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 $X \backslash F_\sigma$?

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I am missing something here. Take $f_n = n\cdot1_{[0,\frac{1}{n}]}$. Then $f_n(x) \to f(x) = 0$ a.e., but $\int f_n = 1$, and $\int f = 0$. –  copper.hat Oct 29 '12 at 6:06
@copper.hat The "We know" is an assumption, I presume. –  Alex Becker Oct 29 '12 at 6:07
@AlexBecker: Thanks, I missed that interpretation completely... –  copper.hat Oct 29 '12 at 6:08
+1 for the interesting problem. –  Davide Giraudo Oct 29 '12 at 10:06

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.

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Giraudo: Thanks, but I have some questions: 1- Why do you assume $f \geq 0$? since $f_n \geq 0, sup |f-f_n|$ can happen when $f \leq 0\ \& f_n \geq 0$ so is it correct to change $\limsup_{n\to +\infty}\int_X|f-f_n|d\mu\leq \int_{X\setminus E_k}fd\mu$ to $\limsup_{n\to +\infty}\int_X|f-f_n|d\mu\leq \int_{X\setminus E_k}2|f|d\mu$? –  Anita Oct 29 '12 at 12:15
$f\geq 0$ as $f_n(x)\geq 0$ and $f_n\to f$ almost everywhere. –  Davide Giraudo Oct 29 '12 at 12:16
The latter part of your proof can be simplified by noting that if $f$ is integrable (and $X$ has finite measure), then the measure $\nu A = \int_A f$ is absolutely continuous, from which it follows that $\lim_k \int_{E_k^C} f = 0$. –  copper.hat Oct 29 '12 at 16:24
@copper.hat Yes, you are right. But I was in the spirit to give a proof which only uses more basic facts about Lebesgue integral. –  Davide Giraudo Oct 29 '12 at 16:30
@DavideGiraudo: are you assuming $|f_n|\geq |f|$? –  Anita Oct 30 '12 at 16:07

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 $n$.

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I agree (for sake of completeness, you should say that $\int_Ef_n\to \int_Ef$ for all $E$ measurable). –  Davide Giraudo Oct 29 '12 at 16:32
Good suggestion! –  copper.hat Oct 29 '12 at 16:49