Let $f: S^{n-1}\to \mathbb{R}$ be a continuous function ($S^{n-1}\subset \mathbb{R}^n$ is the unit sphere), $f(a)>0$ and $f(b)<0$ for certain points $a,b\in S^{n-1}$. By continuity these inequalities hold in $B\_r(a)\cap S^{n-1}$ and $B\_r(b)\cap S^{n-1}$ for a small radius $r$. Let $v:=\mathcal{H}^{n-2}(\partial (B\_r(a)\cap S^{n-1}))=\mathcal{H}^{n-2}(\partial (B\_r(b)\cap S^{n-1}))$ (taking the boundary with respect to the $S^{n-1}$-topology; the $n-2$-dimensional Hausdorff measure $v$ is roughly the surface area of the $n-2$-sphere of radius $r$ for small $r$).


Is it true that $\mathcal{H}^{n-2}(\{f=0\}) \ge v$, and how would one prove this? It seems geometrically obvious; the picture I have in mind looks as follows: Imagine the 2-sphere and two points $a,b$ on it. The zero set of $f$ is at least a curve which surrounds both $a$ and $b$, and the least length for such a curve is $v$, which is the length of the boundary of the neighbourhoods of $a$ and $b$ where $f$ is known to be positive/negative.



Estimating the Hausdorff measure of a subset of the sphere
Yes. Here is an elementary proof.


Construct a distance non-increasing retraction $p:S^{n-1}\setminus B\_r(a)\setminus B\_r(b)\to \partial B\_r(a)$. (For example: divide the sphere into two-hemispheres by the hyperplane $H$ of symmetry between $a$ and $b$. In the hemisphere containing $a$, let $p$ be the radial projection, namely $p(x)$ is the intersection of the geodesic segment $[ax]$ with $\partial B\_r(a)$. In the other hyperplane, let $p$ be the composition of the reflection in $H$ and the above radial projection. This map is distance non-increasing since the radial projection to $\partial B\_r(a)$ is distance non-increasing away from the $r$-pall centered at the point opposite to $a$ on the sphere, and this ball is contained in the second hemisphere.)


For each $x\in \partial B\_r(a)$, the pre-image $p^{-1}(x)$ contains a continuous path connecting $\partial B\_r(a)$ to $\partial B\_r(b)$ and therefore a point of the set $\{f=0\}$. Hence $p(\{f=0\})=\partial B\_r(a)$. Since $p$ does not decrease distances, it does not increase the Hausdorff measure and the desired inequality follows.


