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Let $K\subset \mathbb R^2$ be a compact fractal of Hausdorff dimension $1<d<2$. I want to define a natural measure on $K$.

One option would be to use the so-called Hausdorff measure $\mathcal H^d$. Where, for every $A\subset K$, $$ \mathcal H^d(A) = \lim_{\delta\to 0} \inf \left\{\sum _{i=1}^{\infty }(\operatorname {diam} U_{i})^{d}:\bigcup _{i=1}^{\infty }U_{i}\supseteq S,\ \operatorname {diam} U_{i}<\delta \right\}.$$

Alternatively, given $ε >0$, one could define $K_\varepsilon =\{z\in\mathbb R^2, \operatorname{dist}(z, K) \leq \varepsilon\}$, and define $$\mu_\varepsilon(\mathrm d x) = \frac{\mathbf{1}_{K_\varepsilon} (x)}{\lvert K_\varepsilon\rvert} \mathrm{d}x, $$ where $\mathbf{1}_{K_\varepsilon} $ is the indicator function of $K_\varepsilon$ and $\lvert K_\varepsilon\rvert$ is the Lebesgue measure of $K_\varepsilon$.

Question: Is there any relation between the weak${}^*$ accumulation points of $\{\mu_\varepsilon\}_{\varepsilon >0}$ (as $\varepsilon \to 0)$ and the Hausdorff measure $\mathcal H^d$?

I could not find any book/paper that addresses this question. I am particularly interested in the case where $K$ is a Julia set $J_c = \partial\{z\in\mathbb C; p_c^n(z) \not \to \infty\ \ \text{as }n\to\infty\},$ where $p_c(z) = z^2 +c$, for some $c$ in the Mandelbrot set (it is ok to assume $c$ hyperbolic).

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  • $\begingroup$ Which definition of fractal are you using? $\endgroup$ Commented Apr 27, 2023 at 5:07
  • $\begingroup$ Good question. I do not really have a def for fractal. I understand that this may generate some "degenerate cases" like a Koch snowflake + line. I am interested in the case that $K$ is a Julia set $J_c = \partial\{x\in\mathbb C; p_c^n(0) \not \to \infty\ \text{as }n\to\infty\}$ where $p_c(z) = z^2 +c$, for some $c$ in the Mandelbrot set. $\endgroup$ Commented Apr 27, 2023 at 12:49
  • $\begingroup$ Your question is far too "general" formulated. There is no straightforward connection at this level. There might be possibilities in some specific cases, for instance via harmonic measure. I guess it is necessary to distinguish whether $\mu_0$ in relation to the Hausdorff dimension, is continuous or singular. Extensive research is required for this. But this is only my personal opinion. $\endgroup$ Commented May 1, 2023 at 10:31
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    $\begingroup$ @al-Hwarizmi Do you have any example in mind where the accumulation point would not be absolutely continuous to the Hausdorff measure? $\endgroup$ Commented May 1, 2023 at 14:20
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    $\begingroup$ I worked on this some 25 years ago, and am not more the expert, but the Cantor set is such an example. $\endgroup$ Commented May 2, 2023 at 7:11

1 Answer 1

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The weak${}^*$ accumulation points of the family $\{\mu_\varepsilon\}_{\varepsilon >0}$ are related to the Hausdorff measure $\mathcal H^d$ of $K$ through the Frostman's lemma.

Frostman's lemma states that:
Let $c,C,C'$ non negative constants and $0<r\leq \text{diam}(K)$.
For any compact set $K\subset \mathbb R^d$ and any Borel measure $\mu$ on $K$ that satisfies:
$\mu(B(x,r))\leq Cr^d,\mu(B(x,r))\leq Cr^d$, $\forall x\in K$ and $r>0$,
$\exists$ a Borel probability measure $\nu$ on $K$ such that
$\nu(B(x,r))\leq C′r^d,\nu(B(x,r))\leq C′r^d$, $\forall x\in K$
and such that $\nu(B(x,r))\geq cr^d,\nu(B(x,r))\geq cr^d$, $\forall x\in K$.

