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:

- 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 $a_1$).
- 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.