Today I encountered a problem concerning the Hausdorff measure on the sphere.
Let $n\ge 3$ and $\Bbb S^{n-1}$ denote the $n-1$-dimensional unit sphere in $\Bbb R^{n}$ namely, $$\Bbb S^{n-1}= \{x \in \Bbb R^{n}: \|x\|=1\}$$
For $v\in \Bbb S^{n-1}$ fixed we define $$\Bbb S_v^{n-2}=\Bbb S^{n-1}\cap (\Bbb R v)^\bot$$ where $(\Bbb R v)^\bot$ is the orthogonal of $v$ in $\Bbb R^{n}$.
Question: How can one prove the $n-2$-Hausdorff of $\Bbb S_v^{n-2}$ equals the surface measure of $\Bbb S^{n-2}$ (which is the $n-2$-dimensional unit sphere in $\Bbb R^{n-1}$ ) that $$\mathcal{H}^{n-2}(\Bbb S_v^{n-2})=|\Bbb S^{n-2}|$$
This may not help but I would like to remind the following formula: $$ |\Bbb S^{n-1}| =\frac{2\pi^{n/2}}{\Gamma(n/2)}. $$
Definition: Given $E\subset \Bbb R^n$ and $s\in \Bbb R_+$ by definition the $s$-dimensional Hausdorff measure of $E$ is, $$\mathcal{H}^{s}(E)=\lim_{\delta\to 0^+}\mathcal{H}_\delta^{s}(E)$$ with $$\mathcal{H}_\delta^{s}(E)=\inf\left\{\sum_{i=0}^{\infty}\left(diam(U_i)\right)^s:~~ E\subset\bigcup_{i=0}^{\infty}U_i, ~~~ diam(U_i)\le \delta\right\}$$ where $(U_i)_{i}$ is an open covering of $E$ and $diam(U_i)=\sup\left\{d(x,y):~~ x,y\in U_i\right\}$