Let $\mu$ be a finite measure on $\mathbb{R}$. Define the measures $(\mu\_n)\_{n\geq 1}$ by $\mu\_{n+1}=\mu\ast \mu\_n$ and $\mu\_1=\mu$


Is there a singular (with respect to the Lebesgue measure) continuous measure $\mu$ on $\mathbb{R}$ such that for all $n\geq 2$, $\mu\_n$ is also singular continuous ?



Self convolutions of singular continuous measure
Here is a simple explicit example of a measure $\mu$ with all convolution powers singular: let $\mu$ be the distribution of
$$
\sum\_{k=1}^\infty X\_k 2^{-k!}, 
$$
where the $X\_k$ are IID taking values $0$ and $1$ with equal probability. The support of $\mu$ is the set $A$ of points whose binary expansion has non-zero digits only at places $k!,k\in\mathbb{N}$. Using this, it follows easily that for each $n$, the $n$-fold arithmetic sum $A^{+n}=A+\ldots+A$ has zero Lebesgue measure (in fact it has zero Hausdorff dimension), since the binary expansion of points in $A^{+n}$ must have zeros at all places not of the form $k!+j$ for some $0\le j\le n$. Since the convolution power $\mu^{\*n}$ is supported on $A^{+n}$, it must be purely singular. On the other hand, $\mu$ has no atoms so neither does $\mu^{\*n}$. This example doesn't use the Fourier transform at all.


Let me also give complete details of why the Cantor-Lebesgue measure $\mu$ on the middle-thirds Cantor set is an example of a measure all of whose self-convolutions are purely singular (but continuous, since $\mu$ itself already has no atoms).


Recall that $\mu$ can be realized as the distribution of the random sum $\sum\_{i=1}^\infty X\_i 3^{-i}$, where $X\_i$ are independent random variables taking the values $0$ and $2$ with equal probability. In particular, 
$$
\widehat{\mu}(\xi) =\prod\_{i=1}^\infty \cos(2\pi 3^{-i}\xi)
$$
from which it easily follows that $\widehat{\mu}(\xi)$ doesn't decay to $0$ as $\xi\to\infty$ (this also follows from the fact that $\mu$ is invariant under multiplication by $3$ on the circle). Hence, if $\mu^{\*n}$ denotes the $n$-fold self-convolution, then $\widehat{\mu^{\*n}}(\xi)=\widehat{\mu}(\xi)^n$ also doesn't decay to $0$, so $\mu^{\*n}$ cannot be absolutely continuous by the Riemann-Lebesgue Lemma.


Now, recall that a Borel probability measure $\nu$ on the real line is called *self-similar* if there are contracting similarities $f\_1,\ldots,f\_m$ and a probability vector $(p\_1,\ldots,p\_m)$ such that
$$
\nu(A)= \sum\_{j=1}^m p\_i\, \nu(f\_j^{-1}(A)).
$$
Given the tuples $(f\_1,\ldots,f\_m)$ and $(p\_1,\ldots,p\_m)$ there is exactly one Borel probability measure satisfying the above identity; this follows from seeing the right-hand side as a contracting operator on an appropriate Banach space (actually we only care about uniqueness, so we only need to have a contraction, no completeness is required).


If $\nu$ is a self-similar measure as above, then the absolutely continuous and singular parts of $\nu$ are seen to satisfy the same self-similarity relation. By the uniqueness of $\nu$, one of these parts has to be trivial. So self-similar measures are either absolutely continuous or purely singular with respect to Lebesgue.


To conclude, I claim that $\mu^{\*n}$ is a self-similar measure for every $n$. Since we already established that $\mu^{\*n}$ cannot be absolutely continuous, this will show that it must be purely singular. 


By definition, $\mu^{\*n}$ is the distribution of the random sum
$$
\sum\_{i=1}^\infty Y\_i^{(n)} 3^{-i},
$$
where the $Y\_j^{(n)}$ are IID with the distribution of the sum of $n$ independent realizations of $X\_i$ (defined above). In particular, the distribution of $Y\_i^{(n)}$ has the form $\sum\_{j=0}^{n} p\_j^{(n)} \delta\_{2j}$ for some probability vector $p\_j$. It can then be easily checked that $\mu^{\*n}$ is the self-similar measure corresponding to the contractions $(x/3+2j)\_{j=0}^{n}$ an the probability vector $(p\_j)\_{j=0}^n$.


Let me mention that it follows from work of Garsia in the 60s that not only $\mu^{\*n}$ is singular for all $n$, but in fact for each $n$ there is a Borel set $A\_n$ of Hausdorff dimension strictly less than $1$ such that $\mu^{\*n}(A\_n)=1$ (the topological support of $\mu^{\*n}$ is an interval for all $n\ge 2$). On the other hand, the Hausdorff dimension of $A\_n$ must tend to $1$; this follows from recent work of M. Hochman.


