Let $f$ be integrable over $E$.
(i) If $\{E_n\}$ is an ascending countable collection of measurable subsets of $E$, then $\int_{\cup E_n} f=\lim\limits_{n \to \infty} \int_{E_n}f $.
(ii) If $\{E_n\}$ is a descending countable collection of measurable subsets of $E$, then $\int_{\cap E_n} f = \lim\limits_{n \to \infty} \int_{E_n} f$.
Let $E_0=\emptyset$. Then let $F_n=E_n \setminus E_{n-1}$. Now we have $E=\cup E_n=\cup F_n$ and $F_n$'s are disjoint measurable sets. I wanted to apply the following theorem $\int_E f = \sum\limits_{n=1}^{\infty}\int_{F_n} f$. However, I'm stuck on incorporating the continuity of measure into a proof about the continuity of integration.
I'm stuck on proving part (1) as part (2) will follow from taking complements of things from part (1).
Best Answer
Suppose first that $f \geqslant 0$. Suppose we have $E_1 \subseteq E_2 \subseteq \ldots$ as in your question. Consider $\chi_{E_i} f$, we have $$\int_E \chi_{E_i} f = \int_{E_i} f$$(when I learned this stuff, this was the definition of integrating over a subspace). It's easy to see $\chi_{E_i} f \uparrow \chi_{\cup E_i} f$, so by Monotone Convergence we deduce $$\int_{\cup E_i} f = \int_E \chi_{\cup E_i} f = \lim_i \int_E \chi_{E_i} f = \lim_i \int_{E_i} f$$ For the general case, apply this argument to both $f_+, f_-$.