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In Folland 1.5, the conversation on Cantor sets comes up. He then describes the generalized Cantor set:

Given $\{a_j\}\subset (0,1)$ let $K_0=[0,1]$, $K_1$ be $K_0$ minus the middle $a_1$th interval, and let $K_j$ be $K_{j-1}$ minus the middle $a_j$th interval of each interval that makes up $K_{j-1}$.

We then see that $m(C)=\prod_{1}^\infty (1-a_j)$, since $C=\cap K_j$. I get all of this, but what I am having a hard time resolving is: if $a_j=c$ then $m(C)=0$. I understand why from the formula, but for instance if we apply the same logic that is done to produce the standard cantor set ($a_j=1/3$) to get the Smith-Volterra-Cantor set, we arrive at a positive measured set (but isn't a_j=1/4?)

Is the generalized Cantor set really generalized? There is a difference between removing the middle $4th$ and removing [1/4,2/4) and so on.

Bernard
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CitizenSnips
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1 Answers1

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In the Smith-Volterra-Cantor set, the lengths of the intervals that get removed are multiplied by $1/4$ in each step. But, that's not what $a_j$ represents. Instead, $a_j$ represents the fraction of each of the remaining intervals that gets removed in the $j$th step.

So in the first step of the Smith-Volterra-Cantor set, you remove the middle $1/4$, and $a_1=1/4$. But in the next step, you remove intervals of length $1/4^2$ from each of the two remaining intervals. Those remaining intervals each have length $3/8$, so the fraction we're removing is $\frac{1/4^2}{3/8}=\frac{1}{6}$. That's $a_2$.

So the values of $a_j$ are changing and aren't always $1/4$. It turns out that they are shrinking fast enough that the product $\prod_{1}^\infty (1-a_j)$ ends up converging to a value greater than $0$, and so in the end $C$ has positive measure.

Eric Wofsey
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