Question

If \(f\) is a nonnegative Lebesgue measurable function and \(\{E_n\}_n^\infty\) is a sequence of Lebesgue measurable sets with \(E_1\subset E_2.....\), then \[\large \int_{\cup_{n=1}^{\infty} E_n} f d\lambda = \lim_{n\rightarrow \infty}\int_{E_n} f d\lambda\]

Answer 1

@eliassaab please and thank you.

Answer 2

Please read this
http://en.wikipedia.org/wiki/Monotone_convergence_theorem

Answer 3

Using the above let
\[
E=\cup_{n=1}^\infty E_n
\\
f_n = f 1_{E_n}\\
f_E= f 1_E
\]
then
\[
f_n \le f_{n+1} \le \cdots \le f_E\\
\int_{E_n} f d\lambda =\int_Ef_n d\lambda

Now apply the above and you are done.

Answer 4

then \[ f_n \le f_{n+1} \le \cdots \le f_E\\ \int_{E_n} f d\lambda =\int_Ef_n d\lambda \]Now apply the above and you are done.

Answer 5

is \(f1_E\) the characteristic rep?

Answer 6

\( f 1_E\) is f on E and 0 outside E

Answer 7

Right

Answer 8

Sorry but can you walk.me through how you got from the inequality tho the integral

Answer 9

f_n is, covering pointwise?

Answer 10

Converging*

Answer 11

Also it it increasing

Answer 12

Did you read and understand this
http://en.wikipedia.org/wiki/Monotone_convergence_theorem

Answer 13

So we are using the pointwise version.of mct?

Answer 14

The mct is only for pointwise convergence

Answer 15

ok I had to get off my tablet, this place is impossible with a tablet...
Ok we are using the Lebesgue MCT. We have a \(f_n\) which converges to \(f1_E\) in a increasing manner, and \(f1_E\) is the same thing as \(f_E\), so by MCT we have that \(\lim_{n\rightarrow \infty}\int f_nd\lambda=\int f d\lambda\). Am I right so far?

Answer 16

Yes

Answer 17

so the part im confused about, is when do we drop the limit and where do my bounds come from?

Answer 18

I imagine they have something to do with each other.

Answer 19

What bounds are you referring to?

Answer 20

Actually
\[
\lim_{n\rightarrow \infty}\int f_nd\lambda=\int_E f d\lambda
\]

Answer 21

how do i get from \(\lim_{n\rightarrow \infty}\int f_nd\lambda=\int f d\lambda\)
to
\(\int_{E_n} f d\lambda =\int_Ef_n d\lambda\)

Answer 22

ok and we get the E_n on the left by the way you constructed f_n?

Answer 23

yes

Answer 24

ok I get it:) thank you so much, I'm going to go try and explain that to someone else, and that should solidify it:)

Answer 25

That is a good idea.

Answer 26

Is this correct?

Answer 27

\[
f_n \to f \quad \text {pointwise on E
}
\]

Answer 28

The rest is OK