A set is null exactly when every fine cover has arbitrarily cheap countable subfamilies covering it up to a null remainder

Assume first that $E$ is null, and let $\epsilon >0$. Choose an open set $U\supseteq E$ with $\lambda (U)<\epsilon$. Restrict $\mathcal{V}$ to intervals lying in $U$; the restricted family is still a fine cover of $E$. Applying The Vitali covering theorem for fine covers on the real line gives a countable disjoint subfamily $(J_n)$ of $\mathcal{V}$ with $\lambda \text{\hspace{0.17em}⁣}(E\setminus {\bigcup }_n{J}_n)=0.$ Because the $J_n$ lie in $U$ and are disjoint, ${\sum }_{n\ge 1}\mathrm{\mid }{J}_n\mathrm{\mid }\le \lambda (U)<\epsilon \mathrm{.}$ Thus $(J_n)$ is the required cheap countable subfamily.

Conversely, assume condition 2. Apply it to the fine cover consisting of all closed intervals with $\epsilon =1\mathrm{/}(2m)$ for each $m\ge 1$. Then there is a countable family $(I_{m,n}{)}_{n\ge 1}$ of closed intervals such that $\lambda \text{\hspace{0.17em}⁣}(E\setminus {\bigcup }_{n\ge 1}{I}_{m,n})=0\text{and}{\sum }_{n\ge 1}\mathrm{\mid }{I}_{m,n}\mathrm{\mid }<\frac{1}{2m}\mathrm{.}$ Let $N_m:=E\setminus {\bigcup }_{n\ge 1}{I}_{m,n}$. Since $N_m$ is null, cover $N_m$ by closed intervals $(K_{m,j}{)}_{j\ge 1}$ of total length $<1\mathrm{/}(2m)$. Then the combined family $(I_{m,n}{)}_{n\ge 1}\cup (K_{m,j}{)}_{j\ge 1}$ covers $E$ and has total length $<1\mathrm{/}m$. Hence $E$ has elementary measure zero by Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover). Equivalently, $E$ is Lebesgue null by A subset of $\mathbb{R}$ has Lebesgue outer measure zero if and only if it has measure zero in the sense of countable closed-interval covers.

Steps 1.1 and 1.2 prove the equivalence.