Past exam of the options II course of the University of Oxford / 2013 / D4 / b

Given $ϵ>0$, cover $E_1$ by open intervals of total length less than $ϵ/2$, and cover $E_2$ similarly. Combining the two countable covers gives a cover of their union of total length less than $ϵ$. Thus $E_1\cup E_2$ is a null set.

For each $n\ge 1$, choose a countable interval cover of $E_n$ of total length below $ϵ/2^n$. The union of these covers is countable and covers ${\bigcup }_n{E}_n$, with total length less than ${\sum }_nϵ/2^n=ϵ$. As $ϵ$ is arbitrary, a countable union of null sets is null.