Timeline for Where does the Lebesgue differentiation theorem fail?
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| Jan 31, 2017 at 17:29 | comment | added | Christian Remling | @PabloShmerkin: Also, the Lebesgue differentiation theorem (which is what the question is about) fails for the collection of rectangles with sides parallel to the coordinate axes, see Stein, Singular integrals ..., I.5.3. | |
| Jan 31, 2017 at 17:26 | comment | added | Christian Remling | @PabloShmerkin: I'm not very happy with your wording here. Your comments are relevant and much appreciated, but what you say is almost exactly what I wrote myself in my update, so I find it a bit harsh to summarize this by saying my answer is incorrect. | |
| Jan 31, 2017 at 13:20 | comment | added | Pablo Shmerkin | I agree with Squark, I don't think this answer is correct. The Lebesgue density theorem fails for the collection of all rectangles. It holds for the collection of rectangles with axis-parallel sides, so it is not a matter of eccentricity alone (I think it is an open problem to determine for which sets of directions $A$, the Lebesgue density theorem holds for all rectangles with long side in direction $A$, but it definitely holds if $A$ is finite) | |
| Jan 30, 2017 at 22:41 | comment | added | Vanessa | Your new construction is still isomorphic to the usual $L^1$ metric via the transformation $y' = y^{\frac{1}{3}}$. Moreover, any construction that has bounded eccentricity in some neighborhood of any point except for a set of zero measure cannot work (of course we can also start playing with measure, but I don't know how). | |
| Jan 30, 2017 at 22:02 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jan 30, 2017 at 21:08 | comment | added | Vanessa | OK, and what is $f(x,y)$? | |
| Jan 30, 2017 at 21:02 | history | undeleted | Christian Remling | ||
| Jan 30, 2017 at 21:00 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jan 30, 2017 at 20:56 | history | deleted | Christian Remling | via Vote | |
| Jan 30, 2017 at 20:49 | comment | added | Christian Remling | @Squark: No, I wanted a metric whose balls are rectangles with unbounded eccentricity, see my new version please. | |
| Jan 30, 2017 at 20:47 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jan 30, 2017 at 20:47 | comment | added | Vanessa | (also, I realized that what I really need to know is "Is there a compact Polish space s.t. for any metrization there is a counterexample?", but I guess it's too late to rephrase the question) | |
| Jan 30, 2017 at 20:44 | comment | added | Vanessa | Thanks for answering, Christian! Regarding $[0,1]^\omega$: I meant that I want a counterexample with any measure. Regarding $\mathbb{R}^2$: the equation you wrote is not a metric since the "distance" between $(-1,0)$ and $(+1,0)$ vanishes. I guess you meant to e.g. take the half-plane $x \geq 0$? But this is isomorphic to the usual $L^1$ metric (transform by $x' = x^2$)? | |
| Jan 30, 2017 at 16:20 | history | answered | Christian Remling | CC BY-SA 3.0 |