The Riemann integral over a Jordan set is independent of the bounding rectangle

Extend the zero extension on $Q_i$ further by zero to $Q$. Cut $Q$ at the lower and upper endpoint of $Q_i$ in each coordinate. The strict containment in [L1] and nondegeneracy of $Q_i$ make every cut strictly interior. On every added nondegenerate subrectangle the restriction is zero away from the finitely many coordinate faces of $Q_i$; only shared boundary points may retain nonzero values.

Every bounded piece of a coordinate hyperplane has content zero: subdivide its bounded $(m-1)$-dimensional coordinate ranges into cubes of side at most $1/\iota(N)$, and thicken the fixed coordinate by the same amount. The number of cubes grows at most as a fixed multiple of $\iota(N)^{m-1}$, so their total $m$-volume is at most a fixed multiple of $1/\iota(N)$, which can be made arbitrarily small (Measure zero and content zero in $\mathbb{R}^m$ by countable and finite cube covers, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$). Finite unions preserve this estimate. Thus the exceptional face set $H$ from step 1.1 is Jordan measurable with content zero, and [L3] gives $\int 1_H=0$.

On each added subrectangle the extended function is bounded and is zero off $H$, so its discontinuities lie in the null set $H$. It is integrable by [L3]. If $|f|\le B$, then $|h|\le B1_H$; monotonicity and the absolute-value estimate in [L2] give $\left|\int h\right|\le\int|h|\le B\int1_H=0$. Hence every added subrectangle has integral $0$.

Repeated coordinate-slice additivity [L2] now says that the extension is integrable on $Q$ exactly when it is integrable on $Q_i$, and its integral equals the $Q_i$-integral because every added integral is $0$. Applying this to $i=1,2$ gives the same integrability decision and value in both rectangles.