Finite Hardy–Krause variation implies bounded Riemann integrability

Let $\rho (x)=\mathbf{1}-x$ and $g=f\circ \rho$. Reflection sends each upper face of $f$ to the corresponding lower face of $g$, sends each product grid to a product grid, and reverses the sign of an alternating increment by at most a factor $(-1{)}^{\mathrm{\mid }u\mathrm{\mid }}$. Its absolute value is unchanged. Hence $V_{HK,0}(g)=V_{HK,1}(f)<\mathrm{\infty }$.

For $x\in [0,1{]}^d$, put $v(x)=V_{HK,0}(g;[0,x])$. Every grid on a smaller face rectangle extends to a grid on the corresponding whole face by inserting the smaller rectangle's upper endpoints as knots; the old cells remain and the added absolute-increment terms are nonnegative. Hence each face variation on $[0,x]$ is at most the corresponding whole-face variation, and enlarging an upper endpoint cannot decrease it. Thus $0\le v(x)\le V_{HK,0}(g)$ and $v$ is nondecreasing in each coordinate.

Fix a coordinate $j$, fix the other coordinates $x_{-j}$, and let $0\le a<b\le 1$. Write $x^a,x^b$ for the vectors with $j$th coordinate $a,b$. Terms in $v$ whose varied-coordinate set omits $j$ cancel in $v(x^b)-v(x^a)$. For a term whose set is $A\cup \{j\}$, additivity of Vitali variation at the split $[0,a]\cup [a,b]$ shows that its difference is the Vitali variation on the face rectangle $R_A$ with $j$ interval $[a,b]$, coordinates in $A$ intervals $[0,x_i]$, and all remaining coordinates fixed at $0$. Additivity follows directly by inserting $a$ into every grid: refinement and the triangle inequality give one inequality, while joining grids whose sums approach the two suprema gives the reverse inequality. Therefore $v(x^b)-v(x^a)={\sum }_{A\subseteq [d]\setminus \{j\}}{V}_V(g_{A\cup \{j\},0};R_A)$.

Expanding the right side of the finite-difference identity by inclusion–exclusion gives $g(x^b)-g(x^a)={\sum }_{A\subseteq [d]\setminus \{j\}}{\mathrm{\Delta }}_{A\cup \{j\}}g(R_A)$: every vertex with a coordinate outside $A\cup \{j\}$ set to $0$ cancels with the same vertex from the term with that coordinate added to $A$, leaving only the two vertices with all other coordinates equal to $x_{-j}$. Each one-cell grid shows $V_V(g_{A\cup \{j\},0};R_A)\ge \mathrm{\mid }{\mathrm{\Delta }}_{A\cup \{j\}}g(R_A)\mathrm{\mid }$. The triangle inequality and step 3.1 now give $v(x^b)-v(x^a)\ge \mathrm{\mid }g(x^b)-g(x^a)\mathrm{\mid }$. Consequently both $v$ and $w:=v-g$ are nondecreasing in every coordinate.

The function $v$ is bounded by step 2.1. The function $w$ is finite-valued and coordinatewise nondecreasing, so $w(\mathbf{0})\le w(x)\le w(\mathbf{1})$ for every $x$ and is bounded as well. By [F3], $v$ and $w$ are Riemann integrable; since $g=v-w$, [F4] makes $g$ Riemann integrable. Reflection preserves boundedness and maps each grid cell to a cell of the same volume with the same infimum and supremum values for $f=g\circ \rho$, so f has the same arbitrarily small Darboux gaps as g. The criterion [F5] makes $f$ Riemann integrable.