$\lim_{n\rightarrow\infty}\int f_n d\mu = \int fd\mu$ over finite measure space with conditions which are similar to uniformly integrability 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.
 A: 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.
A: 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*}
