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Aug 31, 2018 at 11:22 answer added Stefano Gogioso timeline score: 1
Aug 31, 2018 at 3:48 comment added BigM Let's rule that out. Id rather to find counterexamples for non atomic measures and also none discrete .
Aug 31, 2018 at 2:09 comment added Christian Remling $X=\{ x\}$ gives such examples.
Aug 30, 2018 at 20:52 comment added BigM In fact, for some nice Riemannian manifold our double integral is bounded. My limited personal experience suggests its bounded for general care.
Aug 30, 2018 at 20:41 comment added Aryeh Kontorovich Ah ok thanks. I somehow only thought the large values of $d(x,y)$ could be problematic -- ignored the small ones.
Aug 30, 2018 at 20:16 comment added BigM No.it doesnt.d(x,y) can be very small , consequently log^2 can be very large. In many cases integal becomes bounded but even in R^n it's not obvious, at least to me, why for an arbitrary measure the double integral should be bounded. it's trivial that its bounded bellow though. :)
Aug 30, 2018 at 20:07 comment added Aryeh Kontorovich If $U$ is compact, then in particular it's bounded and so $d(x,y)$ is upper-bounded by some $M<\infty$. Doesn't that make the integral trivially finite?
Aug 30, 2018 at 17:20 history asked BigM CC BY-SA 4.0