The graph of a continuous function on a closed nondegenerate rectangle in $\mathbb{R}^m$ has content zero in $\mathbb{R}^{m+1}$

Given $\varepsilon>0$, choose a uniform coordinate grid with cell widths at most $\delta$, where uniform continuity makes the oscillation of $f$ on each cell below a vertical amount $\eta$. Since $Q$ is nondegenerate, the grid may be chosen so that the number $N_\delta$ of cells satisfies $N_\delta\delta^m\le C_Q$ for a constant depending only on $Q$.

One horizontal cube footprint of side $\delta$ covers each domain cell. Above it, stack $(m+1)$-cubes of side $\delta$ across the graph's vertical range. Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$ bounds their number by $\eta/\delta+2$, so all stacks together have volume at most $N_\delta(\eta\delta^m+2\delta^{m+1})\le C_Q(\eta+2\delta)$.

Summing over the finitely many domain cells gives total covering volume at most a rectangle-dependent constant times $\eta+\delta$. Choose $\eta$ and then $\delta$ to make this below $\varepsilon$.

This finite cube cover proves content zero in the sense of Measure zero and content zero in $\mathbb{R}^m$ by countable and finite cube covers.