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$C(n)$ is defined as the set that remains after removing from $[0, 1]$ an open interval of length 1/n centered at 1/2, then an open interval of length $1/n^2$ from the center of each of the two remaining intervals, then open intervals of length $1/n^3$ from the centers of each of the remaining $4=2^2$ intervals, and so on.

How would I show that the measure of $C(n)$ is $\frac{n-3}{n-2}$?

Any hints greatly appreciated: By drawing out examples for small n, I would think that the measure is 0 since we are continuously removing open intervals and keeping the endpoints. I also see that C(3) = 0 since C(3) is the Cantor set.

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up vote 5 down vote accepted

After $k\geqslant0$ steps of the procedure, there remains $2^k$ disjoint intervals (and not $k+1$, as you seem to believe) and one removes an interval of length $1/n^{k+1}$ from each of them. If $n\gt2$, there is no overlap of the pieces that are removed hence the total length of the removed intervals is $$ \sum_{k=0}^{+\infty}\frac{2^k}{n^{k+1}}=\frac1{n-2}. $$

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