To be concise, I am wondering whether there are natural examples of probability measures $\mu$ compactly supported on the real line which satisfy $\mu(I) \lesssim l_n^\alpha$ for all intervals $I$ with $|I| = l_n$, for some $0 \leq \alpha \leq 1$ and a certain sequence $l_1 \geq l_2 \geq \dots \to 0$, but for which the general inequality $\mu(I) \lesssim |I|^\alpha$ fails.

Often, in order to lower bound the Hausdorff dimension of a certain set $X$ by a constant $\alpha$, one applies Frostman's lemma, finding a natural probability measure $\mu$ supported on $X$ with $\mu(I) \lesssim |I|^\alpha$. Often, one constructs $X$ as a fractal, as a limit of sets $X_n$, each a union of intervals of length $l_n$ (i.e. $X_1 \supset X_2 \supset \dots \to X$). Often, one obtains $\mu(I) \lesssim l_n^\alpha$ for all intervals $I$ of length $l_n$, and then tries to 'cheat out' the general bound from these estimates. My question is whether there are natural counterexamples of 'fake' $\alpha$ dimensional probability measures for which one can obtain the bound $\mu(I) \lesssim l_n^\alpha$ for all intervals $I$ of length $l_n$, where $l_1 \geq l_2 \geq \dots$ is a sequence converging to zero, but for which the bound $\mu(I) \lesssim |I|^\alpha$ does not hold for all intervals.

  • $\begingroup$ This reminds me of the faithfulness notion for Hausdorff dimension. You might want to look here, especially Theorem 1. $\endgroup$
    – zhoraster
    Jul 16, 2018 at 11:56

1 Answer 1


Later on in my research I answered this question in the negative:

Consider a sequence of dyadic scales $\{ r_k \}$, such that $r_k/r_{k+1} \geq 4$. Consider a sequence of dyadic lengths $\{ r_k \}$, and construct a set $X$ by a Cantor-type decomposition, defined as $\lim_{k \to \infty} X_k$ where $X_k$ is a union of side-length $r_k$ intervals. We set $X_0 = [0,1]$, and $r_0 = 1$. Given $X_k$, which is a union of sidelength $r_k$ intervals, we define $X_{k+1}$ arbitrarily, such that for any sidelength $r_k$ interval in $X_k$, $X_{k+1}$ contains a single sidelength $r_{k+1}^{1/2}$ interval. If we let $N_k$ to be the covering number of $X_k$ by $r_k$ intervals, then $N_0 = 1$, and $N_{k+1} = N_k(r_k/r_{k+1}^{1/2})$. Thus

$$ N_k = \frac{\left( r_1 \dots r_{k-1} \right)^{1/2}}{r_k^{1/2}} $$

If we set $\mu(I) = 1/N_k$ for each side-length $r_k$ interval $I$ selected in $X_k$, and $\mu(I) = 0$ otherwise, then $\mu$ satisfies the mass distribution principle and thus extends to a Borel probability measure. If $l_k = r_k^{1/2}$, then for any side-length $l_{k+1}^{1/2}$ interval, either $\mu(I) = 0$ or

$$ \mu(I) = \frac{l_{k+1}}{(r_1 \dots r_k)^{1/2}} \leq \frac{l_{k+1}}{r_k^{k/2}}. $$

If $r_{k+1} \leq r_k^{4k}$, then $l_{k+1}^{1/4} = r_{k+1}^{1/8} \leq r_k^{k/2}$, so $\mu(I) \leq l_k^{3/4}$ for each index $k$ and side-length $l_k$ interval $I$. Thus at the scales $l_k$, the set $X$ looks like it has dimension at least $3/4$. But as $k \to \infty$,

$$ H^{1/2}_{r_k}(X) \leq N_k \cdot r_k^{1/2} \leq (r_1 \dots r_{k-1})^{1/2} \to 0 $$

Thus $\dim(X) \leq 1/2$, so we cannot possibly obtain a general bound $\mu(I) \leq_\varepsilon l^{3/4-\varepsilon}$ for all scales $l$.


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