= 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 define
$$
f(x)=1_K(x),\qquad s_j(x)=1_{S_j}(\widehat x_j)\eta(x_j).
$$
If $x\in K$, every projection $\widehat x_j$ belongs to $S_j$. If $x\notin K$, $f(x)=0$. Thus for every $j$, $f(x)\leq1_{S_j}(\widehat x_j)=\int s_j(\widehat x_j,t)\,dt$, so the introductory estimate applies. <Tonelli theorem> gives $\|s_j\|_1=\lambda_{d-1}(S_j)$, and
$$
\|1_K\|_{d/(d-1)}=\lambda_d(K)^{(d-1)/d}.
$$
Substitute these values and raise to the $d$th power:
$$
\boxed{\lambda_d(K)^{d-1}\leq\prod_{j=1}^d\lambda_{d-1}(S_j)}.
$$
This proves the geometric <Loomis--Whitney inequality>, including zero-volume sets or zero-measure projections, without any division by their measures.
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