Home » Mathematics » Fractals » Hausdorff dimension of certain sets

# Hausdorff dimension of certain sets

## 1. Generalized Cantor set:

Fix $0<\alpha<1$. Choose $\eta$ (small) and $N\geq 2\in\mathbb{N}$ (large) such that $N\eta^{\alpha}=1$. Fix $0\leq a_1.

Let $K_1$ be the union of $N$ intervals $[a_i,a_i+\eta]$ of length $\eta$. Let $\mu_1$ be the natural measure (normalized one dimensional Lebesgue measure restricted to $K_1$) arises from $K_1$.

Let $K_2$ be the union of $N^2$ intervals $[a_i+a_j\eta,a_i+a_j\eta+\eta^2]$ of length $\eta^2$ and $\mu_2$ be its natural measure.

Similarly for each $j,$ let $K_j$ be the union of $N^j$ intervals of length $\eta^j$ and $\mu_j$ be its natural measure.

We consider the generalized Cantor set $E=\cap K_j$.

(Exercise: Then $E$ is compact perfect and nowhere dense set.)

Also note that,

• $\{\mu_j\}_j$ weakly converges to a measure $\mu$ supported in $E$.
• $\mathbb{R}^n$ is locally compact and $C_c(\mathbb{R}^n)$, the space of compactly supported continuous function on $\mathbb{R}^n$ is a separable topological vector space.
• By the application of Riesz representation theorem, space of continuous function vanishing at infinity is same as the space of Radon measures. (All the above $\mu_j$‘s are Radon measure, that is locally finite and inner regular measure)
• By the application of Banach-Alaoglu theorem, $\{\mu_j\}$ is a weakly convergent sequence.
• $dim_{\mathcal{H}}E=\alpha$ (enough to prove $0<\mathcal{H}^{\alpha}(E)<\infty$).
• By the definition of $\mathcal{H}^{\alpha}$, for $\delta>0$ there exists large $j$ such that $\eta^j<\delta$. Then$\mathcal{H}_{\delta}^{\alpha}(E)$
$\ \ \ \leq \sum_{1}^{N^j}(\eta^j)^{\alpha}$
$=N^j\eta^{j\alpha}=1$

and hence $\mathcal{H}^{\alpha}(E)\leq 1 <\infty$.

• Since $E$ is compact, if $\{E_i\}_i$ is a Borel cover of $E$ then we can assume that $\{E_i\}$ is a finite Borel cover. Also since $E$ is totally disconnected, $\{E_i\}$ can be assumed to be finite Borel cover, with $E_i$‘s disjoint open intervals and their end points are in the complement of $E$.Let $\{E_i\}$ be such a cover. Then for each $i$ let $p_i$ be the smallest integer such that $E_i$ has at least one interval of length $\eta^{p_i}$. Note that there cannot be more than $2N-2$ consecutive intervals of length $\eta^{p_i}$ in $E_i$ (otherwise, it contradicts the choice of $p_i$ to be the smallest). Let $k_i$ be the number of consecutive intervals of length $\eta^{p_i}$ in $E_i$. Then $1\leq k_i\leq 2N-2$. Hence $k_i\eta^{p_i}\leq d(E_i)=$diameter of $E_i$.

We know that each interval of length $\eta^{p_i}$ has $N$ intervals of length $\eta^{p_i+1}$ and $N^2$ intervals of length $\eta^{p_i+2}$. Similarly each interval of length $\eta^{p_i}$ has $N^{m-p_i}$ intervals of length $\eta^{m}$ for $m>p_i$. Thus $k_i$ intervals of length $\eta^{p_i}$ has $k_iN^{m-p_i}$ intervals of length $\eta^{m}$. There are totally $N^m$ intervals of length $\eta^m$ and since $E_i$‘s are disjoint and finite, we have $\sum k_iN^{m-p_i}=N^m$ for large $m>p_i$ (for all $i$).

Also we have $N\eta^{\alpha}=1$. Hence

$(2N-2)^{\alpha-1}$
$\ \ \ \leq \eta^{m\alpha}\sum_i k_i^{\alpha-1}k_iN^{m-p_i}$
$\ \ \ \ =\sum_ik_i^{\alpha}\eta^{p_i\alpha}$
$\ \ \ \ \leq \sum_id(E_i).$

Thus $0<\mathcal{H}^{\alpha}(E)$.

(to be updated)

This blog is loosely based on the notes prepared while giving a lecture series during January 2017-April 2017 at IISER - Bhopal. The definitions, the notations, proofs, theorems are from the references: 1) W. F. Donoghue; Distributions and Fourier Transforms, Acad. Press, 1969. MR3363413 2) Pertti Mattila, Geometry of sets and measures in Euclidean spaces: Fractals and rectifiability, Cambridge Studies in Advanced Mathematics, vol. 44, Cambridge University Press, Cambridge, 1995.  MR1333890 3) P. Mattila. Fourier Analysis and Hausdorff Dimension, Cambridge University Press, Cambridge, 2015. 4) Wolff T.H.: Lectures on Harmonic Analysis. University Lecture Series, 29. Amer. Math. Soc., Providence, RI (2003)