Two questions:
(1.) Construct a subset of $[0,1]$ in the same manner as the Cantor set, except that at the $k$-th stage, each interval removed has length $\delta 3^{-k}$, $0<\delta <1$. Show that the resulting set is perfect, has measure $1-\delta$, and contains no intervals.
Showing that it's perfect is not difficult. The resulting set is an intersection of closed intervals, so is closed. Any point in the set is a limit point of endpoints of intervals (obviously I have more details in my written solution, but that's the idea.) But I don't see how the measure is $1-\delta$... At stage $k$, we remove $2^{k-1}$ intervals of length $\delta 3^{-k}$, so the measure of the resulting set is $$1-\sum\limits_{k=0}^{\infty}{2^{k}\cdot\dfrac{\delta}{3^{k+1}}} = 1-\dfrac{\delta}{3}\cdot\sum\limits_{k=0}^{\infty}{\dfrac{2^k}{3^k}} = 1-\dfrac{2\delta}{3} .$$
Thoughts?
(2.) Construct a Cantor-type set subset of $[0,1]$ by removing from each interval remaining at the $k$-th stage a subinterval of relative length $\theta_k$, $0<\theta_k<1$. Show that the remainder has measure zero if and only if $\sum{\theta_k}=\infty$. (Use the fact that for $a_k>0$, $\prod\limits_{k=1}^{\infty}{a_k}$ converges, in the sense that $\lim\limits_{N\to\infty}{\prod\limits_{k=1}^N{a_k}}$ exists and is not zero, if only if $\sum\limits_{k=1}^{\infty}{\log{a_k}}$ converges. )
I don't really understand the construction. Suppose we have our $\theta_1$ and we remove an interval of that length. So then we choose $\theta_2$ so that $\theta_2 < \dfrac{1-\theta_1}{2}$, or something like that? Or... ? The phrasing is weird to me. Secondly, the hint does not help at all; it is complete opaque to me, so any insight you can offer would be great.