# Dominated by convergent sequence of functions

Suppose, $f_n,g_n$ are measurable functions on a measure space $(\Omega,\mathcal{A},\mu)$, satisfying $|f_n|\le g_n\forall n$. Given, $f_n\xrightarrow{a.e.}f$, $g_n\xrightarrow{a.e.}g$ and $\int g_n\rightarrow \int g <\infty$

I have to show that $\int f_n \rightarrow \int f$

Using Fatou's lemma, I can show that $\liminf\int f_n = \int f$ but I am stuck with the $\limsup$ direction.

Can someone please point me in the right direction? Any help regarding this appreciated.

First note that for all sufficiently large $n$, $f_n$ are integrable. Consequently, $f$ is integrable.
Now $\left|f_n\right| \leq g_n$ implies that $-g_n \leq f_n \leq g_n$. By Fatou's lemma, \begin{align*} \int f d \mu - \int (-g) d\mu = & \int \lim_n(f_n - (-g_n)) d \mu \\ \leq &\liminf_n \int(f_n - (-g_n)) d \mu = \liminf_n \int f_n d\mu + \int g d\mu \end{align*} and \begin{align*} \int g d \mu - \int f d\mu = & \int \lim_n(g_n - f_n) d \mu \\ \leq &\liminf_n \int(g_n - f_n) d \mu = \int g d\mu - \limsup_n\int f_n d\mu \end{align*} Therefore $$\limsup_n \int f_n d\mu \leq \int f d \mu \leq \liminf_n \int f_n d\mu.$$
• Ah! I just realized that I've messed up $\liminf(-x_n)=-\limsup x_n$ in my solution. Nov 6 '15 at 3:42