 Solution

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.