Need help proving that $\int_{\bigcup_{n=1}^{\infty}E_n}f{\rm d}\lambda = \lim_{n\to\infty}\int_{E_n}f{\rm d}\lambda$

Question

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.

Answer 1

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.

Answer 2

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

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

2. $$\int_{\Omega} f 1_{E}\,d\lambda = \int_{E} f \,d\lambda$$

Now prove that

3. $$\lim f1_{E_n} = f1_E$$

4. $$f1_{E_n} \le f1_{E_{n+1}}$$

Next

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

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