A finite rectangle cover admits grid control with arbitrarily small volume excess

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 $\mathrm{vol}(Q)\le V<V+\eta$, 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 $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$. 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 ${\sum }_j\mathrm{vol}(R_j^{+})<V+\eta$.

The constructed grid has the asserted control.