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I'm not fully acquainted with measure theory, so a detailed explanation may be needed here.

From what I already understand, the Lebesgue measure on Cantor set (denote it by: CL-measure) $C$ gives probability $\frac{1}{2}$ to each of the the digits $0,2$ for each $x \in C$ expansion in ternary base.

Now, is there any difference in terms of measure of subsets of $C$ when the probabilities are $p,1-p$? To further focus my question, how is the power law (namely $c_1r^\alpha \leq \mu(B(x,r)) \leq c_2r^\alpha$ for all $ x \in \operatorname{supp}\mu$) affected by this change? I have seen some remarks about it holding for CL-measure when $\alpha = \frac{\log2}{\log3}$.

I have found very sparse information about this, and I'll be happy if someone knows a good source to look at.

Davide Giraudo
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Pavel
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2 Answers2

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The answer to your first question, "is there any difference in terms of measure of subsets of $C$ when the probabilities are $p$, $1-p$?" is: Yes, in fact there's about as much difference as one could possibly imagine. As $p$ varies, these measures concentrate on pairwise disjoint subsets of $C$.

In more detail: Let $\mu_p$ be the measure on the Cantor set that you obtain by giving the digits $0$ and $2$ the probabilities $p$ and $1-p$, respectively. (More precisely, this defines a measure on $\{0,2\}$; use it to define the product measure on the set of all infinite sequences of $0$'s and $2$'s; and finally identify those sequences with points of the Cantor set $C$ via base-$3$ expansions.) Let $A_p$ be the set of those $x\in C$ whose ternary expansion has asymptotically a fraction $p$ of $0$'s; that is, if we let $Z_x(n)$ be the number of zeros among the first $n$ digits in the ternary expansion of $x$, then $x\in A_p$ iff $\lim_{n\to\infty}(Z_x(n)/n)=p$. Notice that the sets $A_p$ for different values of $p$ are pairwise disjoint. Furthermore, the strong law of large numbers implies that $\mu_p(A_p)=1$. So the measures $\mu_p$, for different values of $p$, concentrate on disjoint sets.

Andreas Blass
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To further focus my question, how is the power law (...) affected by this change?

Yes it is.

To show why, let us first recall the heuristics leading to a power law for the Cantor set $C$ with respect to the Lebesgue measure, that is, when $p=\frac12$. For $x$ in $C$ and $n\geqslant1$, the elements of $C$ in $B(x,3^{-n})$ are roughly those whose $n$ first digits of their ternary expansion coincide with those in the expansion of $x$. This happens with probability $2^{-n}$ hence $\mu(B(x,3^{-n}))\approx2^{-n}$. And indeed, this approximate equivalent reads $\mu(B(x,r))\approx r^\alpha$ with $\alpha=\log2/\log3$.

When the probabilities of the digits $0$ and $2$ are $p$ and $1-p$, the first part of the reasoning subsists, that is, the elements of $C$ in $B(x,3^{-n})$ are roughly those whose $n$ first digits of their ternary expansion coincide with those in the expansion of $x$. But now this happens with probability $p^{N_n(x)}(1-p)^{n-N_n(x)}$, where $N_n(x)$ denotes the number of $0$ in the $n$ first digits of the ternary expansion of $x$, and, for every $p\ne\frac12$, this probability depends on $N_n(x)$. Some consequences follow:

  • The Hausdorff exponent $\alpha$ of $C$ at $x$, defined by the property that $\mu_p(B(x,r))\approx r^\alpha$ when $r\to0$, may not exist, and indeed it does not exist at those points $x$ such that $N_n(x)/n$ does not converge.

  • Every exponent $\alpha$ between $-\log p/\log 3$ and $-\log(1-p)/\log3$ corresponds to an infinite set of points $x$, those such that $N_n(x)/n$ converges to $\nu(\alpha)$ where, for every $a$ in this interval, $\nu(a)$ is defined by the identity $$ p^{\nu(a)}(1-p)^{1-\nu(a)}3^a=1. $$

  • Almost surely with respect to $\mu_p$, $\alpha=\alpha_p$, where $$ \alpha_p=-(p\log p+(1-p)\log(1-p))/\log3, $$ hence $\alpha_p$ can also be defined by the identity $\nu(\alpha_p)=p$.

To sum up, with respect to $\mu_p$, the Hausdorff exponent of $C$ is $\alpha_p$ almost everywhere while the subsets of $C$ where the Hausdorff exponent has any fixed value between $-\log p/\log 3$ and $-\log(1-p)/\log3$ and the subset of $C$ where the Hausdorff exponent does not exist, are all nonempty, uncountable and negligible for $\mu_p$.

Did
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  • Just some things that are not totally clear yet: why can $N_n(x)/n$ diverge? $N_n(x) \leq n$ always. Also could you elaborate a little more on points (2) and (3)? Especially how those identities were computed. Thanks for your help. – Pavel Jul 11 '13 at 12:02
  • Because the sequence of general term $N_n(x)/n$, while always between $0$ and $1$, can oscillate indefinitely in this interval. Then it does not converge. – Did Jul 11 '13 at 17:28
  • Thanks, but what about the other question? Sorry for the annoyance. – Pavel Jul 12 '13 at 20:12
  • You might want to make more precise what it is you fail to grasp here. – Did Jul 12 '13 at 20:26
  • In (2) I'm not sure why the exponent $\alpha$ should necessary be between those two values. Is it about the convexity of the binomial distribution or something of that kind? Next, the identity $ p^{\nu(a)}(1-p)^{1-\nu(a)}3^a=1$ - what does the $3^a$ signify? Any finally, in the equality that defines $\alpha_p$ - why does it equal to the sum of the aforementioned (in (2)) values? – Pavel Jul 12 '13 at 20:45
  • I do not understand, this is already explained in the point you mention. Assume that $N_n(x)/n\to\nu$, then the ball of radius $\rho(n)\approx3^{-n}$ has probability $\pi(n)\approx[p^\nu(1-p)^{1-\nu}]^n$. The exponent $\alpha$ at $x$ is such that $\mu(B(x,r))\approx r^\alpha$, that is $\pi(n)\approx\rho(n)^\alpha$, which is equivalent to $p^\nu(1-p)^{1-\nu}=3^{-\alpha}$. QED. – Did Jul 12 '13 at 21:23