Lebesgue's criterion in ${\mathbb{R}}^m$: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null

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 ${\mathrm{\ell }}_{s,j}=b_{s,j}-a_{s,j}$. For every $\delta >0$, [L5] supplies naturals $N_{s,j}\ge 1$ with ${\mathrm{\ell }}_{s,j}\le \iota (N_{s,j})\delta \le {\mathrm{\ell }}_{s,j}+\delta$. 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 ${\prod }_{j<m}\iota (N_{s,j})\delta \le {\prod }_{j<m}({\mathrm{\ell }}_{s,j}+\delta )\mathrm{.}$ The finite sum of the expressions on the right is a polynomial in $\delta$ whose value at $0$ is ${\sum }_s\mathrm{vol}(R_s)$. Continuity at $0$ therefore permits a common $\delta >0$ for which the resulting finite cube cover of ${\bigcup }_s{R}_s$ has total volume below ${\sum }_s\mathrm{vol}(R_s)+\gamma$. 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 $\epsilon >0$, choose $\alpha >0$ with $\alpha \mathrm{vol}Q<\epsilon \mathrm{/}2$, 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 $r\ge 1$ and $\eta >0$, and choose a grid $P$ whose Darboux gap is below $\eta \mathrm{/}(2r)$. Let $H_P\subseteq Q$ be the finite union of the pieces of the coordinate hyperplanes forming cell boundaries. Every point of $S_{1\mathrm{/}r}\setminus H_P$ lies in the interior of a unique cell whose oscillation is at least $1\mathrm{/}r$. Thus the total volume of these high-oscillation cells is below $\eta \mathrm{/}2$.

Cover $S_{\alpha }$ by finitely many cubes and enlarge them so that their interiors still cover $S_{\alpha }$, keeping their total volume below $\epsilon \mathrm{/}(8(B+1))$. 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 $\epsilon \mathrm{/}(4(B+1))$.

The set $S_{1\mathrm{/}r}$ 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 $K$.

Take the high-oscillation cells from step 1.3 together with the degenerate rectangles forming $H_P$ in step 2.2. This finite rectangle family covers $S_{1\mathrm{/}r}$ by step 2.2, and its total volume is below $\eta \mathrm{/}2$ by step 1.3 because the hyperplane pieces have volume zero. Apply the finite rectangle-to-cube claim of step 1.1 to this family with $\gamma =\eta \mathrm{/}2$ to obtain a finite cube cover of $S_{1\mathrm{/}r}$ of total volume below $\eta$. Since $\eta >0$ was arbitrary, $S_{1\mathrm{/}r}$ 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 $O$.

By [L1] and [L6], $D={\bigcup }_{r\ge 1}{S}_{1\mathrm{/}r}$. Fix $\epsilon >0$. Step 3.2 gives each $S_{1\mathrm{/}r}$ a finite cube cover of arbitrarily small total volume. Enlarge each cube slightly to a cube with rational lower coordinates and rational side length whose interior contains the original cube. Density of the rationals and continuity of the finite sum of volumes ensure that, starting with volume below $\epsilon 2^{-r-1}$, this can be done with total volume below $\epsilon 2^{-r}$. This is a finite selection for each fixed $r$.

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

Among the natural-number codes in [L7] for finite rational-cube lists whose interiors cover $S_{1\mathrm{/}r}$ and whose total volume is below $\epsilon 2^{-r}$, take the least code. Step 4.2 shows this set of codes is nonempty. This rule uniquely defines the list for every $r$ without countable choice. Pad each list by zero-volume cubes and enumerate all pairs of indices by the explicit pairing bijection. The resulting sequence covers $D$. Each finite volume sum is bounded by the sum of the budgets $\epsilon 2^{-r}$, hence by $\epsilon$; [L8] gives convergence with sum at most $\epsilon$. Thus $D$ is null. Together with step 5.1 this proves both directions.