Does the convolution of the $1/3$-Cantor distribution with itself has a non trivial absolute-continuous part? This question is motivated by this postings 1 and more recently by 2
Let  $\mu_{1/3}$ be the Devil's staircase distribution. This distribution is supported on the $1/3$-Cantor set $C$ and so, it is singular with respect to the Lebesgue distribution $\lambda$. There are several ways to construct this distribution (see for the Wikipedia article). A probabilistic way to do it is by considering and sequence $(X_n:n\in\mathbb{N})$ of Bernoulli 0-1 random variables with parameter $1/2$ and define $Y=\sum_{n\in\mathbb{N}}\frac{2X_k}{3^k}$. The law of $Y$ is $\mu_{1/3}$.
It is known that $C+C=\{x+y:(x,y)\in C\times C\}=[0,2]$, and
in the older posting I referenced above, it is shown that $\mu_{1/3}*\mu_{1/3}\not\ll\lambda$. My question is: in the Lebesgue decomposition
$$\mu_{1/3}*\mu_{1/3}=f\cdot\lambda + \nu_a$$
where $\mu_a$ is singular with respect to $\lambda$, is $\int f\,d\lambda>0$?
More generally, is there $n\in\mathbb{N}$ for which $\mu^{*n}_{1/3}$ has a nontrivial absolutely continuous part?
Any hints, solutions or references will be appreciated.
 A: This measure $\nu_2:=\mu_{1/3}*\mu_{1/3}$ is purely singular, and there are several known proofs. It is easier to discuss  $\nu$, the push-forward of $\nu_2$
under $x \mapsto x/2$, since $\nu$ is the distribution of the sum of the random series
$Z=\sum Z_n 3^{-n}$ where $Z_n$ are i.i.d., and take the values $0,1,2$ with probabilities $(1/4,1/2,1/4)$ respectively. Indeed, each $Z_n$ is a sum $X_n+X_n*$ of two independent Bernoulli variables. Like Lebesgue measure, $\nu_1$ is invariant and ergodic for the map $x \mapsto 3x$ mod $1$, and two ergodic measures for the same map are always singular by the pointwise ergodic theorem (each measure assigns probability 1 to the corresponding generic points.) Concretely, by the law of large numbers, $\nu$ assigns full measure to points where the asymptotic frequency of the digit 1 in the base 3 expansion is $1/2$, while Lebesgue measure assigns these points measure zero.
There is also a nice observation, (first proved by Mauldin and Simon in a special case), that   any self-similar measure is either absolutely continuous or purely singular. Indeed, in the Lebesgue decomposition, both parts are self-similar with the same weights and contractions, so the result follows from uniqueness of the self similar probab. measure given the weights and contractions.
Another classical fact, going back to Besicovich and Eggleston,  is that $\nu$ gives full measure to a set of Hausdorff dimension  $h_3((1/4,1/2,1/4)<1$, where $h_3$ is the base 3 entropy. See Example 1.5.6 in [1]
or the book [2].
More generally, the $n$-fold convolution $\nu_n$ of $\mu_3$ is also purely singular. We know from the Riemann-Lebesgue lemma that this convolution is not absolutely continuous. But the pushforward $\nu_n^*$ of $\nu_n$ to the circle by $x \mapsto x \mod 1$ is an invariant ergodic measure, so if it is not Lebesgue, it must be singular to Lebesgue.
[1] Fractals in Probability and Analysis
https://www.cambridge.org/core/books/fractals-in-probability-and-analysis/D8CBD4181FDC20C387E22939DA2F6168
https://www.math.stonybrook.edu/~bishop/classes/math324.F15/book1Dec15.pdf
[2] Cajar, H., 2006. Billingsley dimension in probability spaces (Vol. 892). Springer.
