Density function and cumulative distribution function of a random variable Suppose that $X_n$ is a sequence of random variables with probability function:
$$P(X_n=1) = P(X_n=2)=P(X_n=4)=P(X_n=5)=0.25$$ and   
$$Y= \sum_{n=1}^\infty\frac{X_{n}}{5^n}$$
I want to show that:


*

*The random variable Y hasn't density function

*Cumulative distribution function of Y is continuous.
Thanks for your hints!
 A: Presumably the sequence $(X_n)_n$ is independent. Assume that the distribution of $(X_n)_n$ with respect to the probability measure $\mathbb P$ is as described in the question and that a second probability measure $\mathbb Q$ is such that $(X_n)_n$ is still i.i.d. but now each $X_n$ is uniform on $\{1,2,3,4,5\}$. Let $Z_n=5-X_n$. Every $Z_n$ is uniform on $\{0,1,3,4\}$ for $\mathbb P$ and uniform on $\{0,1,2,3,4\}$ for $\mathbb Q$. Furthermore,
$$
Y=\sum_{n=1}^{+\infty}\frac{5-Z_n}{5^n}=\frac54-T,\qquad T=\sum_{n=1}^{+\infty}\frac{Z_n}{5^n}.
$$
Note that $0\leqslant T\leqslant1$ almost surely for $\mathbb P$ and for $\mathbb Q$, and that each $Z_n$ is $\sigma(T)$-measurable, for example, $Z_1=\lfloor5T\rfloor$, $Z_2=\lfloor25T-5Z_1\rfloor$, that is, $Z_2=\lfloor25T-5\lfloor5T\rfloor\rfloor$, and so on. Abbreviate this as $(Z_n)_n=(u_n(T))_n$ and consider the set
$$
B=\{t\in(0,1);u_n(t)=2\ \text{for infinitely many}\ n\geqslant1\}.
$$
Then $\mathbb P(T\in B)=0$ and $\mathbb Q(T\in B)=1$ hence $\mathbb P$ and $\mathbb Q$ are mutually singular. Since $T$ is uniform on $(0,1)$ with respect to $\mathbb Q$, $\mathbb P(T\in A)=1$ where $A=(0,1)\setminus B$ has Lebesgue measure zero. In particular, the distribution of $T$ with respect to $\mathbb P$ has no absolutely continuous part, and the same is true for the distribution of $Y$ with respect to $\mathbb P$. (Note that, by contrast, the distribution of $Y$ with respect to $\mathbb Q$ is purely absolutely continuous, uniform on $(\frac14,\frac54)$.)
On the other hand, the distribution of $Y$ has no atom with respect to $\mathbb P$ since, for each $N\geqslant1$, every jump of the CDF of $\sum\limits_{n=1}^{N}\frac{Z_n}{5^n}$ has size at most $\frac1{4^N}$. In other words, for every $y$, $\mathbb P(Y=y)=0$.
Take-home message: Every probability measure on $(\mathbb R,\mathcal B(\mathbb R))$ is a sum $\mu_d+\mu_c+\mu_s$ where each $\mu_*$ is a subprobability, $\mu_d$ is purely discrete, $\mu_c$ is absolutely continuous and $\mu_s$ is the remaining part, that is, a measure with neither density nor atom.
