Part a) of the question is as follows: "Suppose that $E \subset \mathbb{R}^d$ is a measurable set and that $f, f_n$ are measurable functions on $E$ satisfying $f_n \to f$ a. e. on $E$. Suppose that $\lim_{n \to \infty} \int_E f_n = \int_E f$. Prove that if $\int_E f < \infty$, then $\lim_{n \to \infty} \int_A f_n = \int_A f$ for every measurable $A \subset E$.
I proved that, but for part b) the question wants me to find a counterexample if $\int_E f = \infty$. I have no idea what such a function would look like. Does anyone have a good counterexample?
Okay, here is a simple counter example. Let us take $E = [0,2]$.
On the interval $[0,1)$, define $$f_n (x)= n\chi_{[0,\frac{1}{n}]}(x)\quad \text{ and }\quad f \equiv 0.$$
$f_n \rightarrow f$ pointwise a.e. and this is the classic example for the strict inequality in Fatou's Lemma $$1 = \lim_n \int_0^1 f_n > \int_0^1 f = 0$$
And now on the interval $[1,2]$ define both $f_n$ and $ f $ to be some crazy function $g$ such that $\int_1^2 g = \infty$.
Together, we have $f_n \rightarrow f$ pointwise a.e. and $\int_0^2 f = \int_1^2 g = \infty$. However take the measurable set $A = [0,1)$ we get $$1=\lim_n \int_0^1 f_n > \int_0^1 f = 0.$$