$0\le f_n\le g_n\le h_n$ $f_n$ and $h_n$ are Lebesgue integrable then $g_n$ is also.

Question

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 $$\int_Eg_ndm\rightarrow \int_E gdm $$

Answer 1

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 $\int_E g_n \rightarrow 1$.