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?

Solution 1:

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.