= Solution
{c}
Set $C_0=[0,1]$. Inductively, replace each component interval of $C_n$ by its two closed end intervals, each of length $\ell_{n+1}:=q_{n+1}/2^{n+1}$. The current component length is $\ell_n=q_n/2^n$. Because $q_{n+1}<q_n$, the removed central gap has positive length
$$
\ell_n-2\ell_{n+1}=\frac{q_n-q_{n+1}}{2^n}.
$$
This is a <generalized Cantor set> construction. The three required properties are checked in parts <3/a/i>, <3/a/ii> and <3/a/iii>.
The <sets> are <closed>, <measurable>, decreasing, and contained in $[0,1]$, which has finite <Lebesgue measure>. By <continuity from above of a measure>,
$$
\boxed{m(C)=m\!\left(\bigcap_{n=0}^\infty C_n\right)=\lim_{n\to\infty}q_n.}
$$
This limit exists because $(q_n)$ is a bounded decreasing <monotone sequence>. To exclude a nondegenerate <interval> $J\subseteq C$, note that its <connectedness> forces it into a single component interval of every $C_n$: crossing two components would cross a removed gap. Its length would therefore be at most $2^{-n}$ for every $n$, which is impossible for a positive-length <interval>. Thus \b[$C$ contains no nondegenerate <interval>], even when its <Lebesgue measure> is positive.
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