Prove that Random Variable is not absolutely continuous with respect to Lebesgue measure The random variable is as follows: 
$X = \sum_n \frac{\beta_{n}}{3^{n}}$, where $\beta_n$ is $0$ or $2$ with probability $\frac{1}{2}$. 
 A: Note that $X \in C$ where $C$ is the Cantor set, as the allowable ternary digits are either $0$ or $2$. The Cantor set has Lebesgue measure $0$. (See https://en.wikipedia.org/wiki/Cantor_set#Measure_and_probability). Hence $X$ is not absolutely continuous with respect to the Lebesgue measure, as $P(X \in C) = 1 > 0$.
A: Strictly speaking, it is the random variable's distribution, rather than the random variable itself, that is not absolutely continuous. And you probably want to add that $\beta_1, \beta_2, \beta_3, \ldots$ are independent.
Notice that if $\beta_1=2$ then $2/3\le X\le 1$ and if $\beta_1=0$ then $0\le X\le 1/3.$
Thus $X$ is constrained to lie within a set of Lebesgue measure $2/3.$
Similarly, see if you can show that, because $\beta_2$ must be $0$ or $2,$ $X$ is constrained to lie within $2/3$ of $2/3$ of the interval $[0,1].$
And then because $\beta_3$ is either $0$ or $2,$ it is within $2/3$ of $2/3$ of $2/3$ of the interval.
And so on. So the support of the distribution has measure no more than $(2/3)^n,$ for $n=1,2,3,\ldots.$
