TITLE: Prove that $\int_{E}f =\lim \int_{E}f_{n}$
QUESTION [9 upvotes]: I'm doing exercise in Real Analysis of Folland, and got stuck on this problem. I try to use Fatou lemma but can't come to the conclusion. Can anyone help me. I really appreciate.


Consider a measure space $(X, M, \mu)$.Suppose $\{f_{n}\} \in L^{+}, f_{n} \rightarrow f$ pointwise and $\int f = \lim \int f_{n} < \infty$. Then $\int_{E} f = \lim \int_{E} f_{n}$ for all $E \in M$. However, this need not be true if $\int f = \lim \int f_{n} = \infty$


Thanks so much.

REPLY [18 votes]: By Fatou's lemma, we have
$$\int_E f \, d\mu \leq \liminf_{n \to \infty} \int_E f_n \, d\mu . \tag{1}$$
Similarly,
$$\int_{X \backslash E} f\, d\mu  \leq \liminf_{n \to \infty} \int_{X \backslash E} f_n\, d\mu . \tag{2}$$
Since $\int_X f\, d\mu = \lim_{n \to \infty} \int_X f_n \, d\mu $, we get
$$\begin{align*} \int_{E} f\, d\mu  &= \int_X f \, d\mu - \int_{X \backslash E} f\, d\mu  \\ &\stackrel{(2)}{\geq} \lim_{n \to \infty} \int_X f_n \, d\mu - \liminf_{n \to \infty} \int_{X \backslash E} f_n \, d\mu \\ &= \limsup_{n \to \infty} \left( \int_X f_n \, d\mu - \int_{X \backslash E} f_n \, d\mu \right) = \limsup_{n \to \infty} \int_E f_n\, d\mu .  \tag{3}\end{align*}$$
Combining $(1)$ and $(3)$ shows
$$\int_E f \, d\mu \leq \liminf_{n \to \infty} \int_E f_n \, d\mu \leq \limsup_{n \to \infty} \int_E f_n\, d\mu  \leq \int_E f\, d\mu .$$
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