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

Given: The Axiom of Countable Choice, the case $n=1$ of Lebesgue outer measure, and a subset $A\subseteq \mathbb{R}$.

Assuming countable choice, ${\lambda }^{\mathrm{c}\mathrm{l}}(E)={\lambda }_n^{\ast }(E)$, where ${\lambda }^{\mathrm{c}\mathrm{l}}(E)$ is the infimum of ${\sum }_{k=0}^{\mathrm{\infty }}\mathrm{vol}[u^k,v^k]$ over countable covers of $E$ by closed rectangles (Countable covers by closed boxes, by open boxes and by closed cubes all compute Lebesgue outer measure, Lebesgue outer measure on ${\mathbb{R}}^n$).

$A$ has measure zero, equivalently $A$ is null, when for every real $\epsilon >0$ there are sequences $(a_k{)}_{k\ge 0}$ and $(b_k{)}_{k\ge 0}$ of reals with $a_k\le b_k$ for every $k\ge 0$, such that $A\subseteq {\bigcup }_{k\ge 0}[a_k,b_k]$ and ${\sum }_{k=0}^{\mathrm{\infty }}(b_k-a_k)$ converges with sum $\le \epsilon$ (Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover), Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).

For a fixed $\epsilon >0$, ${\sum }_{k=0}^{\mathrm{\infty }}(b_k-a_k)$ converges with sum $\le \epsilon$ if and only if ${\sum }_{k<n}(b_k-a_k)\le \epsilon$ for every $n\in \mathbb{N}$ (Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover)).

The nonnegative extended sum of a sequence in $[0,+\mathrm{\infty }]$ is ${\sum }_{k=0}^{\mathrm{\infty }}{a}_k:={\sup }_{n\in \mathbb{N}}{s}_n$, the supremum of its nondecreasing partial sums (Series in the nonnegative extended real line).

Under the standard identification ${\mathbb{R}}^1\cong \mathbb{R}$, the rectangle $[a,b]$ of ${\mathbb{R}}^1$ is the interval $[a_0,b_0]$ and its volume is its length (Axis-parallel rectangles in ${\mathbb{R}}^m$ and their volume).