This lemma gives a way to construct a Borel probability measure on a compact set $K$ from any Borel measure on $K$ that satisfies a that scaling condition. The measure constructed in this way has the same scaling property as the original measure, but it is normalized to be a probability measure.

In the case of a compact fractal $K$ of Hausdorff dimension $1<d<2$, we can apply Frostman's lemma to the Hausdorff measure $\mathcal H^d$ to obtain a Borel probability measure $\nu$ on $K$ such that:
$\nu(B(x,r))\leq Cr^d,\nu(B(x,r))\leq Cr^d$ for some $C>0$ and all $x\in K$ and $0<r\leq \operatorname{diam}(K)$, and such that $\nu(B(x,r))\geq cr^d,\nu(B(x,r))\geq cr^d$, for some $c>0$ and all $x\in K$ and $0<r\leq \operatorname{diam}(K)$.

Now, let $\{\mu_\varepsilon\}_{\varepsilon>0}$ be the family of measures defined as: $$\mu_{\varepsilon}(\mathrm{d}x)=\dfrac{\mathbf{1}_{K_\varepsilon}(x)}{|K_{\varepsilon}|}\mathrm{d}x,$$ where $K_{\varepsilon}=\{z\in\mathbb{R}^2, \operatorname{dist}(z, K) \leq \varepsilon\}$ and $\lvert K_\varepsilon\rvert$ is the Lebesgue measure of $K_\varepsilon$.

We claim that the weak${}^*$ accumulation points of the family $\{\mu_\varepsilon\}_{\varepsilon >0}$ are all absolutely continuous with respect to the measure $\nu$ constructed by Frostman's lemma.

To see this, note that for any $\varepsilon_1<\varepsilon_2$, we have $\mu_{\varepsilon_1}(A) \leq \mu_{\varepsilon_2}(A)$ for any Borel set $A\subset K$. This follows from the fact that $K_{\varepsilon_1} \subseteq K_{\epsilon_2}$ and $\lvert K_{\varepsilon_1}\rvert \geq \lvert K_{\varepsilon_2}\rvert$, so $\mu_{\varepsilon_1}(A) \leq \frac{\mathbf{1}{K{\varepsilon_2}}(A)}{\lvert K_{\varepsilon_2}\rvert} \leq \mu_{\varepsilon_2}(A)$.
Now, let $\mu$ be any weak$^*$ accumulation point of $\{\mu_\varepsilon\}_{\varepsilon>0}$. We claim that $\mu$ is absolutely continuous with respect to $\mathcal{H}^d$. To see this, suppose that $A\subset K$ is a Borel set with $\mathcal{H}^d(A)=0$. We want to show that $\mu(A)=0$. Since $\mathcal{H}^d(A)=0$, we have $\mathcal{H}^d(A_{\delta})=0$ for any $\delta>0$, where $A_\delta=\{x\in\mathbb{R}^2 : \operatorname{dist}(x,A)<\delta\}$. Note that $A_\delta$ is an open set and $A_{\delta_1}\subseteq A_{\delta_2}$ if $\delta_1<\delta_2$.
Since $\mu$ is a weak$^*$ accumulation point of $\{\mu_\varepsilon\}_{\varepsilon>0}$, there exists a subsequence $\varepsilon_n$ such that $\mu{\varepsilon_n}$ converges weakly to $\mu$. By the Portmanteau theorem, we have $$\liminf_{n\to\infty}\mu_{\varepsilon_n}​​​(A_\delta​)\geq \mu(A_\delta​)\geq \limsup_{n\to\infty}\mu_{\varepsilon_n}​​​(A_\delta​).$$ Taking $\delta=\frac{1}{m}$ and letting $m\to\infty$, we get $\displaystyle\liminf_{⁡n\to\infty}\mu_{\varepsilon_n}(A_{1/m})\geq \mu(A)\geq \limsup_{⁡n\to\infty}\mu_{\varepsilon_n}(A_{1/m})$ Since $A_{1/m}$ is an open set and $\mu_{\varepsilon_n}(A_{1/m})\leq \frac{1}{\lvert K_{\varepsilon_n}\rvert}\lvert A_{1/m}\rvert$, we have $$\limsup_{⁡n\to\infty}\mu_{\varepsilon_n}(A_{1/m})\leq \limsup_{⁡n\to\infty}\frac{|A_{1/m}|}{|K_{\varepsilon_n}|}=0$$ Hence, $\mu(A)=0$, which shows that $\mu$ is absolutely continuous with respect to $\mathcal{H}^d$.

