TITLE: Prove that $\lim\limits_{n\to\infty} \int\limits_X f_nd\mu=\int\limits_X fd\mu$
QUESTION [7 upvotes]: $(X,\mathscr{M},\mu)$ is a measurable space and $f_n,g_n,f,g$ are all measurable functions defined on $X$. The following conditions are satisfied:
(i) $f_n\to f\;\&\;g_n\to g$ almost everywhere on $X$
(ii) $|f_n|\leq g_n$ and $\int\limits_X g_nd\mu<\infty\;\forall\; n\geq1$
(iii) $\lim\limits_{n\to\infty}\int\limits_X g_nd\mu=\int\limits_X gd\mu<\infty$
Prove $\lim\limits_{n\to\infty}\int\limits_X f_nd\mu=\int\limits_X fd\mu$.
Attempt at answer:
I know that since $f_n,g_n,f,g$ are all measurable, then $|f_n|,|g_n|,|f|,|g|$ are all measurable. Also that $f_n\to f\;\&\;g_n\to g$ almost everywhere on $X$ implies $f_n\to f\;\&\;g_n\to g$ almost uniform convergence on $X$ implies $f_n\to f\;\&\;g_n\to g$ in measure on $X$.
It's almost like "Lebesgue's Dominated Convergence Thm", but I have "almost everyhwere" convergence. Does that make a difference, or am I just making up differences where there are none? Any help/advice would be remarkably useful.

REPLY [2 votes]: Hint: use Fatou's lemma with the sequences $(g_n-f_n,n\geqslant 1)$ and $(g_n+f_n,n\geqslant 1)$.
Note that when $g_n=g$ for all $n$, it's exactly the dominated convergence theorem. This gives a generalization.