TITLE: integral of a measure zero set
QUESTION [5 upvotes]: 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 $X \backslash F_\sigma$?

REPLY [2 votes]: 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 $n$.