Question

Real analysis, ill ask in the comments.

Answer 1

Suppose that \(\{A_n\}_n^\infty \subset \mathcal{M}\), where \(\mathcal{M}\) is the \(\sigma-\)algebra of Lebegue measurable sets. Also suppose \(....A_3\subset A_2\subset A_1\).
Prove that \(\lambda\large(\cap_{n=1}^\infty A_n)\) = \(\lim_{n\rightarrow \infty}(A_n)\)

Answer 2

this is what I know, and I think I should use.
given the same conditions, but with \(A_1\subset A_2\subset A_3...\), then
\[\lambda\large(\cup_{n=1}^{\infty}A_n)=\lim_{n\rightarrow\infty}\lambda(A_n)\]

Answer 3

that should say
prove that \(\lambda\large(\cap_{n=1}^\infty A_n)\) = \(\lim_{n\rightarrow \infty}\lambda(A_n)\)

Answer 4

I feel that DeMorgan's law is in there some place, but I'm not sure hot to apply it.

Answer 5

The statement is not true in general. Take this example
\[
A_n = [ n, +\infty[\\
then \\
A_{n+1} \subset A_n\\
A=\cap_n A_n =\emptyset\\
\lambda (A)=0 < \lim_{n\to} \lambda\left ( A_n \right)=\infty


\]

Answer 6

It is true however if you suppose that one of your decreasing sets has finite measure.

Answer 7

\[
\lim_{n\to \infty}\lambda\left ( A_n \right)=\infty
\]
How do you get this?

Answer 8

So \(\lambda([\infty,\infty]) = \infty\)?

Answer 9

Yes

Answer 10

the measure of each of the sets is infinite

Answer 11

Wouldn't \((\infty,\infty)=\emptyset\)?

Answer 12

For the case of the finite case. We can suppose that all of them have finite measure. You can read page 34 on
http://www.ams.org/bookstore/pspdf/stml-48-prev.pdf
Where all the sets are inside a finite interval.

Answer 13

Actually for each n on my example
\[
\lambda(A_n) = +\infty
\]
Hence their limit is
\[
+\infty
\]

Answer 14

May I ask how exactly you get \[
\lambda(A_n) = \infty
\]Even in the case where \(n\) is finite? What are you using?

Answer 15

How much do you think \[ \lambda (A_n) \] is?

Answer 16

Well, my best guess is \[
\lim_{m\to \infty}m-n
\]based on the resource you gave me.

Answer 17

For each integer k, we have
\[
[ n, n+k] \subset A_n\\
k=\lambda([n,n+k]) \le \lambda(A_n)\\
\]
Would this argument satisfy you?

Answer 18

What is your background in Math @wio?

Answer 19

I'm not unsatisfied, I just want to see a more clear definition.

Answer 20

If a quantity is bigger than any number k given in advance, what could this quantity be?

Answer 21

Have you taken a real analysis course about Lebesgue integration?

Answer 22

No, I just think that \(\infty-\infty\) is involved here, so it is tricky.

Answer 23

ok sorry it said \(\lambda(A_1)<\infty\)

Answer 24

sorry, its 4am

Answer 25

In this case, the reference I gave you above on page 34 gives you the proof.

Answer 26

For the case of the finite case. We can suppose that all of them have finite measure. You can read page 34 on http://www.ams.org/bookstore/pspdf/stml-48-prev.pdf Where all the sets are inside a finite interval.

Answer 27

ahh great, ty.

Answer 28

Suppose:\[
\lim_{y\to\infty}f(x,y) =\infty
\]Does this mean: \[
\lim_{x\to\infty}\lim_{y\to\infty}f(x,y) =\infty
\]Or does it depend on \(f\)? What if \(f(x,y) = y-x\)?

Answer 29

I find this question somewhat interesting, even if not completely relevant.

Answer 30

YW

Answer 31

hey @eliassaab, on the first line for the decreasing proof, where do they get the 1 - from?

