Suppose $a\_n > 0$ is a sequence of real numbers in $l^2 \setminus l^1$. i.e. $\sum a\_n^2 < \infty$ but $\sum a\_n = \infty$.


If $B\_n$ are an infinite sequence of independent Bernoulli random variables with parameter $\frac{1}{2}$ then the infinite sum $\sum a\_n (-1)^{B\_n}$ exists almost surely and converges in $L^2$ and almost surely to some random variable, say $A$ (standard result from Martingale theory, although you can prove it directly).


$A$ has the property for any $u < v$, $P(u < A < v) > 0$, so it is "almost continuous", but it's not obvious to me that the distribution can't have atoms. i.e. values $u$ such that $P(A = u) > 0$.


So my question is:


1. Does a sequence $a\_n$ exist with $P(A = 0) > 0$? (if a sequence with any atom exists, one with $0$ being an atom exists because you can just shift $a\_1$).
2. For the specific sequence $a\_n = \frac{1}{n}$ does $A$ have any atoms?


I conjecture the answer to both is no, but I'm struggling with the details of proving it.



Can a probability distribution from summing alternating signs have atoms?
The answer is no.


For any $T \subseteq \mathbb N$, let  

$$ A(T) = \sum\_{n \in T} a\_n (-1)^{B\_n}$$
Now $A = A(T) + A(\mathbb N \backslash T)$ where $A(T)$ and $A(\mathbb N \backslash T)$ are independent, and so if $A$ had atoms both $A(T)$ and $A(\mathbb N \backslash T)$ would have atoms. Thus it suffices to find $T$ 
such that $A(T)$ has no atoms.


In particular, take an increasing sequence $T = \{t\_1, t\_2, \ldots\}$ such that $0 < a\_{t\_{j+1}} < a\_{t\_j}/2$. Then the function $f\_T(b\_1, b\_2, \ldots) = \sum\_{j} a\_{t\_j} b\_j$ on $\{-1,1\}^\mathbb N$ is one-to-one. For any real $c$ there is at most one sequence of values of $B\_t$, $t \in T$ that makes $A(T) = c$, and thus $P(A(T)=c) = 0$. 


