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

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