Self convolutions of singular continuous measure 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 ?
 A: Yes, indeed. You can build a one dimensional family of non-atomic singular measures on $[0,1]$ simply by pushing the Bernoully measure with parameter $p$ with the help of the dyadic decomposition of real numbers.
Start with $X_i$ iid random variables such that $P(X_i = 1) = p = 1 - P(X_i = 0)$ and define
$$
Y_p = \sum_{k = 1}^\infty {X_i\over 2^i}
$$
The law of $Y_{1/2}$ is the Lebesgue measure. For other values of $p$ however, it is singular with respect to the Lebesgue measure because it is supported by the set $\{ \sum {x_i \over 2^i} \mid {1\over n} \sum_k x_k \rightarrow p\}$ by the strong law of large numbers and these sets are disjoints for different values of $p$.
The same remark shows that $Y_p + Y'_p$ is singular with respect to the Lebesgue measure if $Y_p$ and $Y'_p$  are independent with the same law (associated to $X_i$ and $X'_i$ iid Bernoulli) and $E(Y_p+Y'_p) = 2p \neq {1}$. And so on. So just take $p\neq 2$ and the law of $Y_p$ is such that all its products are non-atomic singular.
A: 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.
A: You can also get this (the existence of such an example, that is) cheaply from some abstract nonsense, though I would certainly prefer a direct example.
If you put the weak $*$ topology on the probability measures, then the generic measure is purely singular continuous. In other words, the set of purely sc measures contains a dense $G_{\delta}$. In fact, here I need the more specific information that both $\{\mu: \mu_{ac}=0\}$ and $\{ \mu: \mu_{pp}=0\}$ are $G_{\delta}$'s (and dense, which is obvious), which is how the statement I just quoted is usually proved. See also Thms 1.1, 1.2 here.
Since $f_n: \mu \mapsto \mu * \mu * \ldots * \mu$ is continuous, the set $\{ \mu: f_n(\mu)_{xx}=0 \}$ still contains a dense $G_{\delta}$ set, for both $xx=ac$ and $xx=pp$. So $f_n(\mu)$ is purely singular continuous on at least a dense $G_{\delta}$ of $\mu$'s.
Since we have only countably many conditions, what you want also happens on at least a dense $G_{\delta}$ set.
A: A pisot number is an algebraic integer all of whose conjugates are < 1 in magnitude.  If $X_i$ are i.i.d, symmetric,  $\pm 1$ random variables and $\lambda$ is a pisot number look at $\sum \frac {X_i} {\lambda^i}$ whose characteristic function is $\prod cos(\frac {2 \pi t} {\lambda^i})$.  You can show that this does not go to zero at $\infty$.  The key is that for a pisot number $|||\lambda^i||| \rightarrow 0$ exponentially fast, where that norm is distance to the nearest  integer, so look in a $t =  \lambda^N$.   If it does not go 0 no power of it does either, so no power is absolutely continuous.  As to it's singularity, I think it follow from the law of pure types, but it is easily proven in case $\lambda > 2$ in which case $\sum^N \frac {X_i} {\lambda^i}$ consists of $2^N$ distinct equally weighted points.
