TITLE: Isometrically-invariant measures and dilation of the Cantor set
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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.