In conclusion, we have shown that any weak$^*$ accumulation point of $\{\mu_\varepsilon\}_{\varepsilon>0}$ is absolutely continuous with respect to $\mathcal{H}^d$.
However, we don't know if there exists any weak$^*$ accumulation point in the first place.
In the case of the Julia set $J_c$, it is known that for any $c$ in the Mandelbrot set, the Hausdorff dimension of $J_c$ is strictly between $1$ and $2$, so we can apply the above result to obtain that any weak$^*$ accumulation point of $\{\mu_\varepsilon\}_{\varepsilon>0}$ on $J_c$ is absolutely continuous with respect to $\mathcal{H}^d$. However, it is not clear if there exists any weak$^*$ accumulation point in this case either.

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  • $\begingroup$ Hi, Thank you for your answer. A few things that made me a bit confused. You say that $\mathcal H^d (A) = 0$, implies that $\mathcal H^d(A_\delta )$, every $\delta >0.$ Imagine that $d= 2,$ and $A=\{0\}.$ Then, $H^2(A) = 0$, but $H^2 (A_\delta) = \pi \delta^2>0$, for every $\delta >0$. Moreover, the equation $$\liminf_{n\to\infty}\mu_{\varepsilon_n}​​​(A_\delta​)\geq \mu(A_\delta​)\geq \limsup_{n\to\infty}\mu_{\varepsilon_n}​​​(A_\delta​).$$ Is the last inequality because, if $\varepsilon <\delta/2$, then $\mu_\varepsilon (A_\delta) = \mu_\varepsilon (\overline{A_{\delta/2}})$? $\endgroup$ Commented Apr 30, 2023 at 15:45
  • $\begingroup$ Finally, why is the case that $$|A_{1/m}|/|K_{\varepsilon_n}| =0 \to 0,$$ as $n\to \infty?$. It seems to me that $|A_{1/m}|/|K_{\varepsilon_n}| \to \infty$. Since $|A_{1/m}|>0 $ and $|K_{\varepsilon_n}|\to 0.$ Also, I believe that an accumulation point of $\{\mu_\varepsilon\}_{\varepsilon>0}$ always exists because, $\mu_\varepsilon$ is a probability measure on $\overline{B_{2 \mathrm{Diam(K)}}(a)}$ for every $0<\varepsilon <1$, where $a\in K$. $\endgroup$ Commented Apr 30, 2023 at 15:56
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    $\begingroup$ @MatheusManzatto you are right in everything you said (sorry but the answer was very long I lost my train of thought several times and I'm not very comfortable typing without seeing the output). Regarding the accumulation point the sequence $\{\mu_\varepsilon\}_{\varepsilon>0}$ is uniformly bounded and equicontinuous, so it has a weak$^*$ accumulation point by the Banach-Alaoglu theorem. $\endgroup$ Commented Apr 30, 2023 at 16:36
  • $\begingroup$ But the existence of a weak* accumulation point for 𝐽𝑐 is not guaranteed. $\endgroup$ Commented May 2, 2023 at 7:13
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    $\begingroup$ @al-Hwarizmi I believe it is. Let $\mu$ be an accumulation point of $\{\mu_\varepsilon\}_{\varepsilon>0}.$ I will show that $\mu (J_c) = 1.$ Given $x\in \mathbb C \setminus J_c,$ since $J_c$ is compact $\delta = \mathrm{dist}(x,J_c) >0.$ This implies that $B_{\delta/3}(x) \subset \mathbb C\setminus J_\varepsilon,$ for every $0<\varepsilon < \delta/3$. Therefore $\mu_\varepsilon (B_{\delta/3}(x)) =0,$ for $0<\varepsilon < \delta/3$. Therefore, $\mu(B_{\delta/3}(x))=0.$ Hence $x\not\in J_c$ implies that $x \not\in \mathrm{supp}(\mu).$ $\endgroup$ Commented May 2, 2023 at 9:17

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