Let $X_n$ be Ber(1/2).
To show $\sum_{n=0}^\infty\frac{X_n}{\gamma^n}$ is singular to the Lebesgue Measure for all $\gamma > 2$.
Of course for $\gamma = 2$ this decomposes into the ordinary Lebesgue Measure over $[0,1]$. This seems like an interesting problem and would seem to give better understanding of measure theoretic probability.
EDIT I think for $\gamma = 3$ or $\gamma=4$ one can use an argument from Cantor Ternary set or quaternary set. But I dont think that's going to solve the problem
Best Answer
Let $\gamma >2$. For all $n \geq 0$,
$$0 \leq \sum_{k=n}^{+\infty} \frac{X_n}{\gamma^n} \leq \sum_{k=n}^{+\infty} \gamma^{-n} = \frac{\gamma^{-n}}{1-\gamma^{-1}}.$$
Hence, $\mu$ is supported by the union $A_n$ of $2^n$ intervals:
$$A_n := \bigcup_{(x_0, \ldots, x_{n-1}) \in \{0, 1\}^n} \left[\frac{x_k}{\gamma^k}, \frac{x_k}{\gamma^k}+\frac{\gamma^{-n}}{1-\gamma^{-1}}\right].$$
Hence, $\mu(A_n) = 1$ for all $n$. In addition, $\text{Leb} (A_n) \leq \frac{2^n \gamma^{-n}}{1-\gamma^{-1}}$.
As a consequence, $\mu \left(\bigcap_{n \geq 0} A_n\right) = 1$ and $\text{Leb} \left(\bigcap_{n \geq 0} A_n\right) = 0$, so that $\mu$ and $\text{Leb}$ are mutually singular.