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let $(X,\mathcal{F}, \mu)$ a finite measure space and $\{f_n\}_1^\infty$ are integrable functions which suffice: For any $\varepsilon > 0$ exists $\delta > 0$ such that for all $A\in \mathcal{F}$, in case $\mu(A) < \delta$ $\Rightarrow$ $|\int_{A} f_n d\mu | < \varepsilon$ $\forall n\in \mathbb{N}$. In addition, $\lim_{n\rightarrow \infty} f_n =^{\mu.almost.surely} f$.

I need to show that $f$ is integrable function and $\lim_{n\rightarrow\infty}\int f_n d\mu = \int fd\mu$

I was trying to use the next answer: If $\mu$ is finite, then $\{f_n\}$ is uniformly integrable iff $\sup_n \int|f_n| d\mu<\infty$ and $\{f_n\}$ is uniformly absolutely continuous.

But I know nothing about $\{f_n\}$ boundedness.

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    $\begingroup$ Did you mean integrals of $f_n$ over the set $A$? Did you mean for all $n$? $\endgroup$ – uniquesolution Jan 12 '18 at 18:05
  • $\begingroup$ You are right, I edited the question. @uniquesolution $\endgroup$ – dan Jan 12 '18 at 18:09
  • $\begingroup$ Is it $\displaystyle\int_{A}|f_{n}|d\mu<\epsilon$ or $\left|\displaystyle\int_{A}f_{n}d\mu\right|<\epsilon$? $\endgroup$ – user284331 Jan 12 '18 at 19:13
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    $\begingroup$ "In addition, $\lim_{n\rightarrow \infty} f_n = f$." I assume you mean point wise. You should say that. $\endgroup$ – zhw. Jan 12 '18 at 20:16
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    $\begingroup$ are you familiar with egoroff's theorem? $\endgroup$ – user363464 Jan 12 '18 at 20:46
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We first observe that if $\left|\displaystyle\int_{A}f_{n}d\mu\right|<\epsilon$ for all $n$ and measurable $A$ with $\mu(A)<\delta$, then $\left|\displaystyle\int_{A\cap\{f_{n}\geq 0\}}f_{n}d\mu\right|<\epsilon$ and $\left|\displaystyle\int_{A\cap\{f_{n}<0\}}f_{n}d\mu\right|<\epsilon$. But $\left|\displaystyle\int_{A\cap\{f_{n}\geq 0\}}f_{n}d\mu\right|=\displaystyle\int_{A\cap\{f_{n}\geq 0\}}f_{n}d\mu$ and $\left|\displaystyle\int_{A\cap\{f_{n}<0\}}f_{n}d\mu\right|=-\displaystyle\int_{A\cap\{f_{n}< 0\}}f_{n}d\mu$, and hence \begin{align*} \displaystyle\int_{A}|f_{n}|d\mu&=\displaystyle\int_{A\cap\{f_{n}\geq 0\}}|f_{n}|d\mu+\displaystyle\int_{A\cap\{f_{n}<0\}}|f_{n}|d\mu\\ &=\displaystyle\int_{A\cap\{f_{n}\geq 0\}}f_{n}d\mu-\displaystyle\int_{A\cap\{f_{n}<0\}}f_{n}d\mu\\ &<2\epsilon. \end{align*}

Now by Egorov Theorem, for that $\delta>0$, choose a measurable set $B$ such that $f_{n}\rightarrow f$ uniformly on $B$ and $A:=X-B$, $\mu(A)<\delta$. By writing that \begin{align*} \int_{B}|f|d\mu\leq\int_{B}|f_{n}-f|d\mu+\int_{B}|f_{n}|d\mu, \end{align*} it is easy to see that $\displaystyle\int_{B}|f|d\mu<\infty$.

On the other hand, we have \begin{align*} \int_{A}|f|d\mu&=\int_{A}\liminf_{n}|f_{n}|d\mu\\ &\leq\liminf_{n}\int_{A}|f_{n}|d\mu\\ &\leq 2\epsilon, \end{align*} so \begin{align*} \int_{X}|f|d\mu=\int_{B}|f|d\mu+\int_{A}|f|d\mu\leq\int_{B}|f|d\mu+2\epsilon<\infty. \end{align*}

For the part $\lim_{n}\displaystyle\int_{X}f_{n}d\mu=\int_{X}fd\mu$:

One controls the limit of integrals on $B$ by uniform convergence. Now use the assumption regarding the integrals of $f_{n}$ on $A$ and the fact that $f$ is integrable and hence absolutely continuous: $\left|\displaystyle\int_{A}fd\mu\right|<\epsilon$ for any $A\in\mathcal{F}$ with $\mu(A)<\delta$: \begin{align*} \left|\int_{A}f_{n}d\mu-\int_{A}fd\mu\right|&\leq\left|\int_{A}f_{n}d\mu\right|+\left|\int_{A}fd\mu\right|\\ &<\epsilon+\epsilon\\ &=2\epsilon. \end{align*}

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  • $\begingroup$ Could you please give some more details for showing that $\lim_n \int_A f_n d\mu = \int_A fd\mu $ I was trying to bound $f_n$ over this set and use DCT but couldn't make it. $\endgroup$ – dan Jan 13 '18 at 15:30
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    $\begingroup$ The set $A$ is small, and so just consult to the assumption to the integrals of $f_{n}$ as stated in the question. $\endgroup$ – user284331 Jan 13 '18 at 17:33
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We need to assume $f$ is finite $\mu$-a.e. Otherwise, set $X = \{1\}$, $\mathcal{F} = 2^{X}$, let $\mu$ be the counting measure on $X$, and set $f_{n}(1) = n$. If $\epsilon > 0$, then picking $\delta < 1$ makes the uniform integrability statement vacuous. Meanwhile, $\lim_{n \to \infty} f_{n}(1) = \infty = f(1)$ everywhere.

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