Consider $2^\omega$, the Cantor space, as a measure space by the uniform measure. That is, define the measure of clopen sets in the Cantor space in the obvious manner, and extend it to countably additive set function by considering the outer measure. I have following question about this space.

  1. What exactly are the measureable sets in this space?
  2. For a set $S$ of the special form $S=[W]$, where $[W]$ is the set of sequences that extends some string in $W$, a set of finite string, I believe $S$ is measurable and $$\mu(S) = \sum_{x\in U} 2^{-|x|},$$ where $U$ is a prefix-free set of finite strings. How can I prove this?

I would be grateful for your help.


Let $\Sigma$ be the set of finite binary strings. For $\sigma,\tau\in\Sigma$ write $\sigma\preceq\tau$ is $\sigma$ is a prefix of $\tau$ and $\sigma\prec\tau$ if $\sigma$ is a proper prefix of $\tau$; if $\sigma\npreceq\tau\npreceq\sigma$, I’ll write $\sigma\parallel\tau$. Note that

$$\begin{cases} [\tau]=[\sigma]&\text{if }\tau=\sigma\;,\\ [\tau]\subsetneqq[\sigma]&\text{if }\sigma\prec\tau\;,\text{ and}\\ [\tau]\cap[\sigma]=\varnothing&\text{if }\sigma\parallel\tau\;. \end{cases}\tag{1}$$

For $W\subseteq\Sigma$ let $W^*=\{\sigma\in W:\neg\exists\tau\in W(\tau\prec\sigma)\}$; you can use $(1)$ to show that $\{[\sigma]:\sigma\in W^*\}$ is a (necessarily countable) partition of $[W]$, and from this it follows immediately that

$$\mu([W])=\sum_{\sigma\in W^*}\mu([\sigma])=\sum_{\sigma\in W^*}2^{-|\sigma|}\;.$$

If $C$ is the middle-thirds Cantor set, and $h:2^\omega\to C$ is the obvious homeomorphism, $S\subseteq 2^\omega$ is measurable iff $h[S]$ is Lebesgue measurable; see this question and this one, for instance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.