The measure generated by the Cantor staircase and the intersection of the Cantor set with its translate

Question

Suppose that $T$ is the shift $\bmod 1$ of the Cantor set by an irrational number $\alpha\in (0,1)$. Consider the measure $\mu$ on the interval $[0,1]$ generated by the Cantor staircase. I'd like to know whether (if it is known) $\mu(T(C)\cap C)=0$.

Answer 1

Yes, $\mu((C+\alpha)\cap C)=0$ for all irrational $\alpha$ and for most rational $\alpha$. This can be inferred from the description of $(C+\alpha)\cap C$ on page 57 of [DH95]. I offer a translation of their terminology ("(1,1)-string sequence", etc). Assume $\alpha$ is not a ternary rational (i.e., does not admit a finite ternary expansion). In the ternary expansion of $\alpha$, the following digits are restrictive:

• all digits $1$
• digits $0$ that are preceded by an odd number of digits $1$
• digits $2$ that are preceded by an even number of digits $1$

Each restrictive digit of $\alpha$ forces the corresponding ternary digit of every element $x\in (C+\alpha)\cap C$ to take a particular value, $0$ or $2$. (As opposed to being free to choose between $0$ or $2$.) Thus, every restrictive digit halves the Cantor measure of the intersection.

It is easy to see that every number that is not a ternary rational has infinitely many restrictive digits.

For a more elaborate investigation of the Hausdorff measure and dimension of the intersection, see [PP13].

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[DH95] Davis, Gregory J.; Hu, Tian You. On the structure of the intersection of two middle third Cantor sets. Publ. Mat. 39 (1995), no. 1, 43–60.

[PP13] Pedersen, Steen; Phillips, Jason D. On intersections of Cantor sets: Hausdorff measure.
Opuscula Math. 33 (2013), no. 3, 575–598.