A bounded set in ${\mathbb{R}}^m$ is Jordan measurable iff its boundary is null, equivalently of content zero

Given: Metric-bounded $E$ in the sense of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space.

If $Q$ is a nondegenerate rectangle with $\bar{E}\subseteq \mathrm{int}Q$, then the relative-domain indicator $1_E:Q\to \mathbb{R}$ is discontinuous exactly at the ambient boundary $\mathrm{\partial }E$. At a boundary point every sufficiently small ambient ball lies in $Q$ and meets both $E$ and its ambient complement, while away from the boundary the indicator is locally constant (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

Indicator integrability is Jordan measurability (A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content), and integrability is equivalent to a null discontinuity set (Lebesgue's criterion in ${\mathbb{R}}^m$: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null).

The boundary is closed: it is the intersection of the closed set $\bar{E}$ with the complement of the open set $\mathrm{int}E$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, The closure of a nonempty $A$ is $\{x:d(x,A)=0\}$, equals $A$ together with its limit points, and is the smallest closed superset, 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). By Jordan inner and outer content and Jordan measurable bounded sets in ${\mathbb{R}}^m$, metric boundedness places $E$ in a closed bounding rectangle $Q$. Since $Q$ is closed and contains $E$, the smallest-closed-superset property gives $\mathrm{\partial }E\subseteq \bar{E}\subseteq Q$. Thus $\mathrm{\partial }E$ is closed and bounded, hence compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, and compact nullity is equivalent to content zero (For compact subsets of ${\mathbb{R}}^m$, measure zero and content zero coincide).