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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?

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1 Answer 1

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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.

What is your background on Measure Theory?

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  • $\begingroup$ Background is very minimal. I really only understand the basics. Could you explain your answer a little more? $\endgroup$
    – bkarthik
    Commented May 29, 2018 at 1:47
  • $\begingroup$ 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 $1/3^n$ $\endgroup$
    – Endov
    Commented May 29, 2018 at 1:53
  • $\begingroup$ 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. $\endgroup$
    – Endov
    Commented May 29, 2018 at 1:55
  • $\begingroup$ This makes much more intuitive sense. Much obliged. $\endgroup$
    – bkarthik
    Commented May 29, 2018 at 1:56
  • $\begingroup$ You're welcome. I have add here that measure theory is more profound then what I just explained, if you ever want to study it a good reference book is Folland's "Real Analysis". $\endgroup$
    – Endov
    Commented May 29, 2018 at 1:58

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