Exchange limit on bounds of Lebesgue integral

lebesgue-integralmeasure-theory

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 $\lim \inf$?

Best Answer

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.

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