$\lim\int_Mf_nd\mu=\int_Mfd\mu\implies\lim\int_Af_nd\mu=\int_Afd\mu$ When $A\subset M$ I am working on the following question that's got me stumped.
Let $A\subseteq M$ be measurable.
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of non-negative integrable functions such that $\lim_{n\to \infty}f_n=f $ a.e. in $M$, and $f$ is integrable.
Futhermore $\lim_{n\to \infty}\int_Mf_nd\lambda=\int_Mfd\lambda.$
Show: $\lim_{n\to \infty}\int_Af_nd\lambda=\int_Afd\lambda.$
My idea is to apply Fatou on $f_n\chi_A$ and $f_n\chi_{M \setminus A}$.
Any help or hints is greatly appreciated!
 A: You may apply the following generalized Dominated Convergence Theorem:
Let $(X,\mathcal{M},\mu)$ be a measure space. Let $(g_{n})$ be a
sequence non-negative integrable functions such that $g_{n}\rightarrow g$
(a.e.), $g$ is integrable, and $\int g_{n}d\mu\rightarrow\int gd\mu$.
Let $(f_{n})$ be a sequence of measurable functions such that $|f_{n}|\leq g_{n}$
and $f_{n}\rightarrow f$ (a.e.) for some measurable function $f$,
then $f$ is integrable and $\int f_{n}d\mu\rightarrow\int fd\mu$.
Proof: The proof is the same as that for Dominated Convergence Theorem.
We apply Fatou's lemma twice. By Fatou Lemma, $\int|f|d\mu\leq\liminf_{n}\int|f_{n}|d\mu\leq\liminf_{n}\int g_{n}d\mu=\int gd\mu<\infty$,
so $f$ is integrable.
Note that $-g_{n}\leq f_{n}\leq g_{n}$, so $g_{n}-f_{n}\geq0$ and
$g_{n}-f_{n}\rightarrow g-f$ (a.e.). By Fatou Lemma,
\begin{eqnarray*}
 &  & \int gd\mu-\int fd\mu\\
 &  & \int(g-f)d\mu\\
 & \leq & \liminf_{n}\int(g_{n}-f_{n})d\mu\\
 & = & \int gd\mu-\limsup_{n}\int f_{n}d\mu.
\end{eqnarray*}
Therefore, $\int fd\mu\geq\limsup_{n}\int f_{n}d\mu$. On the other
hand, $g_{n}+f_{n}\geq0$ and $g_{n}+f_{n}\rightarrow g+f$ (a.e.).
By Fatou Lemma again,
\begin{eqnarray*}
 &  & \int gd\mu+\int fd\mu\\
 & = & \int(g+f)d\mu\\
 & \leq & \liminf_{n}\int(g_{n}+f_{n})d\mu\\
 & = & \int gd\mu+\liminf_{n}\int f_{n}d\mu.
\end{eqnarray*}
Therefore, $\int fd\mu\leq\liminf_{n}\int f_{n}d\mu$. Combining,
we have $\liminf_{n}\int f_{n}d\mu=\limsup_{n}\int f_{n}d\mu=\int fd\mu$
and hence $\int f_{n}d\mu\rightarrow\int fd\mu$.

For your problem, the dominating functions are $f_{n}$ while the
functions being dominated are $f_{n}\chi_{A}$.
