Let $X$, $Y$ be independent, $X$ be discrete, and $Y$ be a singular continuous random variable

Question

Prove that $X+Y$ is singularly continuous. I've know that when $X$ is an absolute continuous random variable, $X+Y$ is absolutely continuous, which can be proved by convolution. But the above question I have know idea.

Answer 1

There is a theorem which says that if $\{f_n\}$ is any sequence of non-negative increasing functions such that $f(x)=\sum f_n(x) <\infty$ for all $x$ then $\sum f_n'(x)=f'(x)$ almost everywhere. If $X$ takes the values $x_1,x_2,\cdots$ with probablities $p_1,p_2,\cdots$ then$F_{X+Y} (z)=\sum p_k F_Y(z-x_j)$. Applying above theorem we get $F'_{X+Y} (z)=0$ almost everywhere. It is trivial to check that $F_{X+Y}$ is continuous so $F_{X+Y}$ is a continuous singular distribution.