$0\le f_n\le g_n\le h_n$ $f_n$ and $h_n$ are Lebesgue integrable then $g_n$ is also. I was trying to solve the following problem but unfortunately I'm unable to solve it. Please help me. Let $f_n,g_n,h_n$ be Lebesgue integrable functions with $0\le f_n\le g_n\le h_n$ for all $n$. Suppose that $f_n\rightarrow f,\ g_n\rightarrow g $ and $h_n\rightarrow h$ on $E$. Show that if the $\int_Ef_ndm\rightarrow \int_E fdm $ and $\int_Eh_ndm\rightarrow \int_E hdm $ then $$\int_Eg_ndm\rightarrow \int_E gdm $$

If $\int_E h dm < \infty$ then you can prove according to my comment above. 
Here is a counter-example if we are allowed to have $\int_E hdm = \infty$:
Define $E=[0,1]$. For $n \in \{1, 2, 3, ...\}$ define: 
\begin{align}
&f_n = 0 \\
&g_n = \left\{ \begin{array}{ll}
0 &\mbox{ , if $x \in [0, 1/(n+5)]$} \\
\frac{1}{\log(n+5)x}  & \mbox{ , otherwise} \\
\end{array}\right. \\
&h_n = \left\{ \begin{array}{ll}
0 &\mbox{ , if $x \in [0, 1/(n+5)]$} \\
\frac{1}{x}  & \mbox{ , otherwise} 
\end{array}
\right.\\
&h = \left\{ \begin{array}{ll}
0 &\mbox{ , if $x =0$} \\
\frac{1}{x}  & \mbox{ , otherwise} 
\end{array}
\right.
\end{align}
Then $0\leq g_n(x) \leq h_n(x)$ for all $x \in E$, all $g_n, h_n$ functions are integrable, and $h_n\rightarrow h$, $\int_E h_n \rightarrow \int_E h$, $g_n\rightarrow 0$.  But $\int_E g_n \rightarrow 1$. 
