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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

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  • $\begingroup$ Correct, thank you. $\endgroup$
    – JustSomeGuy
    Commented Sep 17, 2019 at 14:24

2 Answers 2

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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.

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  • $\begingroup$ Something like this is what I thought of a couple minutes ago. :-) $\endgroup$
    – Iosif Pinelis
    Commented Sep 17, 2019 at 14:57
  • $\begingroup$ @YuvalPeres thank you for this very nice construction. $\endgroup$
    – JustSomeGuy
    Commented Sep 17, 2019 at 18:21
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If $E\ne\emptyset$, then $d(x,E)\le2$ for all $x\in B_1(0)$. So, your integral is $\le\lambda(B_1(0))\ln2<\infty$.

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    $\begingroup$ 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
    Commented Sep 17, 2019 at 14:16
  • $\begingroup$ @Dirk : This is what I am thinking about now. :-) $\endgroup$
    – Iosif Pinelis
    Commented Sep 17, 2019 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
    Commented Sep 17, 2019 at 14:24
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    $\begingroup$ @Iosif Yes, there are. $\endgroup$
    – Yuval Peres
    Commented Sep 17, 2019 at 14:52

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