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I am working on the following problem from Royden's Real Analysis book:

Let $F$ be the subset of [0, 1] constructed in the same manner as the Cantor set except that each of the intervals removed at the $n^{th}$ deletion stage has length $\frac{\alpha}{3^n}$ with $0 < \alpha < 1$. Show that $m(F) = 1 - \alpha$.

When I solve this by removing the open intervals, I get the correct answer. This approach is described here.

However, when I consider the closed intervals that are $kept$ at each stage, I get a different, incorrect answer. I am not seeing why this is the case. At each $n^{th}$ stage I am keeping $2^n$ intervals of length $$\bigg(\frac{ 1- \frac{ \alpha}{3}}{2} \bigg)^n.$$

Therefore at each $n^{th}$ iteration, the measure of the closed intervals is $$C_n = 2^n \cdot \bigg(\frac{ 1- \frac{ \alpha}{3}}{2} \bigg)^n = \bigg( \frac{3- \alpha}{3} \bigg)^n $$ Now, because $F$ is the intersection of all the $C_n$, we have that for all $n$ $$m(F) \leq m(C_n) = \bigg( \frac{3- \alpha}{3} \bigg)^n$$ The last part goes to 0 as $n \to \infty$, which implies that $m(F)$ should be zero. Why does differ from the answer you get when you add up the removed open intervals?

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  • $\begingroup$ @Masacroso: Huh? Are you sure? The $n$th step of the Cantor Set (or of a "fat" Cantor set) usually consists of $2^n$ intervals all of the same length... $\endgroup$ Commented Jan 8, 2020 at 20:45
  • $\begingroup$ They are not the same though, from one iteration to the next. At each stage the length of the intervals that being kept is $( (1- \alpha/3)/2)^n$. So this shrinks as $n$ increases. However, within each iteration they should be the same, as @ArturoMagidin says $\endgroup$ Commented Jan 8, 2020 at 20:52
  • $\begingroup$ @kgupta91: I'm pretty sure you have the length of what you keep wrong. $\endgroup$ Commented Jan 8, 2020 at 20:57

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Your calculation of the length of the intervals you keep is incorrect.

On the first step, you remove an interval of length $\frac{\alpha}{3}$, so you are correct that the remaining two intervals each has length $\frac{1}{2}(1-\frac{\alpha}{3})$.

On the second step, you are removing from each of those an interval of length $\frac{\alpha}{9}$. that means that you are left with $4$ intervals, each of length $$\frac{1}{2}\left(\frac{1}{2}\left(1-\frac{\alpha}{3}\right) - \frac{\alpha}{9}\right) = \frac{1}{4}\left( 1 - \frac{\alpha}{3}-\frac{2\alpha}{9}\right) = \frac{1}{4}\left( 1- \frac{5\alpha}{9}\right).$$ But you claim that the length is $$\frac{1}{4}\left(1 - \frac{\alpha}{3}\right)^2 = \frac{1}{4}\left(1 - \frac{2\alpha}{3} + \frac{\alpha^2}{9}\right) = \frac{1}{4}\left(1 + \frac{\alpha^2-6\alpha}{9}\right).$$

In order for both computations to be equal you need $\alpha^2-6\alpha = -5\alpha$, or $\alpha^2-\alpha=0$; that is, $\alpha=0$ or $\alpha=1$. So your computation of the length of the remaining intervals is correct for the regular Cantor Set, but is off for all generalized Cantor sets with $0\lt\alpha\lt 1$.

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  • $\begingroup$ Yep, makes sense! $\endgroup$ Commented Jan 8, 2020 at 21:26
  • $\begingroup$ @kgupta91: In fact, I suspect it's hard to come up with a "nice" expression for the length of the remaining intervals in this general setting, which is why one usually goes through the calculation the other way... $\endgroup$ Commented Jan 8, 2020 at 21:27
  • $\begingroup$ Yeah, that makes sense. My mistake was thinking that (as is with the case in the original Cantor set) in the general case you are removing the same fraction from the remaining intervals at each step. But this is not the case and is why, as you say, there is no "nice" expression for the general case. $\endgroup$ Commented Jan 9, 2020 at 17:58

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