Let $f(x, y)$ be a continuous, real-valued function on $\mathbb{R}^2$. Suppose that, for every rectangular region $R$ of area 1, the double integral of $f(x, y)$ over $R$ equals 0. Must $f(x, y)$ be identically 0?
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What did you try? How did you put to use the hypothesis that $f$ is continuous? – Did May 21 '13 at 06:25
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My working hypothesis is that is not true. I am trying to construct an example. – user64494 May 21 '13 at 06:30
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Can you think of any counterexamples? – rurouniwallace May 21 '13 at 06:31
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If your tries at showing this is not true fail, I suggest trying to prove it is true. And to do that, a hint is: small rectangles around a point. – Did May 21 '13 at 06:36
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@ZettaSuro Can you? – Did May 21 '13 at 06:36
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1@Did: Rectangles in the question are of area 1. This implies that these cannot be small, is not so? – user64494 May 21 '13 at 06:40
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This was Problem A6 from the 2012 Putnam Competition. The solutions have been published, and the AoPS wiki also has solutions by those who attempted the exam, so there is no need for this question to be on MSE. – Gyu Eun Lee May 21 '13 at 06:41
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Oops... then one must be more careful. Sorry about the noise. – Did May 21 '13 at 06:42
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@proximal: Thank you. I had in mind the following modification: $f$ is locally integrable on $\mathbb{R}^2.$ – user64494 May 21 '13 at 07:24
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Hint: Consider $g(y)=\int_0^1 f(x,y)dx$. Since the intgeral of $f$ over $[0,1]\times[y,y+1]$ equals that over $[0,1]\times[y+\epsilon,y+1+\epsilon]$, we conclude $\int_y^{y+\epsilon} g(y)dy=\int_{y+1}^{y+1+\epsilon}g(y) dy$ and in the limit $g(y+1)=g(y)$, that is $g$ is periodic with period $1$.
Hagen von Eitzen
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It solves the question. I thought about the discontinuous function $f(x,y):=1/\text{denominator of}$ $ x $ if $x$ and $y$ are rationals and 0 otherwise. – user64494 May 21 '13 at 07:07