# A question regarding the absolute continuity of the distribution of infinite coin flips

Here's the question I'm hopelessly on:

Let $X_1, X_2, \dots$ be an $iid$ sequence of Bernoulli random variables on $(\Omega,\mathcal{F},\mathbb{P})$ with $\mathbb{P}(X_i = 1)=1/2$. Let

$$X = 3\sum_{k=1}^\infty 4^{-k}X_k.$$

Show that the distribution function $F(x)$ of $X$ is not absolutely continuous w.r.t the lebesgue measure.

I've compared this to the cantor function, which shares many of the same properties. What I'm trying to show is that $F$ only changes values on a set of Lebesgue measure zero, but without success.

Trying to find a set with lebesgue measure zero but positive probability hasn't bourne fruit either.

-

Claim: For every $n\geqslant0$, let $Y_n=3\sum\limits_{k=1}^{n}4^{-k}X_k$ and $Z_n=3\sum\limits_{k=n+1}^{+\infty}4^{-k}X_k$. Then $Y_n$ can take (at most) $2^n$ values and $0\leqslant Z_n\leqslant4^{-n}$.

Consequence: For every $n\geqslant0$, $X=Y_n+Z_n$ hence $\mathbb P(X\in U_n)=1$ where $U_n$ is the union of (at most) $2^n$ intervals of length $4^{-n}$. The Lebesgue measure of $U_n$ is (at most) $2^n\cdot4^{-n}=2^{-n}\to0$ when $n\to\infty$. Let $U=\bigcap\limits_{n\geqslant0}U_n$. Then $\mathbb P(X\in U)=1$ and the Lebesgue measure of $U$ is zero.

This proves the distribution of $X$ is purely singular.

-

Step 1. Move to a nice setting.

It causes no difference if we realize such an i.i.d. sequence $(X_i)$ on the space

$$(\Omega, \mathcal{F}, \Bbb{P}) = ([0, 1), \mathcal{B}, \lambda),$$

where $\mathcal{B}$ is the Borel field on $[0, 1)$ and $\lambda$ is the Lebesgue measure, such that for each $\omega \in \Omega$

$$X_i(\omega) = i\text{-th digit of } \omega \text{ in base 2 }.$$

This definition has no ambiguity unless $\omega$ is rational, which is contained in a null-set $\Bbb{Q} \cap \Omega$. But for the sake of simplicity, we modify its value on this null-set so that $X_i$ is right-continuous. Then the resulting $(X_i, i \geq 1)$ is a sequence of independent Bernoulli random variables.

Now, as a function on $[0, 1)$, we may extend each $X_i$ to a 1-periodic function $f_i : \Bbb{R} \to \{0, 1\}$. Then it is easy to observe that $f_{i+1}(x) = f_i(2x)$. Thus we may write

$$X_i(\omega) = f_{1}(2^{i-1}\omega).$$

Thus the partial sum

$$S_n (\omega) = 3 \sum_{i=1}^{n} 4^{-i} X_i (\omega) = \sum_{i=1}^{n} 4^{-i} f_1(2^{i-1}\omega)$$

satisfies

$$S_1(\omega) = \frac{3}{4} f_1(\omega), \quad S_{n+1}(\omega) = S_1(\omega) + \frac{1}{4} S_n(2\omega). \tag{1}$$

We also note that the limit $X = \lim_{n\to\infty} S_n$ converges both absolutely and uniformly.

Step 2. Identify some important properties of the limit $X$

Let $C$ be a Cantor-like set obtained by the following procedure: Starting from the unit interval $[0, 1]$, delete the concentric open interval with the length half of the original interval. Repeat this process to each remnant closed intervals. Clearly, $C$ is a closed set with zero measure. Moreover, from its construction it is clear that $C$ has the following self-similarity: $$C = \left( \frac{1}{4}C \right) \cup \left( \frac{3}{4} + \frac{1}{4}C \right). \tag{2}$$

Now we claim that for each $n$,

1. $S_n (\omega)$ is monotone increasing and right continuous.
2. The range of $S_n (\omega)$, which is denoted by $\mathrm{Rng} \, S_n = \{ S_n(\omega) : \omega \in \Omega \}$, is contained in $C$.

Indeed, for $n = 1$ the property 1 is clear. Also, $\mathrm{Rng} \, S_1 = \{0, \frac{3}{4} \}$ is clearly a subset of $C$. Next, assume both the property 1 and 2 hold for $n$. Then by $(1)$, the property 1 still holds for $n+1$ and

$$\mathrm{Rng} \, S_{n+1} = \left( \frac{1}{4} \mathrm{Rng} \, S_{n} \right) \cup \left( \frac{3}{4} + \frac{1}{4} \mathrm{Rng} \, S_{n} \right) \subset \left( \frac{1}{4}C \right) \cup \left( \frac{3}{4} + \frac{1}{4}C \right) = C,$$

from which the claim is confirmed.

Since $C$ is closed and $S_n(\omega)$ converges uniformly to $X$, we find that $X$ is also monotone-increasing and right-continuous, and that the range of $X$ is also contained in $C$. It is also trivial to check that $X(1-) = 1$.

Step 3. Prove the claim.

Let $F$ be the CDF of $X$. That is,

$$F(x) = \Bbb{P}(X \leq x).$$

Clearly we have

$$F(x) = 0 \text{ if } x \leq 0 \quad \text{and} \quad F_X(x) = 1 \text{ if } x \geq 1.$$

This enables us to restrict our attention to the case $x \in (0, 1)$. Since $X$ is monotone increasing and right-continuous,

$$F(x) = \Bbb{P}(X \leq x) = \lambda (X^{-1}[0, x]) = \max \{\omega \in \Omega : X(\omega) \leq x \}.$$

But since $C^c = [0, 1] \setminus C$ is an open set and contains no values of $X$, it is clear that $F$ is locally constant on $C^c$. In particular, we have $F'(x) = 0$ on $C^c$. Since $\lambda(C^c) = 1$, this proves that

$$F'(x) = 0 \quad \text{a.e.}$$

Therefore $F$ is not absolutely continuous with respect to the Lebesgue measure. (In fact, this proves that $F$ is singular.) ////

Please tell me if there is a typo or an error. It's 4:00 a.m. here and I'm half-nodding.

-