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hparty
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 double integral   « on: Jul 28th, 2010, 4:55pm » Quote Modify

given a function f:RxR---> R bounded and Lebesgue measurable, and satisfies

\int {f d(mxm)} =0  over every open disk D in RxR.
How to prove that f must be 0 mxm a.e. ?
( m: Lebesgue measure)

I believe that we need to use Fubini's theorem, but how?

thanks for help
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hparty
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 Re: double integral   « Reply #1 on: Jul 31st, 2010, 1:04pm » Quote Modify

no one has any idea ?!!
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towr
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 Re: double integral   « Reply #2 on: Jul 31st, 2010, 3:35pm » Quote Modify

This kind of thing is not really my forte, unfortunately. And I don't think everyone has come back yet since the forum was down for a month up to two weeks ago.
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nakli
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 Re: double integral   « Reply #3 on: Aug 1st, 2010, 1:12am » Quote Modify

Shouldn't this be in maths section? Or am I supposed to know this much??
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towr
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 Re: double integral   « Reply #4 on: Aug 1st, 2010, 1:18am » Quote Modify

on Aug 1st, 2010, 1:12am, nakli wrote:
 Shouldn't this be in maths section? Or am I supposed to know this much??
It might not be a bad idea to move it there; it might garner the attention of people that have a special interest in math.
Though in general, you shouldn't take where problems are posted as in indication of how much you should know
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ThudnBlunder
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 Re: double integral   « Reply #5 on: Aug 1st, 2010, 1:27am » Quote Modify

on Jul 31st, 2010, 1:04pm, hparty wrote:
 no one has any idea ?!!

Obob probably knows Lebesque integration, but he hasn't visited since we've been back online.

EDIT: At least he hasn't posted.

 « Last Edit: Aug 1st, 2010, 5:01am by ThudnBlunder » IP Logged

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Obob
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 Re: double integral   « Reply #6 on: Aug 7th, 2010, 5:12pm » Quote Modify

Yeah there hasn't been much activity here lately with all the server issues, and I've also discovered the site mathoverflow.net which has lots of pure math questions.

I could be mistaken, but I think the product measure here is misleading.  Open disks aren't rectangular in shape, so they don't play particularly nicely with thinking of R^2 as a product of two copies of R.

Whats really going on is that bounded measurable functions can't oscillate too much.  In a sense, on the average they actually behave quite a bit like you would expect a continuous function to behave.  The particular result I have in mind is the following:

Let f be a function that is locally in L^1(R^k) (i.e. every point in R^k has a neighborhood on which the absolute value of f has finite integral).  Then for almost every x, f(x) equals the limit of the average value of f on balls centered at x as those balls become smaller and smaller.

This is a restatement of theorem 7.10 in Rudin; it is a consequence of the Lebesgue differentiation theorem.  This result immediately implies the result you are asking about.

If instead of open balls you were asking about open rectangles, you would be able to deduce this result in R^2 from the same result for open intervals in R^1 by using Fubini.  But then you still have to show that a bounded function R -> R whose integral on every interval is zero is actually zero a.e., and I can't think of an easy way to show this without using the Lebesgue differentiation theorem.  Moreover, the proof of the differentiation theorem is not by induction on dimension, so this approach doesn't actually make things any easier.

My measure theory is a bit rusty, but this result just screams "differentiation theorem" to me.  I'd be happy to hear any other approaches though.
 « Last Edit: Aug 7th, 2010, 5:18pm by Obob » IP Logged
Michael Dagg
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 Re: double integral   « Reply #7 on: Aug 9th, 2010, 10:00am » Quote Modify

Obob is correct.

This follows immediately from the Lebesgue differentiation
theorem and Vitali's covering theorem in two-dimensions.

The problem is not at all interesting as it's a very weak
statement of what is true.  But I guess it is a "textbook"
problem. The authors develop some properties of a
2-dimensional Lebesgue integral and then, using just those
properties, you try to prove this simple property.  The fact
that the problem adds "bounded" unnecessarily suggests
this too.
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Regards,
Michael Dagg
Obob
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 Re: double integral   « Reply #8 on: Aug 9th, 2010, 11:17am » Quote Modify

Yes, the bounded hypothesis is stronger than what is necessary, which I agree suggests it is a textbook problem.

Are you aware of a simple proof of this statement (or the differentiation theorem) for functions on R, Michael?  Or do you think its an exercise meant to stump the reader (assuming it comes before the differentiation theorem)?
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Michael Dagg
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 Re: double integral   « Reply #9 on: Aug 9th, 2010, 12:36pm » Quote Modify

This problem likely lies within some context of development,
if it is an exercise in some book. It may in fact be the
I doubt it was intended to be given prior to these topics.
My guess is that it is perhaps the first part of a composite
exercise.

There are numerous elementary proofs of Lebesque but thinking
one up without having really worked with the theory is not
likely to happen easily.

There is a elementary proof of Lebesque by M. Botsko in the
Monthly; don't recall the number. There a paper by D. Austin
(~mid 1960s) Proc. AMS that provides fairly elementary geometrical
proof.

The problem in question would follow directly from any of
these elementary proofs but I believe that it was intended to

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Regards,
Michael Dagg
hparty
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 Re: double integral   « Reply #10 on: Aug 9th, 2010, 3:52pm » Quote Modify

"But I guess it is a "textbook"
problem...."

I belive that what you are saying and Obob is correct, but in fact this problem was in a qualifier exam I found it while I was searching online and I thought it would be related to double integration and fubinu's theorem.
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Michael Dagg
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 Re: double integral   « Reply #11 on: Aug 9th, 2010, 7:15pm » Quote Modify

I do have a slight issue with accepting this problem as
being a qual problem (but of course it is possible).

I don't know what you might do with Fubini's since the
differential is square in  m  , so you can integrate as you
wish, but if the integral was square itself then that
could be different, but not that integral.

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Regards,
Michael Dagg
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