Jordan inner and outer content and Jordan measurable bounded sets in $\mathbb{R}^m$

For bounded $E\subseteq\mathbb R^m$, in the metric sense of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, its Jordan outer content is the infimum of $\sum_{r<q}\operatorname{vol}(R_r)$ 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.

Metric boundedness always supplies a nondegenerate bounding rectangle. For nonempty $E$, choose $x_0\in\mathbb R^m$ and $r>0$ with $E\subseteq B(x_0,r)$. Since $|x_j-(x_0)_j|\le d_\infty(x,x_0)\le d_2(x,x_0)<r$ for every coordinate ($\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it, The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$ clause 3, Open ball, closed ball and sphere in a metric space), the nondegenerate box $\prod_{j<m}[(x_0)_j-r,(x_0)_j+r]$ contains $E$. The empty set lies in any fixed nondegenerate rectangle.

Thus the outer family is nonempty and the same bounding rectangle bounds the inner sums; the empty family gives inner sum $0$. Refining all listed endpoints into one grid and splitting the nested finite sums shows every inscribed sum is at most every covering sum (Grid partitions of a rectangle in $\mathbb{R}^m$, their cells, refinements and mesh, Laws of finite sums and finite products). Completeness therefore supplies finite real extrema (Complete ordered field (least-upper-bound property), Every nonempty set bounded below has an infimum, Greatest lower bound (infimum), Suprema and infima are unique).

The set is Jordan measurable when the contents agree, and their common value is its Jordan content. The empty set and every degenerate rectangle have content $0$.