The Riemann Integral in R^m and Jordan Content

Insert one coordinate hyperplane. Every new cell lies in one old cell, so its infimum is no smaller and its supremum no larger. Splitting the affected coordinate sum proves the four inequalities.

Only fine cells meeting an interior hyperplane of $P$ can cross a coarse-cell boundary. For a hyperplane perpendicular to coordinate $j$, those cells lie in a slab of thickness at most $2\delta$; repeated product distributivity bounds their total volume by $2\delta\prod_{r\ne j}(b_r-a_r)$.

Iterating over the finitely many inserted hyperplanes and coordinates proves refinement monotonicity.

Sum this bound over the finitely many fixed interior hyperplanes to define $C_P$. On all other cells refinement changes no coarse bound, while on boundary cells each contribution changes by at most $2B$ times its volume.

This yields the quantitative estimate and completes both assertions.

If the two integrals equal $I$, choose $P_-$ with $L(f,P_-)>I-\varepsilon/2$ and $P_+$ with $U(f,P_+)<I+\varepsilon/2$. A common refinement $P$ has gap below $\varepsilon$.

Conversely, a common refinement shows every lower sum is at most every upper sum, so for every $P$, $0\le\overline{\int_Q}f-\underline{\int_Q}f\le U(f,P)-L(f,P)$. Arbitrarily small gaps force the integral difference to be $0$.

Thus the conditions are equivalent.

If $f$ is Darboux integrable, choose a fixed grid $P_0$ with small gap by [L2]. For any sufficiently fine $P$, refine it with $P_0$; [L3] makes the Darboux bounds of $P$ differ from those of the refinement by arbitrarily little. Since the refined lower and upper sums trap $I$, [L1] makes every tagged sum over $P$ close to $I$.

Conversely, suppose every sufficiently fine tagged sum is close to $I$. By [L5], choose one grid below the convergence mesh threshold and, using [L4], tag each cell near its supremum and then near its infimum. The two tagged sums approximate $U(f,P)$ and $L(f,P)$, so their common closeness to $I$ makes the Darboux gap arbitrarily small.

Apply [L2] in step 1.2. Since the near-upper and near-lower tagged sums are both arbitrarily close to $I$, the common lower/upper integral lies arbitrarily close to $I$ and therefore equals $I$. Both directions give the same value.

With one coordinate, nondegeneracy says $a<b$. A grid is one ordinary partition of $[a,b]$, its cells are its subintervals, and their volumes are their lengths. The iterated cell sum has one index and is the ordinary finite sum.

Therefore the lower, upper, and tagged sums agree term for term; taking extrema or mesh limits gives identical integrability classes and values.

Refine grids good for $f$ and $g$. Cellwise supremum and infimum estimates make the gap of $\alpha f+\beta g$ at most $|\alpha|$ times the gap of $f$ plus $|\beta|$ times that of $g$; tagged-sum linearity identifies the value.

Termwise $f\le g$ gives monotonicity of every tagged sum and hence of integrals. By [L3], the oscillation of $|f|$ on a cell is no larger than that of $f$, so $|f|$ is integrable; $-|f|\le f\le|f|$ then gives the absolute-value estimate.

Insert the cut coordinate into the grid. [L2] splits every Darboux or tagged sum into the two subrectangle sums. Good grids splice conversely, proving integrability on $Q$ exactly when both restrictions are integrable, and proving additivity.

These arguments establish all clauses with positively oriented rectangles.

Given $\varepsilon>0$, use [L2] with oscillation target $\varepsilon/(1+\operatorname{vol}Q)$ and choose a grid whose mesh is below the resulting sup-metric radius.

Every cell then has oscillation below that target. Since cell volumes sum to $\operatorname{vol}Q$, the Darboux gap is below $\varepsilon$.

The multidimensional Riemann criterion proves integrability.

Under the identification, countable cube covers and their volume-series bounds are word for word the countable interval-cover conditions.

The same is true for finite covers and finite sums.

Hence both implications hold for nullity and for content zero.

Subset closure follows because any cover of a set covers every subset.

Given $\varepsilon>0$, choose for each $j$ a cube cover of $E_j$ with total volume at most $\varepsilon2^{-j-1}$. This simultaneous selection uses [L2].

