 Solution

Each coordinate projection $S_j$ is compact, hence a Lebesgue measurable set, because it is the continuous image of the compact set $K$. Its $(d-1)$-dimensional measure is finite. Choose any nonnegative integrable function $\eta$ on $\mathbb{R}$ with integral one, and defineIf $x\in K$, every projection ${\hat{x}}_j$ belongs to $S_j$. If $x\notin K$, $f(x)=0$. Thus for every $j$, $f(x)\le 1_{S_j}({\hat{x}}_j)=\int s_j({\hat{x}}_j,t)dt$, so the introductory estimate applies. Tonelli theorem gives $\Vert s_j{\Vert }_1={\lambda }_{d-1}(S_j)$, andSubstitute these values and raise to the $d$th power:This proves the geometric Loomis--Whitney inequality, including zero-volume sets or zero-measure projections, without any division by their measures.