The Cantor distribution is singular (with respect to lebesgue measure)

If we define the Cantor distribution $$\mu$$ as the distribution that has $$F=$$"Cantor function" as it's cumulative distribution function, how do we show that $$\mu$$ is singular with respect to the Lebesgue measure? If $$\lambda$$ is the Lebesgue measure I have to show that if $$\lambda(A)=0$$ then $$\mu(A)=0$$. For a point-set $$\{x\}\subset\mathbb{R}$$ it does hold since $$\lambda(\{x\})=0$$ and $$\mu(\{x\})=\mu(\bigcap\limits_{n=1}^{\infty}(x-\frac1n,x])=\lim\limits_{N\to\infty}\mu(\bigcap\limits_{n=1}^{N}(x-\frac1n,x])=\lim\limits_{N\to\infty}\mu((x-\frac1N,x])\lim\limits_{N\to\infty}F(x)-F(x-\frac1N)=0$$ since $$F$$ is continuous. But how to show the property for a general $$A$$ $$\lambda$$-measurable?

• Singularity means that there are two disjoint subsets (in this case of $[0,1]$) $A$ and $B$ so that $A \cup B = [0,1]$, and $\lambda(A')=0$, for all $A' \subset A$, and $\mu(B')=0$, for all $B'\subset B$. The evident candidate sets in your example are $A=$ the cantor set, and $B=$ its complement. Apr 21 '21 at 19:45
• @LostStatistician18, I was totally off then. Have I proved that $\mu$ has no atoms at least? I understand $\lambda(C)=0$ but from the definition of the Cantor measure that I have, could I show $\mu([0,1]\setminus C)=0$? Apr 21 '21 at 20:14
• Yes that seems OK! Yes the "hard part" is to show that $\mu(B') = 0$ for all subsets $B' \subset [0,1]/C$. Apr 21 '21 at 20:32

Now that I think more about this, maybe it is not as hard as I thought. Since to obtain the Cantor set we remove at each step a finite number of disjoints open intervals we can say (I am not sure if this is super rigorous) $$[0,1]\setminus C=\bigcup\limits_{n\in\mathbb{N}}(a_n,b_n)$$. So if $$\mu$$ denotes the cantor probability measure (i.e. the probability measure for which the CDF is the cantor function) then since $$F$$ is flat on the intervals $$(a_n,b_n)$$, and the intervals are disjoint, we have$$\mu([0,1]\setminus C)=\mu\left(\bigcup\limits_{n\in\mathbb{N}}(a_n,b_n)\right)=\sum\limits_{n\in\mathbb{N}}\mu(a_n,b_n)=\sum\limits_{n\in\mathbb{N}}\underbrace{F(b_n)-F(a_n)}_{=0}=0.$$ So we have $$\mu(C)=1$$. However we know that for the Lebesgue measure, $$\lambda(C)=0$$ so we conclude these measures are mutually singular.

To prove two measures $$\mu,\lambda$$ defined on $$\Omega$$ are singular, we need to prove the existence of a measurable set $$A$$ such that $$\lambda(X)=0, \forall X\subset A \text{ ; } \mu(Y)=0, \forall Y\subset \overline A$$ However, by monotonicity of measure, $$\mu(A)\leq \mu(B) \text{ if } A\subset B$$ $$\text{Hence } \mu(B)=0 \implies \mu(A)=0 \forall A\subset B$$ Thus, it is enough if we prove that $$\lambda(A)=0,\mu(\overline A)=0$$

We consider $$A$$ to be the Cantor set $$\mathcal{C}$$, $$\Omega$$ to be $$[0,1]$$. $$\mu$$ is the CDF of the Cantor distribution while $$\lambda$$ is the Lebesgue measure.

1. Proving $$\lambda(\mathcal{C})=0$$

This is quite trivial. It is a well-known fact that the Cantor set has Lebesgue measure zero.

1. Proving $$\mu([0,1]\backslash \mathcal{C})=0$$

We proceed to prove a few lemmas.

Lemma 1:$$x\in [0,1]\backslash \mathcal{C} \text{ iff } x \in[0.x_1...x_n100...\text{ , }0.x_1...x_n200...) \text{, where the interval is in ternary and }$$ $$\{x_i\}_{i\leq n}\in \{0,2\}$$

Proof:

(i) A real number belongs to the Cantor set iff its ternary expansion has only 0s and 2s. Hence, any real number not in the Cantor set must have at least one 1 in its ternary expansion. Let the $$r^{th}$$ digit in $$x$$ be the first 1 in its expansion. Then, $$0.0.x_1...x_{r-1}100... \leq x < 0.x_1...x_{r-1}200...$$ (ii) Let $$x \in[0.x_1...x_n100...\text{ , }0.x_1...x_n200...)$$. Then it is evident that $$x$$ has at least one 1 (in the $$(n+1)^{th}$$ position) in its ternary expansion. Thus, $$x\notin \mathcal{C} \implies x\in[0,1]\backslash \mathcal{C}$$

[Note: Henceforth, it is implied that $$\{x_i\}_{i\leq n}\in \{0,2\}$$. Only the interval is mentioned for readability.]

Lemma 1 implies $$[0,1] \backslash \mathcal{C}$$ consists of intervals of the form $$[0.x_1...x_n100...\text{ , }0.x_1...x_n200...)$$

Lemma 2: $$[0,1] \backslash \mathcal{C}$$ consists of countably many intervals

Proof:

Consider the construction of the Cantor set. At the $$n^{th}$$ stage, $$2^{n-1}$$ intervals are removed. Hence the total number of intervals removed from $$\mathcal{C}, \text{ i.e., present in } [0,1]\backslash \mathcal{C}$$, is countable.

Combining Lemma 1 and Lemma 2, we prove that $$[0,1]\backslash \mathcal{C}$$ consists of countably many intervals of the form $$[0.x_1...x_n100...\text{ , }0.x_1...x_n200...)$$. Now consider $$\mu \Big( [0.x_1...x_n100...\text{ , }0.x_1...x_n200...) \Big)$$. Since, the interval contains no element belonging to the Cantor set, it has zero measure. $$\therefore \mu \Big( [0.x_1...x_n100...\text{ , }0.x_1...x_n200...) \Big) = 0$$ Thus $$[0,1]\backslash \mathcal{C}$$ consists of a countable set of such disjoint intervals, each of which have measure zero. Hence $$\mu \left( \bigcup_{i\in \mathbb{N}} (a_i,b_i) \right) = \sum_{i\in \mathbb{N}} \mu\Big( (a_i,b_i) \Big) = \sum_{i\in \mathbb{N}} (0)=0$$ $$\implies \mu \big([0,1]\backslash \mathcal{C} \big)=0$$

Hence proved that the Cantor distribution and the Lebesgue measure are mutually singular.