Prove $\lim_{n\to\infty}\int_{E_n}f(x)d\mu=\int_E f(x)d\mu\,\,\,\,\,\,\text{ where }E=\cup_{n=1}^\infty E_n$

(a) Let $$E_1\subset E_2\subset ... \subset E_n\subset ... \subset I_0=:[a,b]$$

be a sequence of measurable sets and let $f:[a,b]\to\mathbb{R}$ be a non-negative integrable function. Show that $$\lim_{n\to\infty}\int_{E_n}f(x)d\mu=\int_E f(x)d\mu\,\,\,\,\,\,\text{ where }E=\cup_{n=1}^\infty E_n$$

(b) Is the above statement true without the hypothesis "$f$ is non-negative"? Justify your answer.

By Levi's theorem I know we can deal with non-decreasing measurable functions, but I don't know how to deal with this.

Any help would be greatly appreciated. Thanks in advance.

The sequence of functions $f_n(x) = f(x) I\{x \in E_n\}$ is a monotone sequence of functions that converges pointwise to $f(x)$. The result now follows from the monotone convergence theorem (you don't even need integrability).
If $f$ is integrable, then it is not necessary to assume nonnegativity. The statement then follows by a similar argument from the dominated convergence theorem.
If you drop both the assumption on the integrability and the nonnegativity of $f$, the statement is no longer correct. Consider the function $f(x) = \frac{1}{x}$ on $[-1, 1]$ with $E_n = [-1, -1/n] \cup \{0\} \cup [1/n, 1]$.
• By monotone convergence theorem we write $\lim_{n\to\infty}\int f_n d\mu=\int fd\mu\Rightarrow \lim_{n\to\infty}\int f I(x\in E_n) d\mu=\int fd\mu\Rightarrow \lim_{n\to\infty}\int_{E_n} f d\mu=\int_E fd\mu$ Is it right?