We are given that
$ f_n \rightarrow f, g_n \rightarrow g, g_n \in L^1, g \in L^1, |f_n| \leq g_n, \int_X g_n \rightarrow \int g $
We are to show that $ \int_X f_n \rightarrow \int_X f $.
We note that $ |f_n| \leq g_n $ implies that $ g_n + f_n \geq 0, g_n - f_n \geq 0 $.
By an application of Fatou,
$ \liminf_n \int_X (g_n - f_n) \geq \int_X \liminf_n (g_n - f_n) = \int_X (g-f) $,
and since $ g_n \in L^1 \Rightarrow f_n \in L^1, \liminf_n \int_X (g_n - f_n) = \liminf_n (\int_X g_n - \int_X f_n) = \int_X g \ - \ \limsup_n \int_X f_n $.
Now, since g is integrable, we can subtract $ \int_X g $ from both sides (this is why the assumption that $ g \in L^1 $ is necessary), yielding
$ \limsup_n \int_X f_n \leq \int_X f $
Another application of Fatou yields
$ \liminf_n \int_X (g_n + f_n) \geq \int_X \liminf_n (g_n + f_n) = \int_X (g+f) $, yielding
$ \liminf_n \int_X f_n \geq \int_X f $
Combining inequalities gives the result. We remark that the integrability requirement on g is necessary, as
$ g_n = g = 1, f_n = \chi_{[n,\infty)} $ shows.
Note: In the above, we have used (somewhat flippantly) the result
$ - \infty < \lim_n a_n = a < \infty \Rightarrow \liminf_n (a_n+b_n) = \liminf_n a_n + \liminf_n b_n $
The proof is not difficult, since any subsequence of the $ \left\{ a_n \right\} $ converges to a, but equality is not true in general.
