Exchange limit on bounds of Lebesgue integral Let $(E_n)_{n \in \mathbb{N}}$ be a sequence of measurable sets such that $\lim_{n \to \infty} E_n =E$ for some measurable set $E$. When does it hold that
$$ \lim_{n \to \infty}\int_{E_n}f d \mu = \int_{E}f d\mu$$
What if we replace $\lim$ with $\lim \sup$ or $\lim \inf$?
 A: Take the Lebesgue measured space $(\mathbb R, \mu)$, $E_n= (-\infty, 1/n)$ and
$$f(x)=\begin{cases}
0 & x \le 0\\
1/x & x>0
\end{cases}$$
Then $\lim\limits_{n \to \infty}E_n =(-\infty , 0]$. And
$$\infty = \lim_{n \to \infty}\int_{E_n}f d \mu \neq \int_{E}f d\mu =0$$
A: Let $f$ be an integrable function. i.e. $f\in L^1(\Bbb R)$. Then we have $|f\chi_{E_n}|\le |f|$ for all $n\in\Bbb N$, hence by Dominated Convergence Theorem we have
$$\begin{align}
\int_Ef\,d\mu &= \int f\chi_E\,d\mu \\
&= \lim_{n\to\infty} \int f\chi_{E_n} \,d\mu \\
&= \lim_{n\to\infty} \int_{E_n} f\,d\mu
\end{align}$$
provided that $\chi_{E_n}\to\chi_E$ almost everywhere. 

The pointwise convergence a.e. of $\chi_{E_n}$ can be a result of letting $$E=\bigcup_n E_n, $$
  for example, or even a weaker hypothesis like $E=\liminf_{n\to\infty} E_n$, i.e.
  $$
E = \bigcup_{n=1}^\infty \bigcap_{k\ge n} E_k.
$$

If we only assume that $f\in L^1_{\text{loc}}(\Bbb R)$ then the previous convergent result still holds, provided that we assume further that $E,E_n$ are contained in some bounded set $B\subset \Bbb R$ (because we can apply the above argument by viewing $f\in L^1(B)$).
If $f$ is not even locally integrable, then we can find counterexamples so that $\int_Ef\,d\mu \ne \lim_{n\to\infty} \int_{E_n} f\,d\mu$.


Alternatively, we can consider the pseudo-distance function 
  $$
d(A,B) := \mu(A\Delta B)
$$
  defined on $\mathcal A$, the family of Lebesgue measurable sets of $\Bbb R$. Define the equivalent relation $A\sim B$ iff $d(A,B)=0$, then the set of equivalent classes $\mathcal A/\sim$ forms a (complete) metric space. We can now interpret $E = \lim_{n\to\infty} E_n$ as 
  $$
d(E,E_n) \to 0
$$
  in $\mathcal A/\sim$. 

Now, any $f\in L^1(\Bbb R)$ induces an absolutely continuous signed measure on $\Bbb R$. Hence for any $\varepsilon>0$ we can find a number $\delta>0$ such that $\mu(A)<\delta$ implies
$$
\int_A |f|\,d\mu < \varepsilon.
$$
Thus for sufficiently large $n$, we have
$$
\left| \int_E f\,d\mu - \int_{E_n} f\,d\mu \right|\le \int_{E\Delta E_n} |f|\,d\mu < \varepsilon
$$
since $\mu(E\Delta E_n)\to 0$. This shows that $\lim_{n \to \infty}\int_{E_n}f d \mu = \int_{E}f d\mu$.

Similarly, if we instead assume $f\in L^\infty (\Bbb R)$ then for sufficiently large $n$ we have
$$
\left| \int_E f\,d\mu - \int_{E_n} f\,d\mu \right|\le \int_{E\Delta E_n} |f|\,d\mu \le \mu(E\Delta E_n) ||f||_\infty \to 0
$$
as $d(E,E_n)\to 0$, which implies $\lim_{n \to \infty}\int_{E_n}f d \mu = \int_{E}f d\mu$ also.
