Isometrically-invariant measures and dilation of the Cantor set

Question

Let $C$ be the Cantor middle-thirds set. Let $\mu$ be a finitely-additive isometrically-invariant measure on all subsets of $\mathbb R$. Then $\mu(3C)=2\mu(C)$, where $aB = \{ ax : x \in B \}$. Thus if $a$ is a power of $3$, $\mu(aC) = a^{\log_3 2} \mu(C)$.

Question 1: Is it the case for all $a\in (0,\infty)$ that $\mu(aC)=a^{\log_3 2} \mu(C)$, if $\mu$ is a finitely-additive isometrically-invariant measure on $\mathcal P\mathbb R$?

The remaining questions are predicated on an affirmative answer, though currently I'm suspecting a negative answer to 1.

Question 2: Is this still true if isometric-invariance is replaced by translation-invariance?

Question 3: Has there been any work on using the the relationship between $\mu(aC)$ and $\mu(C)$ for invariant measures $\mu$ on $\mathbb R^n$ (perhaps not defined on all of the powerset, but just on the Borel sets) to define a dimension and comparing it to Hausdorff and box dimensions?

Answer 1

Negative to the first question, so the others are moot.

Let $G_1$ be the isometries of $\mathbb R$.

First note that no finite number of translates (or reflections, but that doesn't add anything) of $(1/2)C$ covers $C$. This can be seen by playing around with base three expansions.

Let $I$ be the ideal in $\mathcal P(\mathbb R)$ generated by $(1/2)C$ and its translates. Let $\mathcal B$ be the quotient boolean algebra $\mathcal P(\mathbb R)/I$. Because $C$ isn't covered by translates of $(1/2)C$, $[C]\ne 0$. The action of $G_1$ on $\mathbb R$ induces an action of $G_1$ on $\mathcal B$. Since $G_1$ is supramenable (i.e., doesn't have nonempty paradoxical subsets), it follows from a theorem of Mycielski ("Finitely additive invariant measures. I", Colloq. Math. 42 (1979), 309–318) that there is a finitely additive measure $\mu_1$ on $\mathcal B$ invariant under $G_1$ with $\mu_1([C])=1$. Let $\mu(A)=\mu_1([A])$ for $A\subseteq\mathbb R$. Then $\mu(C)=1$ and $\mu((1/2)C)=0$ and the answer to Question 1 is negative for $a=1/2$.

This uses the Axiom of Choice.

Answer 2

This is too long for a comment. The following measure (defined on Borel sets) might be a counterexample to Question 1: let $\mathcal{I}_N$ be the collection of all left-closed, right-open intervals of length $3^{-k}$ for some $k\in\mathbb{N}, k\ge N$, and define $$ \mu(A) = \lim_{N\to\infty} \inf\left\{ \sum_i |I_i|^{\log_3 2}: A\subset \bigcup_i I_i, I_i\in\mathcal{I}_N \right\}. $$

This is well-known to be a Borel-regular measure on the Borel $\sigma$-algebra (is constructed using the "Type 2" method from Rogers' classical book "Hausdorff measures"), and it's obviously isometry-invariant. It is clear that $\mu(C)=1$. Since we are covering only with intervals of length $3^{-k}$, it seems to me the scaling law should break down ($\mu(aC)$ should be strictly larger than $a^{\log_3 2}$ for $1<a<3$). I can't prove this at the moment though.

Regarding Question 3, in addition to my comment another problem is that for some sets $E$ there exist two (again, Borel) isometry-invariant measures $\mu,\nu$ which give $E$ positive and finite mass, and distinct numbers $s,t>0$ such that $\mu(a E)=a^s \mu(E)$ and $\nu(a E)=a^t(E)$ for all $a>0$. For example, we may choose a set $E$ of different Hausdorff and packing dimensions, and which has positive and finite Hausdorff/packing measure in their respective dimensions. It is not immediately clear to me if it is possible to get 3 or more different scaling exponents.

Essentially the approach in Question 3 is implicitly used in the definition of Hausdorff and packing dimensions, but it only works if one restricts the measures to the family Hausdorff and packing measures, respectively.

I know very little about finitely additive measures on $\mathcal{P}(\mathbb{R})$ so I don't have much to say on the actual questions.