Answer 32

I think in their proof, the measure of the whole interval is 1. To do your proof
replace 1 by the measure of your first set A_1 and take the compleent with respect to it/

Answer 33

ahh, i see.

Answer 34

very strange, they are saying the compliment is 1-... and they do that all over that book, but I never see them say we are dealing with some special interval. Maybe they assume it up front at some point in the text.

Answer 35

its like they use it as a place holder for the compliment....

Answer 36

You should be using \(|U| - |A|\) I suppose?

Answer 37

well for me A_1^c will be R/A_1, which has infinite measure...

Answer 38

maybe i need to sleep, its 430:)

Answer 39

gn

Answer 40

ill read anything you guys say in the morning.

Answer 41

But you may be able to use \(|U|-|U|=0\) at some point.

Answer 42

@eliassaab do you see where it says they are using some interval with length 1? I dont see them say that anywhere so I cant see why they use 1.

Answer 43

when you said
"I think in their proof, the measure of the whole interval is 1. To do your proof replace 1 by the measure of your first set A_1 and take the compleent with respect to it/"
why would I use my A_1 instead of 1, and not R-union....

Answer 44

Take
\[
X=A_1\\
E_n= X \backslash A_n\\
X\backslash \cap A_n= \cup E_n\\
\lambda(X\backslash \cap A_n)= \lambda(\cup E_n)\\
\lambda(X) -\lambda(\cap A_n))=\lim_n \lambda(E_n)=\lim_n( \lambda(X) -\lambda(A_n))\\

\lambda(X) -\lambda(\cap A_n))=\lambda(X) -\lim_n\lambda(A_n)\\

-\lambda(\cap A_n))= -\lim_n\lambda(A_n)\\
\lambda(\cap A_n))=\lim_n\lambda(A_n)\\

\]

Answer 45

ok ill think about this one, ty

Answer 46

To be able to cancel \(\lambda(X)\) we need it to be finite.

Answer 47

yw

Answer 48

I also used the fact that \(E_n\) is increasing for inclusion.

Answer 49

where did you use that? sorry this is all new to me.

Answer 50

The fact that \( E_n\) is increasing

Answer 51

also also \(\lim_{n\rightarrow\infty}(\lambda(X)-\lambda(A_n))=\lambda(X)-\lim_{n\rightarrow \infty}\lambda(A_n)\) can we say this because lambda(X) does not depend on n?

Answer 52

yes

Answer 53

and finally, \[\lambda(A-B)=\lambda(A)-\lambda(B)\]
I imagine this is true for all l-measurable sets, and that its easy to prove?

Answer 54

nm i answered that last question

Answer 55

\[
X=A_1\\
E_n= X \backslash A_n\\
X\backslash \cap A_n= \cup E_n\\
\lambda(X\backslash \cap A_n)= \lambda(\cup E_n)\\
\text{Simce } E_n \\
\text { is increasing, The measure of their union is the limit of their
individual measures}
\\
\lambda(X) -\lambda(\cap A_n))=\lim_n \lambda(E_n)=\lim_n( \lambda(X) -\lambda(A_n)\quad \text { (since X has finite measure}
)\\

\lambda(X) -\lambda(\cap A_n))=\lambda(X) -\lim_n\lambda(A_n)
\\\ \text { (since X has finite measure we can cancel and get}
\\

-\lambda(\cap A_n))= -\lim_n\lambda(A_n)\\
\lambda(\cap A_n))=\lim_n\lambda(A_n)\\



\]

Answer 56

Are you now happy with the explanation?

Answer 57

i think i understand everything now:)

Answer 58

Excellent

Answer 59

ty, sorry im a little slow to this. we just got introduced to measure about 6 days ago

Answer 60

You are fine. This is normal. When you get used to it, it would become easy.

Answer 61

i hope so:) the cantor set blew my mind.