From the whole space to sunsets: how to prove this integral limit?

Question

Given $(X,\mu)$ a measure space, $\{f_n\}$ is a sequence of measurable and non-negative functions defined on $X$ such that $f_n\to f$ pointwise and $$\int_X f\mathrm d \mu=\lim_{n\to\infty}\int_X f_n\mathrm d\mu.$$ Prove that for all measurable subset $E$, $$\int_E f\mathrm d \mu=\lim_{n\to\infty}\int_E f_n\mathrm d\mu.$$

Tools available to me: Lebesgue monotone/dominated convergence theorem and Fatou's lemma, and I believe it isn't convenient to apply dominated convergence theorem here.

If I should start with simple function case, then should I let $f$ or $f_n$ or both be simple? Unfortunately, all these tricks seem to mess up.

I don't think this problem is hard. But anyhow I'm struggling with it for more than an hour and have hardly made any progress (forgive my stupidity, but it just happens every now and then). So, frankly, I need help. Maybe even a tiny hint can shed enough light on it. Thanks in advance.

Answer 1

Tip: Using \, inbetween $f$ and $d\mu$ enhances visual effect.

By Fatou's Lemma, $\int_E f\,d\mu \leq \liminf_{n \to \infty} \int_E f_n\,d\mu$, and likewise for $E^c$. Observe that $$\begin{align*} \int f\,d\mu - \int_E f\,d\mu &= \int_{E^c} f\,d\mu \leq \liminf_{n \to \infty} \int_{E^c} f_n\,d\mu = \liminf_{n \to \infty} \left(\int f_n\,d\mu - \int_E f_n\,d\mu\right) \\ &= \int f\,d\mu - \limsup_{n \to \infty} \int_E f_n\,d\mu \end{align*}$$ Therefore $\limsup_{n \to \infty} \int_E f_n\,d\mu \leq \int_E f\,d\mu \leq \liminf_{n \to \infty} \int_E f_n\,d\mu$, as desired. Also, the above equation makes sense only if $\int f\,d\mu < \infty$.

Answer 2

The claim is not necessarily true if $\int_Xf=\infty$. For example, let $X=\mathbb{R}$ with Lebesgue measure, and $f_n(x)=n^2\chi_{(0,\frac{1}{n}]}(x)+\chi_{[1,n]}(x)$. Then $f_n\to f$ pointwise, where $f(x)=0$ if $x<1$, and $f(x)=1$ for $x\geq 1$.

Also, $$ \int_{\mathbb{R}}f(x)\;dx=2n-1\to\infty=\int_{\mathbb{R}}f(x)\;dx $$ but if $E=[0,1]$ then $$ \int_Ef_n(x)\;dx=n\to\infty$$ while $$ \int_Ef(x)\;dx=0 $$