Past exam of the options I course of the University of Oxford / 2005 / D / 1

A set $E\subseteq [0,1]$ is null when for every $\epsilon >0$ it can be covered by countably many open intervals whose lengths sum to less than $\epsilon$. Equivalently its Lebesgue outer measure is zero. This definition does not presuppose measurability of $E$.

Let $E_n$ be null. Given $\epsilon >0$, cover $E_n$ by intervals of total length less than $\epsilon 2^{-n}$, for $n\ge 1$. The union of all these interval families is countable, covers ${\bigcup }_n{E}_n$, and has total length less than ${\sum }_n\epsilon 2^{-n}=\epsilon$. Thus the countable union is null.

Take the middle-thirds Cantor set $C$. Its $n$th stage is the union of $2^n$ closed intervals of length $3^{-n}$, and contains $C$. Their total length is $(2/3{)}^n\to 0$; slightly enlarging them to open intervals gives covers of arbitrarily small total length. Therefore $C$ is null.

For uncountability map every binary sequence $({\epsilon }_j{)}_{j\ge 1}$ to ${\sum }_{j\ge 1}2{\epsilon }_j{3}^{-j}\in C$. If two sequences first differ at index $k$, their image difference in absolute value is at least $2\cdot 3^{-k}-{\sum }_{j>k}2\cdot 3^{-j}=3^{-k}>0$. Thus the map is injective. The set of binary sequences is uncountable by Cantor's diagonal argument, so $C$ is uncountable.