6
$\begingroup$

Let $E\subset B_1(0)\subset \mathbb{R}^n$ be a compact set s.t. $\lambda(E)=0$, where $\lambda$ is the Lebesgue measure, and $B_1(0)$ is the Euclidean unit ball centered at the origin. Is the following integral finite:

$$\int_{B_1(0)}-\log d(x,E)d\lambda(x)<\infty?$$

Although this question seems trivial, I have failed to find a reference to it or to variations of it in previous discussions. I was not able to come up with a counter-example nor a proof. I also asked in mathstackexchange a variation of it, but didn’t get a sufficient answer.

Thanks ahead

$\endgroup$
  • $\begingroup$ Correct, thank you. $\endgroup$ – BOS Sep 17 at 14:24
7
$\begingroup$

The integral in question is finite for most sets of measure zero, but can diverge to $\infty$ for some sets. An example in one dimension is obtained by constructing a Cantor set where at stage $k$ the middle $1/(k+1)$ proportion is removed from each of the $2^{k-1}$ intervals obtained at stage $k-1$. Thus the $2^k$ intervals obtained at stage $k$ will each have length $2^{-k}/(k+1)$. Therefore, each of the $2^k$ middle intervals removed in the next stage will have length $2^{-k}/[(k+1)(k+2)]$, and each of these will contribute at least $k/2$ times its length to the integral. Summing over $k$ gives a harmonic series which diverges. The example can be lifted to higher dimensions by taking a Cartesian product with a $n-1$ dimensional box.

$\endgroup$
  • $\begingroup$ Something like this is what I thought of a couple minutes ago. :-) $\endgroup$ – Iosif Pinelis Sep 17 at 14:57
  • $\begingroup$ @YuvalPeres thank you for this very nice construction. $\endgroup$ – BOS Sep 17 at 18:21
2
$\begingroup$

If $E\ne\emptyset$, then $d(x,E)\le2$ for all $x\in B_1(0)$. So, your integral is $\le\lambda(B_1(0))\ln2<\infty$.

$\endgroup$
  • 1
    $\begingroup$ You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-\infty$ )… $\endgroup$ – Dirk Sep 17 at 14:16
  • $\begingroup$ @Dirk : This is what I am thinking about now. :-) $\endgroup$ – Iosif Pinelis Sep 17 at 14:18
  • $\begingroup$ The integral is $>-\infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set? $\endgroup$ – Iosif Pinelis Sep 17 at 14:24
  • 1
    $\begingroup$ @Iosif Yes, there are. $\endgroup$ – Yuval Peres Sep 17 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.