# Lebesgue Measure of Cantor Set

For a presentation, I am learning about the Cantor Set and how it is homeomorphic to the p-adic numbers. I was reading section two of this paper. In it it states that the Cantor Set has a vanishing Lebesgue measure.

Wikipedia says: Given a subset ${\displaystyle E\subseteq \mathbb {R} }$, with the length of interval ${\displaystyle I=[a,b]({\text{or }}I=(a,b))}$ given by ${\displaystyle \ell (I)=b-a}$, the Lebesgue outer measure ${\displaystyle \lambda ^{*}(E)}$ is defined as

$${\displaystyle \lambda ^{*}(E)=\operatorname {inf} \left\{\sum _{k=1}^{\infty }\ell (I_{k}):{(I_{k})_{k\in \mathbb {N} }}{\text{ is a sequence of open intervals with }}E\subseteq \bigcup _{k=1}^{\infty }I_{k}\right\}}$$.

Why would this be zero for the Cantor Set?

Show that the complement of the Cantor set in $[0,1]$ has measure 1. You will also have to use that the Cantor set is measurable, hence $\lambda^*([0,1]) = \lambda^*(E\cap[0,1]) + \lambda^*(E^c\cap[0,1])$, in which $E$ is the Cantor set.
• The Lebesgue measure is pretty much a way of extending the notion of "length", which is intuitive for intervals in $\mathbb{R}$, but not for general subsets. In the case of the cantor set the idea is that in the first step you subtract from $[0,1]$ a one interval of length 1/3, from that remaining set you subtract 2 intervals of length 1/9, and so on, so on the nth step you are subtracting $2^{n-1}$ intervals of length $1/3^n$ – Endov May 29 '18 at 1:53
• So in the end the total "length" you subtract from $[0,1]$ is $\sum_{n=1}^\infty 2^{n-1}/3^n = (1/2)\sum_{n=1}^\infty (2/3)^n = 1$, and since the length of $[0,1]$ is 1, the "length"(measure) of the Cantor set is 0. – Endov May 29 '18 at 1:55