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 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 $\int$ $\int$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 $f$ is Lebesgue measurable and bounded, we have $\| f \|_{\infty} \ < \ \infty$ and so for any open disk $D \in \mathbb{R}\times\mathbb{R}$ we'd have $\int_D \ f \ \mathrm{d}(\mu\times\mu) \ \leq \ \|f\|_{\infty} \ \mu\times\mu(D)= 0$ implying that $\| f \|_{\infty} \= \ 0$ which I think is enough to conclude that $f \= \ 0 \ \mu\times\mu$ 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: " $\int_D \ f \ \mathrm{d}(\mu\times\mu) \ \leq \ \|f\|_{\infty} \ \mu\times\mu(D)= 0$ " We have 0 =$\int_D \ f \ \mathrm{d}(\mu\times\mu) \ \leq \ \|f\|_{\infty} \ \mu\times\mu(D)$ so $\|f\|_{\infty} \ \mu\times\mu(D)$ >=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 math---not a good idea! Let me try again. Suppose that $|f|_{\infty} > 0$. Then the set $A$ where $|f| > 0$ has positive Lebesgue measure. Let me designate $\lambda= \mu\times\mu$ for convenience. We can then find a disk $D$ large enough so that $\lambda(D\cap A) > 0$. But then $\int_D f \ \mathrm{d}\lambda= \int_{D\cap A} f \ \mathrm{d}\lambda + \int_{D\backslash A} f \ \mathrm{d}\lambda > 0=$ contradicting the property of $f$ 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
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Re: double integral

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

$f(x)= \lim_{r \to 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} f \, d\mu = 0$

Hope this helps.

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, $f(x)= \lim_{r \to 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} f \, d\mu = 0$
Now that is a nice proof!

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