Example of functions where limit and integral are interchangeable on $\mathbb R$ but not on a measurable subset Let $f_n : \mathbb R \to [0, \infty)$ be Lebesgue-measurable functions for each $n \in \mathbb N$, and suppose $f_n(x) \to f(x)$ for all $x \in \mathbb R$, and $\int_{\mathbb R} f_n d\mu \to \int_{\mathbb R} f d\mu$ as $n \to \infty$ (where $\mu$ is Lebesgue measure on $\mathbb R$). 
If $f$ is integrable, then you can use Fatou's Lemma to show that $\lim_{n \to \infty} \int_E f_n d\mu$ exists and is equal to $\int_E f d\mu$ for every measurable subset $E$ of $\mathbb R$. 
My question is, what if $\int_{\mathbb R} f d\mu = \infty$? The proof of the above result doesn't work anymore, but I'm having trouble finding an actual counterexample. Is there a sequence of functions where $\int_{\mathbb R}f_n d\mu \to \infty$ but on some measurable subset, one can't interchange the limit and integral?

Let
$$
f_n(x)=\chi_{[n,n+1]}(x)+\frac1x\,\chi_{[1/n,1]}(x),\quad n\ge1,
$$
where $\chi_A$ is the characteristic function of the set $A$.
Then $\lim_{n\to\infty}f_n(x)=\dfrac1x\,\chi_{(0,1]}(x)=f(x)$ and
$$
\lim_{n\to\infty}\int_{\Bbb R}f_n(x)\,dx=\lim_{n\to\infty}(1+\log n)=\int_{\Bbb R}f(x)\,dx=\infty.
$$
Now
$$
\int_1^\infty f(x)\,dx=0,\text{ but }\int_1^\infty f_(x)\,dx=1\quad\forall n\ge1.
$$
