Vrftjkry

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 $nge 1$
$$E_n(x) := begin{cases} 1, & text{ if } 2^{n-1}x - [2^{n-1}x] ge 0.5 \ 0, & text{ if } 2^{n-1}x - [2^{n-1}x] < 0.5end{cases}, $$
where $[s]$ denotes the integer part of $sge 0$.

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 $$mu_pleft(left[frac{k}{2^n}, frac{k+1}{2^n}right]right) = p^{n_1(n, k)}(1-p)^{n_0(n, k)}$$, 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 ${k_{x,n}, nge 1}$ 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.