Let $E_n$ be null. Given $\varepsilon>0$, cover $E_n$ by intervals of total length less than $\varepsilon2^{-n}$, for $n\geq1$. The union of all these interval families is <countable>, covers $\bigcup_nE_n$, and has total length less than $\sum_n\varepsilon2^{-n}=\varepsilon$. Thus \b[the <countable union> is null].

Solution

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

 Solution (source code)