Lebesgue measure and spherical measure

Let $$S^{d-1}$$ be the $$d$$-dimensional unit sphere. Let $$A\subset S^{d-1}\cap \{x_d>0\}$$ be a measurable set in the Borel algebra of $$S^{d-1}$$. Set $$C = \{ \lambda a\, | \, \lambda>0, a\in A\}$$ and denote by $$B_r(x)$$ the ball in $$\mathbb{R}^d$$ with radius $$r>0$$ and center $$x$$.

We claim that \begin{align} |B_r(\xi)\cap C| \leq c \, r \,\sigma(B_r(\xi)\cap A) \end{align} holds for all $$\xi \in S^{d-1}\cap \{x_d>0\}, 0 and some constant $$c>0$$ independent of $$r,\xi$$. Here, $$|\cdot|$$ denotes the $$d$$-dimensional Lebesgue measure and $$\sigma$$ denotes the spherical measure.

I think one needs to do some kind of transformation here. I wanted to use $$x\mapsto \frac{x}{|x|}$$, but somehow I cannot recover the spherical measure if I start my integration with the Lebesgue measure. Is there another kind of transformation theorem that I am unaware of?

• What does $B_r(\xi)$ mean here? – kimchi lover Sep 2 '19 at 12:17
• Somehow this smells like an application of the coarea formula to me. Have a look at en.wikipedia.org/wiki/Coarea_formula. Especially the subsection for the spherical integration may be helpful. – humanStampedist Sep 2 '19 at 12:32
• Using the Coarea formula I obtain $|B_r(\xi) \cap C| = \int_{1-r}^{1+r} H^{d-1}(B_r(\xi) \cap C \cap \partial B_y(0)) dy$, where $H^{d-1}$ is the $d-1$-dimensional Hausdorff measure and $\partial B_y(0) = \{ x \in \mathbb{R} | |x|=y \}$. But now I would need to estimate the Hausdorff measure inside the integral with the $H^{d-1}$-measure in the case $y=1$. Is that possible? Maybe only for sufficient small $r$? – user406143 Sep 3 '19 at 9:33
• Since $H^{d-1}(B_r(\xi)\cap C\cap \partial_1B(0))=\sigma(B_r(\xi)\cap A)$ this may be useful? – humanStampedist Sep 3 '19 at 9:42
• Yes, that is true, but I need something like $H^{d-1}(B_r(\xi)\cap C\cap \partial B_y(0)) \leq c \sigma(B_r(\xi)\cap A)$ for $c>0$ independent of $r,y$. That I cannot see. – user406143 Sep 3 '19 at 10:16

\begin{align} |B_r(\xi)\cap C| &< |(B_r(\xi)\cap A)\cdot[1-r, 1+r]| < \int_{1-r}^{1+r}\left( \int_{B_r(\xi)\cap A} r^{d-1}ds \right)dr \\ &= \sigma(B_r(\xi)\cap A)\int_{1-r}^{1+r}r^{d-1}dr = \sigma(B_r(\xi)\cap A)\frac1d[(1+r)^d - (1-r)^d] \\ &\leq \sigma(B_r(\xi)\cap A)[2r + o(r)] \end{align}
So $$B_r(\xi)\cap C$$ is contained in a spherical cylinder $$(B_r(\xi)\cap A)\cdot[1-r, 1+r]$$ and you use spherical coordinates to calculate its volume. Hope this helps!
• Is it really true that $B_r(\xi)\cap C$ is contained in the spherical cylinder $(B_r{\xi}\cap A) \times [1-r,1+r]$? I don't think so. The thing is that if I draw a line through the origin and a boundary point of $B_r(\xi)\cap A$ then this line is not necessarily a tangent to the ball $B_r(\xi)$, right? – user406143 Sep 5 '19 at 8:29
• The line you're talking about wouldn't be tangent to $B_r(\xi)$ indeed, but that's not relevant to the problem. I'd argue as follows: $B_r(\xi)\cap C$ is enclosed in an infinite spherical cylinder $(B_r(\xi)\cap A)\cdot (0, \infty)$ and it's maximal distance from $S^{d-1}$ is $r$, hence it is enclosed in a spherical cylinder $(B_r(\xi)\cap A)\cdot [-1-r, 1+r]$. I see you have written $B_r(\xi)\cap A \times [1-r, 1+r]$, maybe that's where confusion arises. I think of spherical cylinder $A\cdot [x, y]$ as of set $\{ar: a\in A, r\in[x, y] \}$. Maybe 'spherical cone' would be a better name. – mbartczak Sep 5 '19 at 9:47
• Ok, maybe I just don't get it, but why is $(B_r(\xi)\cap C) \subset (0,\infty)\cdot (B_r(\xi)\cap A)$? This was actually my question in the comment above, where I used the line that is not tangent to argue that I doubt this assertion. – user406143 Sep 5 '19 at 10:34
• That's because $B_r(\xi)\cap C = B_r(\xi)\cap (A\cdot (0, \infty)) \subseteq (B_r(\xi)\cdot(0, \infty))\cap (A\cdot (0, \infty)) = (B_r(\xi)\cap A)\cdot(0, \infty)$. Just draw it in d=2 and you will see :). – mbartczak Sep 5 '19 at 10:57