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?

  • 2
    $\begingroup$ 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. $\endgroup$ Apr 21 '21 at 19:45
  • $\begingroup$ @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 $\mu([0,1]\setminus C)=0$? $\endgroup$
    – edamondo
    Apr 21 '21 at 20:14
  • $\begingroup$ Yes that seems OK! Yes the "hard part" is to show that $\mu(B') = 0$ for all subsets $B' \subset [0,1]/C$. $\endgroup$ 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\}$


(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


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.


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