The proof is by contradiction. So let us assume for some $\epsilon>0$ there is no such $\delta({\rm v},\epsilon)>0$. Thus, we have a sequence of spaces $(M^{4}_{j},g_{j},p_{j})$ with ${\rm Vol}(B_{1}(p_{j}))>{\rm v}>0$, $|{\rm Ric}_{M^{4}_{j}}|\leq\delta_{j}\to 0$ and $|\mathcal{V}_{4}(p_{j})-\mathcal{V}_{1/4}(p_{j})|<\delta_{j}\to 0$, but the conclusions of the theorem fail. After passing to a subsequence we can take a limit

Using the almost volume cone implies almost metric cone theorem of [ChCo1], we then have

where $y_{0}\in C(Y)$ is the cone vertex and $Y$ some metric space of diameter $\leq\pi$.

Now using Theorem 1.1, we know that away from a set of codimension $4$ in $C(Y)$, the harmonic radius $r_{h}>0$ is bounded uniformly from below. Assume there is some point $y\in Y$ such that $r_{h}(y)=0$ and consider the ray $\gamma_{y}$ in $C(Y)$ through the point $y$. In that case, it would follow that for every point of $\gamma_{y}$, the harmonic radius $r_{h}=0$ vanishes. The ray $\gamma$ has Hausdorff dimension $1$, and therefore its existence would contradict Theorem 1.1. Thus, we conclude that $r_{h}>0$ and that $Y=(Y,g_{Y})$ is a $C^{1,\alpha}\cap W^{2,q}$ manifold for every $\alpha<1$ and $q<\infty$.

Now by writing the formula for the Ricci tensor in harmonic coordinates and using $|{\rm Ric}_{M^{4}_{j}}|\to 0$, it follows that $C(Y)$ is smooth and Ricci flat away from the vertex. In particular, since $C(Y)$ is a metric cone over $Y$, we must ${\rm Ric}_{Y^{3}}=3g^{Y}$. Since in dimension $3$, constant Ricci curvature implies constant sectional curvature, it follows $Y=S^{3}/\Gamma$ has constant sectional curvature $\equiv 1$. Additionally, we know from the volume bound, ${\rm Vol}(B_{1}(p))>{\rm v}>0$, that the order $|\Gamma|<N({\rm v})$ is uniformly bounded. In particular, we have that $C(Y)=\mathds{R}^{4}/\Gamma$ is an orbifold with an isolated singularity.

It now follows that there exists $r_{0}({\rm v})>0$ such that for $y\in\mathds{R}^{4}/\Gamma$ with $|y|=1$, we have

where $0^{4}\in\mathds{R}^{4}$. In particular, for all $j$ sufficiently large, we have from the standard $\epsilon$-regularity theorem, Theorem 2.3, that for all $x\in A_{\epsilon,2}(p_{j})$, the harmonic radius, $r_{h}(x)>r_{0}({\rm v},\epsilon)=r_{0}({\rm v})\epsilon$ is bounded uniformly from below independent of $j$. Thus, if there exists $\epsilon$ as above, for which there is no $\delta({\rm v},\epsilon)$, it must be (2) that fails to hold.

However, by using again the diffeomorphism statement of Theorem 8.1, we have that for $j$ sufficiently large, there exists diffeomorphisms

such that

For $j$ sufficiently large, this implies that (2) holds; a contradiction. ∎