Is it true that for any locally integrable function $f \in L_{\mathrm{loc}}^1(\mathbb{R}^2)$, if $$ \int_a^b \int_c^d f(x,y) \,\textrm{d}x \,\textrm{d}y=0 \hspace{6mm} \textrm{for all $a,b,c,d$ with $|b-a||d-c|=1$,} $$ then $f=0\,$ a.e.?
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1Problem A-6 from the 2012 Putnam Exam is similar to this, except it gave the additional condition that $f$ was continuous. I don't know if this will be helpful, but the solution to that problem is here. – JimmyK4542 May 14 '16 at 06:15
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Look similarly to this problem http://math.stackexchange.com/questions/219738/showing-that-f-0-a-e-if-for-any-measurable-set-e-int-e-f-0?rq=1 – Truong May 14 '16 at 06:19
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@chuyenvien94: Why does the edit of Santiago deviate from the original intent of the post? – miracle173 May 14 '16 at 06:29
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@miracle173 What was the edit? – almagest May 14 '16 at 06:57
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@miracle173 I prefer to use words where mathematicial symbolism is not necessary, but putting "0" instead of "zero" may be fine here. – Santiago May 14 '16 at 07:19
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Because the additional assumptions on $f$ seem to be too restrictive. – Truong May 14 '16 at 07:38
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Can this question be re-opened now that I've edited it to be more precise? It would probably be interesting and useful for future readers if this question had a correct answer posted. – Julian Newman Feb 27 '18 at 14:06
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How about this attempt at an counter example (If wrong please just remove). $f:(x,y) \mapsto cos 2\pi\, x$. Then $$ \int_{[y_0+1]\times[x_0+1]} f\, dx\,dy = \int\left[\frac{sin 2\pi\, x}{2\pi} \right]^{x_0+1}_{x_0}\, dy = 0 \qquad \forall x_0,y_0 \in \mathbb{R}, $$ yet $f=0$ a.e. is wrong.
be5tan
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The integral of this $f$ over an axis-aligned unit square is zero, but not every unit-area rectangle is a square, much less axis-aligned. :) – Andrew D. Hwang May 14 '16 at 11:40