August 5th, 2010, 05:30 PM  #1 
Newbie Joined: Apr 2010 Posts: 11 Thanks: 0  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 
August 8th, 2010, 12:05 AM  #2 
Senior Member Joined: Jun 2010 Posts: 618 Thanks: 0  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! 
August 8th, 2010, 11:20 AM  #3 
Newbie Joined: Apr 2010 Posts: 11 Thanks: 0  Re: double integral
I don't think that this is true: " " We have 0 = so >=0 ??!! 
August 8th, 2010, 11:59 AM  #4 
Senior Member Joined: Jun 2010 Posts: 618 Thanks: 0  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 mathnot 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 
August 9th, 2010, 01:07 PM  #5  
Newbie Joined: Aug 2010 Posts: 8 Thanks: 0  Re: double integral Quote:
Hope this helps.  
August 9th, 2010, 03:35 PM  #6  
Senior Member Joined: Jun 2010 Posts: 618 Thanks: 0  Re: double integral Quote:
 

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