Enumerate the doubly indexed cubes through a bijection $\mathbb N\to\mathbb N^2$. Every finite partial sum is contained in a finite rectangle of indices and is at most $\sum_j\varepsilon2^{-j-1}\le\varepsilon$.

By [L4], the enumerated nonnegative volume series converges with sum at most $\varepsilon$. The cubes cover $\bigcup_jE_j$, proving nullity.

One implication is [L1]. For the other, fix $\varepsilon>0$ and choose a countable closed-cube cover of $K$ with total volume below $\varepsilon/2$.

Enlarge the $j$-th cube to a larger closed cube whose interior contains it, choosing the added volume below $\varepsilon2^{-j-2}$. The interiors form an open cover and the total volumes of their closed containing cubes are below $\varepsilon$.

Compactness selects finitely many of those interiors. The corresponding finite family of closed enlarged cubes still covers $K$, and its volume sum is at most the entire nonnegative series, hence below $\varepsilon$.

Thus $K$ has content zero.

If $f$ is continuous at $c$, choose a ball on which $|f(x)-f(c)|<\varepsilon/3$; pairwise differences are then below $2\varepsilon/3$, so the ball oscillation is at most $2\varepsilon/3<\varepsilon$ and $\omega_f(c)=0$.

If $\omega_f(c)=0$, choose $r$ with ball oscillation below $\varepsilon$. Holding one point at $c$ gives $|f(x)-f(c)|<\varepsilon$, proving continuity.

If $\omega_f(c)<\varepsilon$, choose $r$ with $\omega_f(A\cap B(c,r))<\varepsilon$. Every $d\in A\cap B(c,r/2)$ has a sufficiently small ball contained in $B(c,r)$, so $\omega_f(d)<\varepsilon$. Thus the sublevel set is relatively open.

Steps 1.1 and 1.2 give the equivalence; step 1.3 makes the complementary superlevel set closed.

Intersect each covering rectangle with $Q$. Each nonempty intersection is a closed coordinate rectangle $R_j\subseteq Q$ with volume no larger than the original rectangle. If some $R_j=Q$, the one-cell grid already has total meeting-cell volume $\operatorname{vol}(Q)\le V<V+\eta$, so assume otherwise. Move every coordinate face of each $R_j$ that is not already a face of $Q$ outward by a positive margin, staying inside $Q$, so that the resulting rectangle $R_j^+$ has volume increase below a prescribed share of $\eta$. Continuity of the finite volume product and finiteness make the total increase below $\eta$; because no $R_j$ equals $Q$, at least one face of every $R_j$ moves, and the finite set of chosen margins has a positive least member.

Insert every endpoint of every $R_j^+$ into the coordinate grids, then refine to mesh smaller than the least margin using For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$. If a closed cell meets $R_j$, each of its coordinate intervals lies inside the corresponding enlarged interval: away from a face of $Q$ this follows from the mesh-margin bound, while at a face of $Q$ there is no cell on the outside. Hence that cell lies in $R_j^+$.

Assign each cell meeting $E$ to one $R_j$ that it meets. By step 2.1 it lies in the aligned rectangle $R_j^+$. Splitting the iterated sums bounds the assigned cells' total volume by $\sum_j\operatorname{vol}(R_j^+) < V+\eta$.

The constructed grid has the asserted control.

Finite rectangle-to-cube claim. Let $R_s=\prod_{j<m}[a_{s,j},b_{s,j}]$, for $s$ in a finite index set, and let $\gamma>0$. Put $\ell_{s,j}=b_{s,j}-a_{s,j}$. For every $\delta>0$, [L5] supplies naturals $N_{s,j}\ge1$ with $\ell_{s,j}\le\iota(N_{s,j})\delta\le\ell_{s,j}+\delta$. Partition the interval from $a_{s,j}$ into $N_{s,j}$ consecutive intervals of length $\delta$, allowing the last one to extend past $b_{s,j}$. Their Cartesian products are closed cubes of side $\delta$ covering $R_s$, and their total volume is $\prod_{j<m}\iota(N_{s,j})\delta \le \prod_{j<m}(\ell_{s,j}+\delta).$ The finite sum of the expressions on the right is a polynomial in $\delta$ whose value at $0$ is $\sum_s\operatorname{vol}(R_s)$. Continuity at $0$ therefore permits a common $\delta>0$ for which the resulting finite cube cover of $\bigcup_sR_s$ has total volume below $\sum_s\operatorname{vol}(R_s)+\gamma$. This includes degenerate rectangles: every zero side contributes a factor $\delta$, so its covering volume tends to $0$.

