Let $f$ be integrable on $E$.
if $\{E_n\}_{n=1}^\infty$ is an ascending countable collection of measurable subsets of $E$, then:
$\int_{\bigcup_{n=1}^\infty E_n} f = lim_{n \rightarrow \infty} \int_{E_n} f$
Also, if if $\{E_n\}_{n=1}^\infty$ is a descending countable collection of measurable subsets of $E$, then:
$\int_{\bigcap_{n=1}^\infty E_n} f = lim_{n \rightarrow \infty} \int_{E_n} f$
I've been trying to prove this for awhile now and seem to have gotten stuck in tunnel vision. I would really appreciate a fresh perspective! Also, i'm pretty bad at this stuff, so the more details / perspectives the better! Thanks!
Best Answer
Assume that $f$ is integrable on $E$: \begin{align*} \int_{\bigcup_{n}E_{n}}f=\int f\chi_{\bigcup_{n}E_{n}}, \end{align*} and $f\chi_{E_{n}}\uparrow f\chi_{\bigcup_{n}E_{n}}$ and apply Monotone Convergence Theorem. Of course, you need to split $f=f^{+}-f^{-}$ and assuming first that $f$ is nonnegative.
$f\chi_{E_{n}}\downarrow f\chi_{\bigcap_{n}E_{n}}$ and use Lebesgue Dominated Convergence Theorem for the other one.
\begin{align*} \int_{\bigcup_{n}E_{n}}f&=\int_{\bigcup_{n}E_{n}}f^{+}-\int_{\bigcup_{n}E_{n}}f^{-}\\ &=\int f^{+}\chi_{\bigcup_{n}E_{n}}-\int f^{-}\chi_{\bigcup_{n}E_{n}}\\ &=\lim_{n}\int f^{+}\chi_{E_{n}}-\lim_{n}\int f^{-}\chi_{E_{n}}\\ &=\lim_{n}\int(f^{+}-f^{-})\chi_{E_{n}}\\ &=\lim_{n}\int_{E_{n}}f \end{align*}