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August 5th, 2010, 05:30 PM   #1
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double integral

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

f(x,y) d(mxm) =0 over every open disk D in RxR.

How to prove that f= 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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August 8th, 2010, 12:05 AM   #2
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Re: double integral

Hello Eliza.

Maybe you could do something along the lines of the following:

Since is Lebesgue measurable and bounded, we have



and so for any open disk we'd have



implying that



which I think is enough to conclude that almost everywhere.

I hope this helps inspire a true solution.

Best of luck!
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August 8th, 2010, 11:20 AM   #3
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Re: double integral

I don't think that this is true:
"

"

We have 0 =

so >=0

??!!
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August 8th, 2010, 11:59 AM   #4
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Re: double integral

Hi again Eliza.

Haha, yes indeed, you caught me! Sorry for the mistake, I woke up in the middle of the night and tried to do math---not a good idea! Let me try again.

Suppose that . Then the set where has positive Lebesgue measure. Let me designate for convenience. We can then find a disk large enough so that . But then



contradicting the property of given. So the essential supremum must be zero. This is hopefully not flawed! Let me know if it is.

Best,

Ormkärr
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August 9th, 2010, 01:07 PM   #5
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Re: double integral

Quote:
Originally Posted by Eliza
given a function f:RxR---> R bounded and Lebesgue measurable, and satisfies

f(x,y) d(mxm) =0 over every open disk D in RxR.

How to prove that f= 0 mxm a.e. ?
( m: Lebesgue measure)
Since f is locally integrable, almost every point is a Lebesgue point. For each Lebesgue point x,



Hope this helps.
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August 9th, 2010, 03:35 PM   #6
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Re: double integral

Quote:
Originally Posted by Otrak
Since f is locally integrable, almost every point is a Lebesgue point. For each Lebesgue point x,

Now that is a nice proof!
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