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$\mathscr{H}^{n-2}(\Sigma)< \infty$ implies $\mathscr{H}^{n-1}(\pi(\Sigma))=0$
Let $\Sigma\subset \mathbb{R}^{n+1}$ be a set with $(n-2)$-dimensional Hausdorff measure finite, i.e. $\mathscr{H}^{n-2}(\Sigma)<\infty$. Let $\pi:\mathbb{R}^{n+1}\to \mathbb{R}^n$ be the ...
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