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 $K$.

Apply the finite rectangle-to-cube claim of step 1.1 to that finite family, with $\gamma =\eta \mathrm{/}2$. Its rectangle-volume sum is below $\eta \mathrm{/}2$, so $S_{1\mathrm{/}r}$ has a finite cube cover 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 $O$.

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

By [L1] and [L6], $D={\bigcup }_{r\ge 1}{S}_{1\mathrm{/}r}$. 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 (${\mathrm{A}\mathrm{C}}_{\omega }$). Together with step 5.1, this proves both directions using cover-nullity only.