Can there be a set whose Hausdorff dimension gradually changes?
For instance, a set of real numbers contained in an interval, whose Hausdorff dimension is 0 at the beginning and 1 closer to the end, and changes without jumps?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Take a sequence of Cesaro fractals $\endgroup$– WojowuCommented Jan 18, 2020 at 21:47
-
$\begingroup$ Related questions: How can dimension depend on the point? AND What results exist for functions with regionally fluctuant fractal dimension? $\endgroup$– Dave L RenfroCommented Jan 19, 2020 at 8:47
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
6
I assume you want a set $A\subseteq [0,1]$ such that $\dim (A\cap [0,x])=x$ for all $x$. We can define $A_1$ by taking the union of a (Borel) subset of dimension $0$ of $[0,1/2]$ with a subset of dimension $1/2$ of $[1/2,1]$.To obtain $A_2$, we then make our sets larger on $[1/4,1/2]$, $[3/4,1]$ and again, we make the dimension on each current interval equal to its left endpoint. Then $A_1\subseteq A_2\subseteq \ldots$, and $A=\bigcup A_n$ works: by monotone convergence, $h^d(I\cap A)=\lim h^d(I\cap A_n)$, so $A\cap [0,x]$ has the right dimension at each $x=k2^{-n}$ and thus everywhere.
answered Jan 18, 2020 at 22:08
-
$\begingroup$ Can we Lebesgue-integrate over such set? $\endgroup$– AnixxCommented Jan 18, 2020 at 23:06
-
10$\begingroup$ @Anixx Each $A_n$ has Hausdorff dimension below $1$, so their (1-dimensional) Lebesgue measure is zero. So $A$ has Lebesgue measure zero. You can certainly integrate functions over null sets, but the results are not too interesting... $\endgroup$– WojowuCommented Jan 18, 2020 at 23:11
-
2$\begingroup$ In fact for all $0<x\le1$ $A\cap[0,x]$ is an $x$-dimensional set of $x$-dimensional null measure, for the same reason. $\endgroup$ Commented Jan 19, 2020 at 20:05
-
$\begingroup$ Cool. I think a similar idea (reversing subintervals of $[k/2^n,(k+1)/2^n]$ as needed, for $n\rightarrow \infty, 0\le k<n$) can also be used to produce a set $A$ with density $x$ at $x$, i.e. $\mu(A\cap[0,x])=x^2/2$. $\endgroup$ Commented Jan 20, 2020 at 8:31
-
2$\begingroup$ @YaakovBaruch Are you sure? Doesn't that contradict Lebesgue Density Theorem? $\endgroup$– WojowuCommented Jan 20, 2020 at 8:46
$\begingroup$
$\endgroup$
For an example of a different sort, you can consider the record set $R$ of a fractional Brownian motion with varying Hurst parameter $H(t)$. In this short paper it is shown that when $H(t)$ is constant, it equals the Hausdorff dimension of $R$. The properties of the fBM with varying Hurst parameter $H(t)$ are given here for measurable functions $H(t)$ taking values in $(\tfrac12,1)$. One expects that the Hausdorff dimension of $R\cap (t-\epsilon,t+\epsilon)$ converges to $H(t)$ as $\epsilon\to 0$.
