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?
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 $0\lt\alpha\lt 1$.