As in Lemma 4.1, choose a lower triangular matrix $A\in GL(n-2)$ such that

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

(4.18)

is an $\epsilon$-splitting and define the measure $\mu^{\prime}$ as in Lemma 4.1. Then as in (4.5), $\mu^{\prime}=\det(A)\mu$.

Since $u^{\prime}$ is an $\epsilon$-splitting, we have the estimates

		$\displaystyle\fint_{B_{2r}(x)}||\omega^{\prime}|-1|\leq C(n)\epsilon\,,$
		$\displaystyle u^{\prime}(B_{r}(x))\subseteq B_{2r}(u^{\prime}(x))\,.$		(4.19)

By the first estimate above,

	$\displaystyle\mu^{\prime}(B_{r}(x))$	$\displaystyle=\Big(\int_{B_{2}(p)}|\omega|\Big)^{-1}\int_{B_{r}(x)}|\omega^{\prime}|\,,$
		$\displaystyle\geq(1-C(n)\epsilon)\frac{{\rm Vol}(B_{r}(x))}{{\rm Vol}(B_{2}(p))}\fint_{B_{r}(x)}|\omega^{\prime}|$
		$\displaystyle\geq(1-C\epsilon)\frac{{\rm Vol}(B_{r}(x))}{{\rm Vol}(B_{3}(x))}\geq C(n)r^{n}\,,$		(4.20)

where in the last step we have used volume monotonicity for the Riemannian measure. On the other hand, by the second estimate of (4.19),

$\displaystyle|u^{\prime}(B_{r}(x))|\leq C(n)r^{n-2}\,.$

(4.21)

Combining these gives the estimate

$\displaystyle|u^{\prime}(B_{r}(x))|\leq C(n)r^{-2}\mu^{\prime}(B_{r}(x))\,.$

(4.22)

To relate these back to the original function $u$, we observe that

	$\displaystyle|u^{\prime}(B_{r}(x))|$	$\displaystyle=\det(A)|u(B_{r}(x))|\,,$
	$\displaystyle\mu^{\prime}(B_{r}(x))$	$\displaystyle=\det(A)|\mu(B_{r}(x))|\,,$		(4.23)

which immediately gives

$\displaystyle|u(B_{r}(x))|\leq C(n)r^{-2}\mu(B_{r}(x))\,.$

(4.24)

This completes the proof. ∎