Need help proving that $\int_{\bigcup_{n=1}^{\infty}E_n}f{\rm d}\lambda = \lim_{n\to\infty}\int_{E_n}f{\rm d}\lambda$ Here is the problem:  If $f$ is a nonnegative Lebesgue measurable function and $\{E_n\}_{n=1}^{\infty}$ is a sequence of Lebesgue measurable sets with $E_1\subset E_2\subset ...$, then 
$\int_{\bigcup_{n=1}^{\infty}E_n}f{\rm d}\lambda = \lim_{n\to\infty}\int_{E_n}f{\rm d}\lambda$.
I know that $\int_{\bigcup \limits_{n=1}^{\infty} E_{n}} f \,d\lambda = \int \limits_{\Bbb R} f \chi_{\bigcup \limits_{n=1}^{\infty} E_{n}} \,d\lambda$.
I'm also thinking that I should apply the Monotone Convergence Theorem.
However, I don't know where to go from here.
 Any help would be appreciated.
 A: Write $\displaystyle\int_{\bigcup_{n=1}^{\infty}E_{n}}fd\lambda=\int_{X}\chi_{\bigcup_{n=1}^{\infty}E_{n}}fd\lambda$ and $\displaystyle\int_{E_{n}}fd\lambda=\int_{X}\chi_{E_{n}}fd\lambda$ and now observe that $f\chi_{E_{n}}\uparrow f\chi_{\bigcup_{n=1}^{\infty}E_{n}}$ so Monotone Convergence Theorem applies here. 
A: What I like to do is think of this in terms of indicator/characteristic functions like here.
I guess the measure space here is $(\Omega, \mathscr F, \lambda)$ with $\Omega = \mathbb R$ and $\mathscr F$ is Lebesgue-measurable sets (rather than Borel sets or something).
Let $E = \bigcup_{n=1}^{\infty} E_n$. Then

*

*$$\int_{\Omega} f1_{E_n}\,d\lambda = \int_{E_n} f\,d\lambda$$


*$$\int_{\Omega} f 1_{E}\,d\lambda = \int_{E} f \,d\lambda$$
Now prove that


*$$\lim f1_{E_n} = f1_E$$


*$$f1_{E_n} \le f1_{E_{n+1}}$$
Next


*By (3) and (4), monotone convergence theorem applies to let us say that

$$\lim \int_{\Omega} f1_{E_n}\,d\lambda = \int_{\Omega} \lim  f1_{E_n}\,d\lambda \tag{1}$$
