Suppose $\{f_n\} \subset L^+$ , $f_n \rightarrow f$ pointwise, and $\int f = \lim \int f_n < \infty$. Then $\int_{E} f = \lim \int_{E} f_n$ for all $E \in \mathcal{M}$. However, this needn't be true if $\int f = \lim \int f_n = \infty$.
This is the exercise 13 on page 52, Real Analysis, by Gerald B. Folland. $L^+$ is the space of all measurable functions from $(X,\mathcal{M}, \mu)$ to $[0, \infty]$.
I can find a counterexample satisfying $\int f = \lim \int f_n = \infty$. Let $X = \Bbb N$, $\mathcal{M} = \mathcal{P}(\Bbb N)$ and $\mu$ be the counting measure. $f_n = \chi_{\{2i+1:i \leq n\} \cup \{2j:n \leq j \leq 2n \}}$. Then $\int f = \lim \int f_n = \infty$. But when $E = \{2k: k \in \Bbb N\}$, Then $\int_{E} f = 0$ and $\lim \int_{E} f_n = \infty$.
But how to show the part "$f_n \rightarrow f$ pointwise, and $\int f = \lim \int f_n < \infty$. Then $\int_{E} f = \lim \int_{E} f_n$ for all $E \in \mathcal{M}$"?
Best Answer
By Fatou's lemma, you have
$$\begin{align} \infty &> \lim_{n\to\infty} \int f_n\\ &= \int f\\ &= \int_E f + \int_{X\setminus E} f\\ &\leqslant \liminf_{n\to\infty} \int_E f_n + \liminf_{n\to\infty} \int_{X\setminus E} f_n\\ &\leqslant \liminf_{n\to\infty} \left(\int_E f_n + \int_{X\setminus E} f_n\right)\\ &= \liminf_{n\to\infty} \int f_n\\ &= \lim_{n\to\infty} \int f_n, \end{align}$$
so the inequalities here are actually equalities, and
$$\int_E f = \liminf_{n\to\infty} \int_E f_n$$
for all $E\in\mathcal{M}$. Now if we had $\limsup \int_E f_n > \int_E f$ for some $E\in\mathcal{M}$, selecting a subsequence with $\lim_{k\to\infty} \int_E f_{n_k} > \int_E f$ would for that subsequence force
$$\lim_{k\to\infty} \int_{X\setminus E} f_{n_k} = \int f - \lim_{k\to\infty} \int_E f_{n_k} < \int f - \int_E f = \int_{X\setminus E} f,$$
contradicting Fatou's lemma.