Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content

Given: A natural number $m\ge 1$, the Axiom of Countable Choice, and a bounded set $E\subseteq {\mathbb{R}}^m$.

Assuming countable choice, ${\lambda }^{\mathrm{c}\mathrm{l}}(E)={\lambda }_m^{\ast }(E)$, 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).

Assuming countable choice, ${\lambda }_m^{\ast }(E)=0$ if and only if $E$ is null in the covering sense of closed-cube covers (A subset of ${\mathbb{R}}^m$ has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers, Measure zero and content zero in ${\mathbb{R}}^m$ by countable and finite cube covers).

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^m)$ is a sigma-algebra, ${\lambda }_m$ is a complete measure on it and is the restriction of ${\lambda }_m^{\ast }$, and every set of Lebesgue outer measure zero is Lebesgue measurable of measure zero (Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume).

Assuming countable choice, every Borel subset of ${\mathbb{R}}^m$ is Lebesgue measurable (Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable).

Every set $R$ with $R^{\circ }\subseteq R\subseteq \bar{R}$ is Lebesgue measurable with ${\lambda }_m(R)={\prod }_{i<m}(b_i-a_i)$, and it gives measure $0$ to all of them whenever $a_i=b_i$ for some $i<m$ (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included).

A box with a degenerate side is Lebesgue measurable of measure $0$ (A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in ${\mathbb{R}}^n$).

For bounded $E\subseteq {\mathbb{R}}^m$ its Jordan outer content is the infimum of ${\sum }_{r<q}\mathrm{vol}(R_r)$ over finite axis-parallel rectangle covers of $E$, its Jordan inner content is the supremum of the same sums over finite families of rectangles contained in $E$ whose interiors are pairwise disjoint, and the set is Jordan measurable when the contents agree (Jordan inner and outer content and Jordan measurable bounded sets in ${\mathbb{R}}^m$, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

A metric-bounded set $E\subseteq {\mathbb{R}}^m$ is Jordan measurable if and only if its boundary $\mathrm{\partial }E$ is null, equivalently has content zero (A bounded set in ${\mathbb{R}}^m$ is Jordan measurable iff its boundary is null, equivalently of content zero).

$[a,b]:=\{x\in {\mathbb{R}}^m:a_j\le x_j\le b_j(j<m)\}$ and $\mathrm{vol}[a,b]:={\prod }_{j<m}(b_j-a_j)$ (Axis-parallel rectangles in ${\mathbb{R}}^m$ and their volume).

The boundary of $A$ is $\mathrm{\partial }A:=\bar{A}\setminus \mathrm{int}(A)$, the interior is open and contained in $A$, and $A\subseteq \bar{A}$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras), it is monotone (Measures are monotone), and it is finitely and countably subadditive (Finite and countable subadditivity of measures).

The nonnegative extended sum of a sequence in $[0,+\mathrm{\infty }]$ is the supremum of its nondecreasing partial sums (Series in the nonnegative extended real line), and for real sequences ${\sum }_{k<n}{a}_k\le {\sum }_{k<n}{b}_k$ whenever $a_k\le b_k$ throughout (Laws of finite sums and finite products, claim 4; Finite sums and finite products, by recursion).