integral of a measure zero set let $X$ be a finite measure space and $\{f_n\}$ be a sequence of nonnegative integrable functions, $f_n \rightarrow f\ a.e.$ on $ X$.
We know that 
$\lim_{n \rightarrow \infty}\int_X f_n d\mu=\int_X fd\mu$ and on any measurable $E_i \subset X$
I should apply Egoroff's theorem to conclude that $\lim_{n \rightarrow \infty}\int_X |f_n-f|d\mu=0$.
My attempt:
I broke the set $X$ to two sets: $F_\sigma$ on which $f_n \rightarrow f$ uniformly based on Egoroff's theorem and $X\backslash F_\sigma$ which is a very small set, i.e. $\mu\{X\backslash F_\sigma\}=\sigma$ and $f_n \nrightarrow f$
I want to show that on each of these sets, the integral is less than $\frac{\epsilon}{2}\ \forall \epsilon$ to finish. 
How can I show this for the set $X \backslash F_\sigma$?
 A: I assume $f$ integrable, otherwise it's not necessarily true, as copper.hat shows.
Fix an integer $k$, and $E_k$ measurable such that $\mu(X\setminus E_k)\leq k^{-1}$ and $\sup_{x\in E_k}|f_n(x)-f(x)|\to 0$. We have, using the hypothesis, 
\begin{align}
\int_X|f-f_n|d\mu&=\int_{E_k}|f-f_n|d\mu+\int_{X\setminus E_k}|f-f_n|d\mu\\
&\leq \mu(E_k)\sup_{x\in E_k}|f(x)-f_n(x)|
+\int_{X\setminus E_k}|f-f_n|d\mu\\
&\leq \mu(X)\sup_{x\in E_k}|f(x)-f_n(x)|
+\int_{X\setminus E_k}|f-f_n|d\mu.
\end{align}
Taking $\limsup_{n\to +\infty}$, we get 
$$\limsup_{n\to +\infty}\int_X|f-f_n|d\mu\leq \limsup_{n\to +\infty}\int_{X\setminus E_k}\left(|f-f_n|-(|f_n|-|f|)\right)d\mu,$$
using the fact that $\int_{X\setminus E_k}f_nd\mu\to \int_{X\setminus E_k}fd\mu$. 
As $0\leq |f-f_n|-(|f_n|-|f|)\leq 2f$, we have for each $k$,
$$\limsup_{n\to +\infty}\int_X|f-f_n|d\mu\leq \int_{X\setminus E_k}fd\mu.$$
Let $s_l$ a simple function such that $0\leq s_l\leq f$ and $\int_X (f-s_l)d\mu\leq l^{—1}$. We can write $s_l(x)=\sum_{j=1}^{N_l}a_{j,l}\chi_{B_{j,l}}$, where $B_{j,l}$ are measurable sets and $a_{j,l}$ positive real numbers. Then for each $k,l$ positive integers,
\begin{align}
\limsup_{n\to +\infty}\int_X|f-f_n|d\mu&\leq\int_{X\setminus E_k}(f-s_l)d\mu+
\int_{X\setminus E_k}s_ld\mu\\
&\leq \int_X(f-s_l)d\mu+\sum_{j=1}^{N_l}a_{j,l}\mu(B_{j,l}\cap E_k^c)\\
&\leq l^{-1}+\sum_{j=1}^{N_l}a_{j,l}\mu(B_{j,l}\cap E_k^c)\\
&\leq l^{-1}+k^{-1}\sum_{j=1}^{N_l}a_{j,l}\mu(B_{j,l}).
\end{align}
Taking the $\limsup_{k\to +\infty}$ then $\limsup_{l\to +\infty}$, we get the result.
A: I do not think this is true unless you assume $f$ is integrable.
Take $X=[0,1]$, Lebesgue measure, $f(x) = \frac{1}{x}$, $f_n = f \cdot 1 _{[\frac{1}{n},1]}$. Then $f_n(x) \to f(x)$ on $(0,1]$, $\int f_n = \log n$, hence $f_n$ are integrable, and $\int_E f_n \to \int_E f$ for all $E$ measurable. However, $\int |f_n-f| = \infty$ for all $n$.
