Let $M^{4}$ satisfy $|{\rm Ric}_{M^{4}}|\leq 3$ and ${\rm Vol}(B_{1}(p))>{\rm v}>0$. Using volume monotonicity, we have for every $x\in B_{1}(p)$ and $r\leq 1$,

$\displaystyle{\rm Vol}(B_{r}(x))\geq\frac{{\rm Vol}_{-1}(B_{r})}{{\rm Vol}_{-1}(B_{2})}{\rm Vol}(B_{2}(x))\geq c(n){\rm Vol}(B_{1}(p))\,r^{4}\geq c{\rm v}\,r^{4}\,.$

(8.66)

Let $\delta({\rm v})$ be as in Theorem 8.12. By Lemma 8.5, we have that for each $x\in B_{1}(p)$, there exists a radius, $r_{\alpha_{x}}=2^{-\alpha_{x}}\in[C({\rm v})\delta^{3},\delta^{2}]$, such that $T^{\delta}_{\alpha_{x}}(x)=0$. Let $\{B_{r_{i}}(x_{i})\}$ be a subcovering such that the balls in $\{B_{r_{i}/4}(x_{i})\}$ are disjoint, where $r_{i}=r_{\alpha_{x_{i}}}$. Since $r_{i}>\bar{r}({\rm v})$, we have by the usual doubling estimates that there are at most $C({\rm v})$ balls in this covering.

Note that, for each ball $B_{r_{i}}(x_{i})$, we can apply Theorem 8.12 in order to get a subset $U_{i}\supseteq B_{r_{i}}(x_{i})$ with bounded diffeomorphism type and uniform boundary control. Now recall in dmiension $4$, the Chern-Guass-Bonnet formula can be written as

$\displaystyle\chi(U_{i})=\frac{1}{32\pi^{2}}\int_{U_{i}}|{\rm Rm}|^{2}-4|{\rm Ric}|^{2}+R^{2}+\int_{\partial U_{i}}\Psi\,,$

(8.67)

where $\Psi=\Psi(A)$ is a function of the second fundamental form. By reorganizing, we obtain the bound

	$\displaystyle\int_{U_{i}}|{\rm Rm}|^{2}$	$\displaystyle\leq 32\pi^{2}|\chi(U_{i})|+4\int_{U_{i}}|{\rm Ric}|^{2}+C\int_{U_{i}}|\Psi|\,,$
		$\displaystyle\leq C({\rm v})\,,$		(8.68)

where we have used the bound on the diffeomorphism type, the Ricci bound, and the second fundamental form bound from Theorem 8.12. By summing over $i$, we get

$\displaystyle\fint_{B_{1}(p)}|{\rm Rm}|^{2}\leq C({\rm v})\sum\int_{U_{i}}|{\rm Rm}|^{2}\leq C({\rm v})\,,$

(8.69)

as claimed. ∎