# Naïve definition of a measure on a fractal

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

• Which definition of fractal are you using? Apr 27 at 5:07
• 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. Apr 27 at 12:49
• 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. May 1 at 10:31
• @al-Hwarizmi Do you have any example in mind where the accumulation point would not be absolutely continuous to the Hausdorff measure? May 1 at 14:20
• I worked on this some 25 years ago, and am not more the expert, but the Cantor set is such an example. May 2 at 7:11

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 $$0.
For any compact set $$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 $$1, we can apply Frostman's lemma to the Hausdorff measure $$\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, 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.

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.

• 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 $\mu_\varepsilon (A_\delta) = \mu_\varepsilon (\overline{A_{\delta/2}})$? Apr 30 at 15:45
• 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 $a\in K$. Apr 30 at 15:56
• @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. Apr 30 at 16:36
• But the existence of a weak* accumulation point for 𝐽𝑐 is not guaranteed. May 2 at 7:13
• @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 $x \not\in \mathrm{supp}(\mu).$ May 2 at 9:17