$\lim\int_Mf_nd\mu=\int_Mfd\mu\implies\lim\int_Af_nd\mu=\int_Afd\mu$ When $A\subset M$ [duplicate]

Solution 1:

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 $\int f_{n}d\mu\rightarrow\int fd\mu$.