The graph of a continuous function on a compact Euclidean set has content zero

If $C=\mathrm{\varnothing }$, then $G=\mathrm{\varnothing }$ and the empty finite cover proves the conclusion. If $m=0$, then ${\mathbb{R}}^0$ has exactly one point, so $C$ has at most one point and $G$ is a single point of ${\mathbb{R}}^1$, covered for every tolerance by one closed cube of small enough side; [F2] gives content zero. Hence assume $C\mathrm{\ne }\mathrm{\varnothing }$ and $m\ge 1$, so that the divisions by $\sqrt{m}$ below are defined.

By [F4], compactness places $C$ in a closed cube $Q$, and [F3] bounds $f(C)$ in an interval. By [F1], for every $\eta >0$ there is $\delta >0$ such that $\mathrm{\mid }f(u)-f(v)\mathrm{\mid }<\eta$ whenever $u,v\in C$ and $\mathrm{\parallel }u-v{\mathrm{\parallel }}_2<\delta$.

Partition a fixed cube slightly larger than $Q$ into coordinate cubes of side $h<\delta \mathrm{/}\sqrt{m}$. Keep only the base cells meeting $C$, and above each such cell keep the vertical $(m+1)$-grid cubes that meet $G$. Values of $f$ over one base cell differ by less than $\eta$, so its retained vertical stack has total height at most $\eta +2h$. If $B$ is the volume of the enlarged base cube, the retained cubes cover $G$ and have total volume at most $B(\eta +2h)$.

Choose $\eta$ and then $h$ so that $B(\eta +2h)<\epsilon$. Step 2.1 is then a finite closed-cube cover of $G$ with total volume below $\epsilon$; by [F2], $G$ has content zero.