## A Nonsingular, Not Absolutely Continuous Probability Measure

In this post, we construct a probability measure on $([0,1], \mathcal{B}([0,1]))$, where $\mathcal{B}([0,1])$ denotes the Borel $\sigma$-algebra on the unit interval, which is neither singular nor absolutely continuous. The example comes from Khoshnevisan’s textbook Probability published by AMS.

Let $\left\{X_{j}\right\}_{j=1}^{\infty}$ be a countable collection of i.i.d. Bernoulli trials with parameter $p = \frac{1}{2}$ on a probability space $(\Omega, \mathcal{A},\mathbb{P})$. Define a random variable $Y := \sum_{j=1}^{\infty}4^{-j}X_{j}$, and let $\mu$ denote the distribution of $Y$, so that $\mu$ is a probability measure on $([0,1],\mathcal{B}([0,1]))$.

Our goal is to show that $\mu$ is not absolutely continuous, is nonsingular, and is not a simple combination of the two classes of measures. We will show that $\mu(\left\{x\right\}) = 0$ for all $x \in [0,1]$ and that there exists a Borel measurable subset $A \subset \mathbb{R}^{\geq 0}$ that has zero Lebesgue measure, but $\mu(A) = 1$.

Proof. For $n \in \mathbb{Z}^{\geq 1}$, define $Y_{n} := \sum_{j=1}^{n}4^{-j}X_{j}$, and let $\mathscr{Y}_{n}$ denote the image of $Y_{n}$. Note that $\left|Y_{n}\right| = 2^{n}$. Since, for $latex n \in \mathbb{Z}^{\geq 1}$,

$\displaystyle \left|Y - Y_{n}\right|=\left|\sum_{j=n+1}^{\infty}4^{-j}X_{j}\right| \leq \sum_{j=n+1}^{\infty}4^{-j} = 4^{-(n+1)}\left(1 - \frac{1}{4}\right)^{-1} = \frac{4^{-n}}{3}$

$Y$ is almost surely within $\frac{4^{-n}}{3}$ of $y \in \mathscr{Y}_{n}$. If, for $n\in\mathbb{Z}^{\geq 1}$, we define

$\displaystyle V_{n} := \bigcup_{j=n}^{\infty}\bigcup_{y \in \mathscr{Y}_{j}}\left[y-\frac{4^{-j}}{3},y+\frac{4^{-j}}{3}\right]$

then $\mu(V_{n}) = \mathbb{P}(Y \in V_{n}) = 1$. Since $(V_{n})_{n=1}^{\infty}$ is a decreasing sequence of sets, $\mu(A) = \lim_{n \rightarrow \infty} \mu(V_{n}) = 1$, where $A := \bigcap_{n=1}^{\infty}V_{n}$.

We now show that $A$ has Lebesgue measure zero. Denote the Lebesgue measure by $m$ and observe that, for $n \in \mathbb{Z}^{\geq 1}$

$\displaystyle m(A) \leq \mu(V_{n}) \leq 2\sum_{k=n}^{\infty}\sum_{y \in \mathscr{Y}_{k}}\frac{4^{-k}}{3} = \frac{2}{3}\sum_{k=n}^{\infty}2^{k}4^{-k} = \frac{2}{3}\sum_{k=n}^{\infty}2^{-k} =\frac{2^{2-n}}{3}$

Letting $n \rightarrow \infty$, we see that $m(A) = 0$.

Lastly, we show that $\mu(\left\{x\right\}) = 0$ for all $x \in [0,1]$. We can express each $x \in [0,1]$ in base-$4$ as $x = \sum_{j=1}^{\infty}4^{-j}x_{j}$, where $x_{j} \in \left\{0,1,2,3\right\}$. Note that, for any $n \in \mathbb{Z}^{\geq 1}$,

$\begin{array}{lcl} \mu(\left\{x\right\}) =\mathbb{P}(Y = x)\leq\mathbb{P}(X_{1} = x_{1},\cdots,X_{n} = x_{n}) & = &\mathbb{P}(X_{1} = x_{1})\cdots \mathbb{P}(X_{n} = x_{n})\\ & = & 2^{-n}\end{array}$

where we use the independence of the $X_{j}$ to obtain the penultimate expression. Letting $n \rightarrow \infty$, we conclude that $\mu(\left\{x\right\}) = 0$. $\Box$