Suppose the discontinuity set $D$ is null. Given $\varepsilon>0$, choose $\alpha>0$ with $\alpha\operatorname{vol}Q<\varepsilon/2$, and put $S_\alpha=\{x:\omega_f(x)\ge\alpha\}$. Then $S_\alpha\subseteq D$ is relatively closed in $Q$, hence compact, and is null.

Conversely, suppose $f$ is integrable. Fix $r\ge1$ and $\eta>0$, and choose a grid $P$ whose Darboux gap is below $\eta/(2r)$. Let $H_P\subseteq Q$ be the finite union of the pieces of the coordinate hyperplanes forming cell boundaries. Every point of $S_{1/r}\setminus H_P$ lies in the interior of a unique cell whose oscillation is at least $1/r$. Thus the total volume of these high-oscillation cells is below $\eta/2$.

Cover $S_\alpha$ by finitely many cubes and enlarge them so that their interiors still cover $S_\alpha$, keeping their total volume below $\varepsilon/(8(B+1))$. Apply [L3] to the union of the enlarged cubes, with the remaining volume budget, to obtain a grid $P_0$ whose cells meeting that union have total volume below $\varepsilon/(4(B+1))$.

The set $S_{1/r}$ is contained in the union of the high-oscillation cells and the finitely many pieces forming $H_P$. Each hyperplane piece is a degenerate rectangle of volume $0$.

Let $O$ be the union of those cube interiors and $K=Q\setminus O$. The set $K$ is relatively closed in compact $Q$, hence compact by [L2]. For every $z\in K$, $\omega_f(z)<\alpha$, so some Euclidean ball about $z$ has oscillation below $\alpha$. Shrink these balls by a factor of two; compactness gives a finite subcover of $K$.

Apply the finite rectangle-to-cube claim of step 1.1 to that finite family, with $\gamma=\eta/2$. Its rectangle-volume sum is below $\eta/2$, so $S_{1/r}$ has a finite cube cover of total volume below $\eta$. Since $\eta>0$ was arbitrary, $S_{1/r}$ has content zero and is null.

Refine $P_0$ to mesh small enough that the fixed norm comparison in [L4] makes every cell meeting a shrunken ball lie inside the corresponding original ball. Every cell not meeting $O$ contains a point of $K$, hence is contained in one of those original oscillation balls; refinement does not increase the total volume of cells meeting $O$.

The Darboux gap is therefore below $\alpha\operatorname{vol}Q+2B\,\varepsilon/(4(B+1))<\varepsilon$. By [L3], $f$ is integrable.

By [L1] and [L6], $D=\bigcup_{r\ge1}S_{1/r}$. Countable-union closure makes $D$ null, with countable choice used exactly through Subsets and countable unions of null subsets of $\mathbb{R}^m$ are null and The Axiom of Countable Choice ($\mathrm{AC}_\omega$). Together with step 5.1, this proves both directions using cover-nullity only.

On a grid cell, the infimum of $1_E$ is $1$ exactly when the cell is contained in $E$, while its supremum is $1$ exactly when the cell meets $E$. Thus lower and upper sums are inscribed and covering grid approximations.

Apply [L3] to each finite outer rectangle approximation to obtain a grid whose cells meeting $E$ have arbitrarily small excess volume. For an inner approximation, shrink each nondegenerate inscribed rectangle by an arbitrarily small volume, insert the shrunken endpoints, and retain the grid cells inside it. Degenerate rectangles contribute zero. Splitting along the aligned endpoints shows that arbitrary Jordan approximations and grid approximations have the same infimum and supremum.

Equality of Jordan contents is therefore equality of the lower and upper integrals on the fixed bounding rectangle, and their common value is $\operatorname{cont}(E)$.

By [L3], choose a closed bounding rectangle $Q_0$ for $E$ and enlarge every coordinate interval by a fixed positive margin to obtain a nondegenerate rectangle $Q$ with $\overline E\subseteq Q_0\subseteq\operatorname{int}Q$. By [L1] and [L2], $E$ is Jordan measurable exactly when $\partial E$ is null.

By [L3], nullity of this compact boundary is equivalent to content zero.

Combining the equivalences proves the criterion.