I need to prove that the distribution of $X=\sum\limits_{n=1}^\infty \frac{X_n}{e^{n}}$ has a Lebesgue measure zero, where $X_n$ is a discrete random variable such that $P(X_n=1)=\frac12$ or $P(X_n=-1)=\frac12$.
My approach is , by proving that $\sum\limits_{n=1}^\infty \frac{X_n}{e^{n}}$ is the distribution of a cantor set.
But as far as i know, the distribution of a cantor set is $\sum\limits_{n=1}^\infty \frac{{2}^{n-1}}{3^{n}}$ .Can anyone help me to relate these two distributions?
I would try the following idea. Denote by $f(t)$ the characteristic function of $X$ i.e. $f(t)=\mathbf{E}e^{itX}$. By using the properties of expectation and the fact that $(e^{ix} + e^{-ix})/2=\cos(x)$, (if i did not make a mistake) we immediately obtain that for any $0<t<\infty$, $f(t)=\prod_{n=1}^\infty \cos(te^{-n})=0$, since $|\cos(x)|\le 1$. This may happen if and only if $X$ has Lebesgue measure zero.