I'm learning about Measure Theory and need some help with this problem:

Let $0 < \alpha < 1$. We construct a set $C_\alpha$ (Cantor type) as follows: In the first step we remove from the interval $[0, 1]$ a "middle" open interval of length $(1 - \alpha)3^{-1}$. In the nth step we remove $2^{n-1}$ open intervals of length $(1 - \alpha)3^{-n}$. Find the Lebesgue measure of $C_\alpha$.

My work and thoughts:

If we remove the set $C_\alpha$ from the close interval $[0, 1]$ we are left with the union of pairwise disjoint intervals. In more traditionally formulaic notation, we can write:

$$[0, 1] \setminus C_\alpha = E_1 \cup E_2 \cup E_3 \cup \ldots$$

Therefore, taking the Lebesgue measure on both side of the preivous equality we get:

$$\mu \left([0, 1] \setminus C_\alpha \right) = \mu \left( \bigcup_{n=1}^{+\infty} E_n \right) = \sum_{n = 1}^{+\infty} \mu(E_n).$$

I need to find a way to express $\mu(E_n)$ and calculate the above series. If I can do so then the result is immediate since:

$$\mu(C_\alpha) = \mu([0, 1]) - \mu([0, 1]\setminus C_\alpha)$$

where $\mu([0, 1]) = 1$.

  • $\begingroup$ What's your $C_\alpha$? The remaining part? If not misunderstanding I think $\mu(E_n)$ is trivially $\frac12(1-\alpha)(2/3)^n$. $\endgroup$
    – Vim
    Mar 1, 2016 at 16:01
  • $\begingroup$ Since $C_\alpha$ is a Cantor type set I made the assumption that it is the remaining part. The problem doesn't say anything else. $\endgroup$
    – glpsx
    Mar 1, 2016 at 16:05

1 Answer 1


As to the way you construct $C_\alpha$, if you denote by $E_n$ (hopefully I'm not misinterpreting it) the removed part during the $n$th step, then clearly $$\mu(E_n)=\frac12(1-\alpha)\left(\frac23\right)^n$$ Hence $$\mu(C_\alpha)=\mu([0,1]\setminus\cup E_n)=1-\sum\mu(E_n)=1-(1-\alpha)=\alpha$$

  • $\begingroup$ You are correct about the expression of $\mu(E_n)$. Taking the infinite sum it just comes down to calculating the geometric series. Thank you for your help, much appreciated. $\endgroup$
    – glpsx
    Mar 1, 2016 at 16:20
  • $\begingroup$ @VonKar you are welcome. $\endgroup$
    – Vim
    Mar 1, 2016 at 16:23
  • $\begingroup$ Cantor set Not countable!! $\endgroup$
    – lee
    Sep 19, 2023 at 2:30

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