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.

Possible duplicate of Not sure how to start proving that $\int_E f,d\lambda$ = $\lim_{n \to \infty} \int_{E_n} f,d\lambda$ — amWhy, Feb 25 '18 at 00:09
How could you not know it was already asked, when you asked the question I link, and ask it again here?: NOT okay. — amWhy, Feb 25 '18 at 00:45
"Here is the following problem: Suppose that f is Lebesgue integrable over E and that ${E_{n}}{n=1}^{\infty}$ is a sequence of lebesgue measurable sets with $E_1$ $\subset$ $E_2$ $\subset$.... and $\cup{n=1}^{\infty} (E_{n})$ = E. Prove that
$\int_E f,d\lambda$ = $\lim_{n \to \infty} \int_{E_n} f,d\lambda$.

Any suggestions/ hints on how to start this problem would be appreciated." Asked already by you here — amWhy, Feb 25 '18 at 00:48
Hold up....these are two different questions. I simply thought you were referring to a different question (that was similar to this question) asked by someone else. — FoxViking, Feb 25 '18 at 00:49
And I don't understand why my answer got downvote, is there mistake? So far I don't see there is. — user284331, Feb 25 '18 at 01:57

score 5 · Accepted Answer · answered Feb 24 '18 at 23:24

5

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.

answered Feb 24 '18 at 23:24

user284331

55,591

$f\chi_{E_{n}}\uparrow f\chi_{\bigcup_{n=1}^{\infty}E_{n}}$ follows since f is a nonnegative Lebesgue measurable function and ${E_n}_{n=1}^{\infty}$ is a monotone nondecreasing sequence of Lebesgue measurable sets? – FoxViking Feb 24 '18 at 23:52
Yes, you are right. – user284331 Feb 24 '18 at 23:54
Alright. Thanks! – FoxViking Feb 24 '18 at 23:56

score 0 · Answer 2 · answered Aug 21 '21 at 09:43

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

$$\lim f1_{E_n} = f1_E$$
$$f1_{E_n} \le f1_{E_{n+1}}$$

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

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

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

2 Answers2