A question on estimates of surface measures

Question

If $\mathcal{H}^s $ is $s$ dimensional Hausdorff measure on $ \mathbb{R}^n$, is the following inequality true for all $ x \in \mathbb{R}^n,\ R,t > 0 $ ? $$ \mathcal{H}^{n-1}(\partial B(x,t)\cap B(0,R)) \leq \mathcal{H}^{n-1}(\partial B(0,R)) $$ If the answer is not affirmative then a concrete counter example of two such open balls would be very helpful.

Answer 1

The definition of $\mathcal H^{s}$ implies that it does not increase under $1$-Lipschitz maps: $\mathcal H^s(f(E))\le \mathcal H^s(E)$ if $f$ satisfies $|f(a)-f(b)|\le |a-b|$ for all $a,b\in E$.

The nearest point projection $\pi:\mathbb R^n \to B(x,t)$ is a $1$-Lipschitz map. (Note that when $y\in B(x,t)$, the nearest point to $y$ is $y$ itself; that is, the projection is the identity map on the target set). It remains to show that $$\partial B(x,t)\cap B(0,R) \subset \pi(\partial B(0,R))\tag1$$ Given a point $y\in \partial B(x,t)\cap B(0,R)$, consider the half-line $\{y+tn: t\ge 0\}$ where $n$ is an outward normal vector to $\partial B(x,t)$ at $y$. This half-line intersects $\partial B(0,R)$ at some point $z$. An easy geometric argument shows that $\pi(z)=y$.