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

Solution

Past exam of the options II for mathematics and philosophy course of the University of Oxford / 2013 / D4 / b / i / Solution

 Solution (source code)