infty$, then we only need to show that the family is absolutely continuous with respect to $\mu$, and is precisely here where I'm stuck.
 A: What you are trying to prove is true but the proof is not that easy. It's actually a theorem, called the Vitali-Hahn-Saks theorem.
Let $\left(  \Omega,\mathfrak{M},\mu\right)  $ be the measure space with $\mu$
finite. By identifying sets that differ by a set of $\mu$ measure zero we can
regard $\mathfrak{M}$ as a closed subset $\mathcal{C}$ of $L_{1}$ through the
mapping
$$
E\in\mathfrak{M}\mapsto\chi_{E}.
$$
Fix $\varepsilon>0$ and for $k\in\mathbb{N}$ define the sets
$$
\mathcal{C}_{k}:=\left\{  \chi_{E}:\,E\in\mathfrak{M},\,\sup
_{n,\,l\geq k}\left\vert \int_{E}f_{n}\,d\mu-\int_{E}f_{l}\,d\mu\right\vert
\leq\varepsilon\right\}  .
$$
We claim that the sets $\mathcal{C}_{k}$ are closed in $L_{1}$. Indeed, if
$\left\{  \chi_{E_{j}}\right\}  \subset\mathcal{C}_{k}$ converges in $L_{1}$
to some function $f$, then, by extracting a subsequence (not relabeled), if
necessary, we may assume that $\chi_{E_{j}}\left(  x\right)  \rightarrow f(x)$
for $\mu$ a.e. $x\in X$. Since $\chi_{E_{j}}(x)$ takes only values $0$ and
$1$, it follows that $f(x)\in\{0,1\}$ for for $\mu$ a.e. $x\in X$. Hence,
$f=\chi_{E_{\infty}}$ for some measurable set $E_{\infty}$. Since  $\chi_{E_{j}}(x)\leq1$ for every $x\in\Omega$ and all
$j$ and $\mu$ is finite, we can apply the Lebesgue dominated convergence theorem to get that for each
$n\in\mathbb{N}$,
$$
\lim_{j\rightarrow\infty}\int_{E_{j}}f_{n}\,d\mu=\int_{E_{\infty}}f_{n}\,d\mu.
$$
Since $\left\{  \chi_{E_{j}}\right\}  \subset\mathcal{C}_{k}$ for any fixed
$n$,$\,l\geq k$\ and for all $j\in\mathbb{N}$, we have
$$
\left\vert \int_{E_{j}}f_{n}\,d\mu-\int_{E_{j}}f_{l}\,d\mu\right\vert
\leq\varepsilon,
$$
and so letting $j\rightarrow\infty$ we obtain that
$$
\left\vert \int_{E_{\infty}}f_{n}\,d\mu-\int_{E_{\infty}}f_{l}\,d\mu
\right\vert \leq\varepsilon,
$$
which shows that $f\in\mathcal{C}_{k}$. Hence, $\mathcal{C}_{k}$ is closed\ in
$L_{1}$.
By the completeness of the reals $\lim_{n\rightarrow\infty}\int_{E}f_{n}
\,d\mu$ exists in $\mathbb{R}$ for all $E\in\mathfrak{M}$, which implies that 
$$
\mathcal{C}=\bigcup_{k=1}\mathcal{C}_{k}.
$$
Applying the Baire category theorem to the complete metric space $\mathcal{C}
$, at least one of the sets\ $\mathcal{F}_{k}$ has nonempty interior. Hence
there exist $\delta_{1}>0$, $k\in\mathbb{N}$, and $\chi_{E_{0}}\in
\mathcal{C}_{k}$\ such that if $\chi_{E}\in\mathcal{C}$ and if
\begin{equation}
\int_{\Omega}\left\vert \chi_{E}-\chi_{E_{0}}\right\vert \,d\mu<\delta
_{1},\label{lp vhs1}
\end{equation}
then $\chi_{E}\in\mathcal{C}_{k}$, that is,
\begin{equation}
\sup_{n,\,l\geq k}\left\vert \int_{E}f_{n}\,d\mu-\int_{E}f_{l}\,d\mu
\right\vert \leq\varepsilon\text{.}\label{lp vhs2}
\end{equation}
Since each single function $f_{n}$ is integrable, we may find $0<\delta
<\delta_{1}$\ such that
\begin{equation}
\int_{E}\left\vert f_{n}\right\vert \,d\mu\leq\varepsilon\label{lp vhas3}
\end{equation}
for all $1\leq n\leq k$\ and for every measurable set $E$ with $\mu\left(
E\right)  \leq\delta$.
Consider a measurable set with $\mu\left(  E\right)  \leq\delta$. Then
we can write $
E=\left(  E\cup E_{0}\right)  \setminus\left(  E_{0}\setminus E\right)$, 
with
$$
\int_{\Omega}\left\vert \chi_{E\cup E_{0}}-\chi_{E_{0}}\right\vert
\,d\mu<\delta_1,\quad\int_{\Omega}\left\vert \chi_{E_{0}\setminus E}-\chi_{E_{0}
}\right\vert \,d\mu<\delta_{1},
$$
and so,
\begin{align*}
\sup_{n,\,l\geq k}\left\vert \int_{E\cup E_{0}}f_{n}\,d\mu-\int_{E\cup E_{0}
}f_{l}\,d\mu\right\vert  &  \leq\varepsilon,\\
\sup_{n,\,l\geq k}\left\vert \int_{E_{0}\setminus E}f_{n}\,d\mu-\int
_{E_{0}\setminus E}f_{l}\,d\mu\right\vert  &  \leq\varepsilon.
\end{align*}
It follows that for any $n\geq k$,
\begin{align*}
\left\vert \int_{E}f_{n}\,d\mu\right\vert \leq &  \left\vert \int_{E}
f_{k}\,d\mu\right\vert +\left\vert \int_{E}f_{n}\,d\mu-\int_{E}f_{k}
\,d\mu\right\vert \\
\leq &  \int_{E}|f_{k}|\,d\mu+\left\vert \int_{E\cup E_{0}}f_{n}\,d\mu
-\int_{E\cup E_{0}}f_{k}\,d\mu\right\vert \\
&  +\left\vert \int_{E_{0}\setminus E}f_{n}\,d\mu-\int_{E_{0}\setminus E}
f_{k}\,d\mu\right\vert \\
\leq &  3\varepsilon.
\end{align*}
In particular, by replacing $E$ first with $E\cap\{f_{n}\geq0\}$ and then with
$E\cap\{f_{n}<0\}$ it follows that $\int_{E}(f_{n})^{+}\,d\mu\leq3\varepsilon$
and that $\int_{E}(f_{n})^{-}\,d\mu\leq3\varepsilon$. Hence, $\int_{E}
|f_{n}|\,d\mu\leq3\varepsilon$.
