Proof. [01YD]

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Proof.

By Theorem 1.11, there exists a lower triangular matrix $A\in GL(n-2)$ such that

\displaystyle u^{\prime}=A\circ u:B_{2r}(x)\to\mathds{R}^{n-2}

(4.5)

is an $\epsilon$ -splitting. Let $dv_{g}$ denote the Riemannian measure and set $\omega^{\prime}\equiv du^{\prime 1}\wedge\cdots\wedge du^{\prime n-2}$ . Define the measure $\mu^{\prime}$ by $\mu^{\prime}=\Big(\int_{B_{2}(p)}|\omega|\Big)^{-1}|\omega^{\prime}|dv_{g}$ . Then

\displaystyle\mu^{\prime}=\det(A)\mu\,.

(4.6)

In particular this gives us

\displaystyle\frac{\mu^{\prime}(B_{2r}(x))}{\mu^{\prime}(B_{r}(x))}=\frac{\mu(B_{2r}(x))}{\mu(B_{r}(x))}\,,

(4.7)

and it is equivalent to show the ratio bound for $\mu^{\prime}$ . Now since $u^{\prime}$ is an $\epsilon$ -splitting we have the estimate

\displaystyle\fint_{B_{2r}(x)}|\,|\omega^{\prime}|-1|\leq C(n)\epsilon\,.

(4.8)

Hence, we also have the estimate

\displaystyle\fint_{B_{r}(x)}|\,|\omega^{\prime}|-1|\leq\frac{{\rm Vol}(B_{2r}(x))}{{\rm Vol}(B_{r}(x))}\fint_{B_{2r}(x)}|\,|\omega^{\prime}|-1|\leq C(n)\epsilon\,,

(4.9)

which of course uses the doubling property for the Riemannian measure. By combining the previous two estimates we get

	$\displaystyle\big(1-C\epsilon\big){\rm Vol}(B_{r}(x))$	$\displaystyle\leq\mu^{\prime}(B_{r}(x))\leq\big(1+C\epsilon\big){\rm Vol}(B_{r}(x))\,$
	$\displaystyle\big(1-C\epsilon\big){\rm Vol}(B_{2r}(x))$	$\displaystyle\leq\mu^{\prime}(B_{2r}(x))\leq\big(1+C\epsilon\big){\rm Vol}(B_{2r}(x))\,.$		(4.10)

Finally, by using the definition of $\mu^{\prime}$ we arrive at:

$\displaystyle\mu^{\prime}(B_{2r}(x))$	$\displaystyle=\Big(\int_{B_{2}(p)}\|\omega\|\,dv_{g}\Big)^{-1}\int_{B_{2r}(x)}\|\omega^{\prime}\|$
	$\displaystyle\leq(1+C(n)\epsilon)\Big(\int_{B_{2}(p)}\|\omega\|\,dv_{g}\Big)^{-1}{\rm Vol}(B_{2r}(x))$
	$\displaystyle\leq C(n)\Big(\int_{B_{2}(p)}\|\omega\|\,dv_{g}\Big)^{-1}{\rm Vol}(B_{r}(x))$	(4.11)
	$\displaystyle\leq C(n)\Big(\int_{B_{2}(p)}\|\omega\|\,dv_{g}\Big)^{-1}\int_{B_{r}(x)}\|\omega^{\prime}\|$
	$\displaystyle=C(n)\mu^{\prime}(B_{r}(x))\,,$	(4.12)

which by (4.7) completes the proof. ∎