# A family of continuous distribution functions with a singular law to the Lebesgue measure

Given a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$, let $$\{X_n: n \ge 1\}$$ be a sequence of i.i.d random variables with the common distribution $$\mathbb{P}(X_1 = 1) = p \text{ and } \mathbb{P}(X_1 = 0) = 1-p.$$ Define a mapping $$D: \Omega \to [0,1]$$ by $$D = \sum_{n = 1} ^ {\infty} \frac{X_n}{2^n}.$$ Obviously $$D$$ is well-defined as a random variable because the series above is convergent almost surely. Denote by $$\mu_p$$ and $$F_p$$ the law and the distribution function of $$D$$ respectively. (therefore, $$\mu_p$$ is a probability measure on $$([0, 1] , \mathcal{B} ([0, 1])))$$.

(i) Prove that $$F_p$$ is a continuous function.

(ii) Prove that $$\mu_p \left( \left\{x \in [0, 1]: \lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^n E_n(x) = p \right\} \right) = 1,$$ where, for each $$x \in [0,1]$$ and $$n\ge 1$$ $$E_n(x) := \begin{cases} 1, & \text{ if } 2^{n-1}x - [2^{n-1}x] \ge 0.5 \\ 0, & \text{ if } 2^{n-1}x - [2^{n-1}x] < 0.5\end{cases},$$ where $$[s]$$ denotes the integer part of $$s\ge 0$$.

From this argue that if $$p_1 \neq p_2$$, then $$\mu_{p_1} \perp \mu_{p_2}$$ , and in particular, if $$p \neq \frac{1} {2}$$ , $$\mu_p \perp \lambda$$ where $$\lambda$$ is the Lebesgue measure on $$\mathbb{R}$$.

For the first part, one can argue that $$F_p$$ is continuous iff $$\mu_p(\{x\}) = 0, \forall x \in [0, 1]$$. On the other hand, $$\mu_p$$ for each interval $$\left[\frac{k}{2^n}, \frac{k+1}{2^n}\right], k = 0, \ldots 2^n$$ can be readily shown to be $$\mu_p\left(\left[\frac{k}{2^n}, \frac{k+1}{2^n}\right]\right) = p^{n_1(n, k)}(1-p)^{n_0(n, k)}$$, where $$n_1(n, k)$$ and $$n_0(n, k)$$ are obtained by writing $$k$$ in the binary basis, padded with zeros on the left until it reaches $$n$$ binary digits, and counting the number of 1s and 0s. For example $$\mu_p([0.25, 0.375]) = \mu_p([\frac{2}{2^3}, \frac{2+1}{2^3}]) = \mu_p([0.010, 0.011]) = p(1-p)^2$$. Now for each $$x \in [0,1]$$ there exists $$\{k_{x,n}, n\ge 1\}$$ such that $$\{x\} = \bigcap_{n=1}^{\infty} [\frac{k_{x,n}}{2^n}, \frac{k_{x,n} +1}{2^n}]$$, which implies $$\mu_p(\{x\}) = \lim_{n \to \infty} p^{n_1(n, k_{x,n})}(1-p)^{n_0(n, k_{x,n})} = 0.$$

As for the second part, I understand that $$\{E_n(x)\}$$ gives the binary representation of $$x \in [0,1]$$, but I cannot wrap my head around the claim of the problem. As a remark, it is stated that "the result in this problem offers a family of examples of continuous distribution functions that do not have probability density functions, i.e., if $$p \neq 1$$, $$F_p$$ is a continuous distribution function but $$\mu_p$$ is singular to the Lebesgue measure."

Any hint is appreciated.

• I don't know if this help. Have you thought of the measure of the set $$\bigg\{x\in[0,1]:\liminf_n\frac{1}{n}\sum_{j=1}^nE_n(x)<\alpha<\beta<\limsup_n\frac{1}{n}\sum_{j=1}^nE_n(x)\bigg\}$$ where $\alpha$, $\beta$ are rationals such that $\alpha<p<\beta$? – ei2kpi Nov 15 '18 at 13:47