Past exam of the integration course of the University of Oxford / 2021 / 3 / a / Solution

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 lengthThis is a generalized Cantor set construction. The three required properties are checked in parts i, ii and iii.

The sets are closed, measurable, decreasing, and contained in $[0,1]$, which has finite Lebesgue measure. By continuity from above of a measure,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 $C$ contains no nondegenerate interval, even when its Lebesgue measure is positive.