[proofplan]
The proof is an immediate application of Bishop-Gromov volume comparison with model sectional curvature $-K$. The comparison says that the ratio between the volume of a geodesic ball in $M$ and the corresponding model volume is nonincreasing in the radius. Applying this monotonicity at the two radii $R$ and $2R$ gives the displayed doubling estimate. The uniform local doubling constant follows by showing that the model ratio $V_{-K}^n(2R)/V_{-K}^n(R)$ remains bounded for $0<R\leq R_0$, using the small-radius asymptotic $V_{-K}^n(r)\sim c_n r^n$.
[/proofplan]
[step:Apply Bishop-Gromov monotonicity at the radii $R$ and $2R$]
Fix $p \in M$ and $R>0$. Since $(M^n,g)$ is complete and satisfies $\operatorname{Ric}_g \geq -(n-1)K g$, Bishop-Gromov volume comparison with model curvature $-K$ applies (citing a result not yet in the wiki: [Bishop-Gromov Volume Comparison Theorem](/theorems/5371)). It states that the function
\begin{align*}
\Theta_p:(0,\infty)&\to (0,\infty)\\
r&\mapsto \frac{\operatorname{vol}_g(B_g(p,r))}{V_{-K}^n(r)}
\end{align*}
is nonincreasing.
Because $R \leq 2R$, monotonicity gives
\begin{align*}
\frac{\operatorname{vol}_g(B_g(p,2R))}{V_{-K}^n(2R)}
\leq
\frac{\operatorname{vol}_g(B_g(p,R))}{V_{-K}^n(R)}.
\end{align*}
Multiplying both sides by $V_{-K}^n(2R)>0$ gives
\begin{align*}
\operatorname{vol}_g(B_g(p,2R))
\leq
\frac{V_{-K}^n(2R)}{V_{-K}^n(R)}
\operatorname{vol}_g(B_g(p,R)).
\end{align*}
[guided]
Fix $p \in M$ and $R>0$. The theorem is designed to be read directly from Bishop-Gromov comparison: we compare the actual volume growth of $(M,g)$ with the volume growth in the simply [connected space](/page/Connected%20Space) form whose sectional curvature is the constant $-K$.
The hypotheses needed for Bishop-Gromov volume comparison are exactly present here. The manifold $(M^n,g)$ is complete, and the Ricci curvature lower bound is
\begin{align*}
\operatorname{Ric}_g(v,v) \geq -(n-1)K\,g(v,v)
\end{align*}
for every tangent vector $v$. Therefore Bishop-Gromov volume comparison with model curvature $-K$ applies (citing a result not yet in the wiki: Bishop-Gromov Volume Comparison Theorem). It says that the map
\begin{align*}
\Theta_p:(0,\infty)&\to (0,\infty)\\
r&\mapsto \frac{\operatorname{vol}_g(B_g(p,r))}{V_{-K}^n(r)}
\end{align*}
is nonincreasing.
Now choose the two radii in the statement. Since $R \leq 2R$, nonincreasingness of $\Theta_p$ gives
\begin{align*}
\Theta_p(2R) \leq \Theta_p(R).
\end{align*}
Writing out the definition of $\Theta_p$ at these two radii gives
\begin{align*}
\frac{\operatorname{vol}_g(B_g(p,2R))}{V_{-K}^n(2R)}
\leq
\frac{\operatorname{vol}_g(B_g(p,R))}{V_{-K}^n(R)}.
\end{align*}
The model volume $V_{-K}^n(2R)$ is positive because it is the volume of a nonempty geodesic ball of positive radius in the model space. Multiplying the inequality by $V_{-K}^n(2R)$ therefore preserves the inequality and yields
\begin{align*}
\operatorname{vol}_g(B_g(p,2R))
\leq
\frac{V_{-K}^n(2R)}{V_{-K}^n(R)}
\operatorname{vol}_g(B_g(p,R)).
\end{align*}
This is the first asserted estimate.
[/guided]
[/step]
[step:Bound the model volume ratio on bounded radius intervals]
Fix $R_0>0$. Define the model radial function
\begin{align*}
s_K:[0,\infty)&\to [0,\infty)\\
t&\mapsto
\begin{cases}
t, & K=0,\\
\frac{\sinh(\sqrt{K}\,t)}{\sqrt{K}}, & K>0.
\end{cases}
\end{align*}
Let $\omega_{n-1}:=\mathcal{H}^{n-1}(S^{n-1})$ denote the Euclidean $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) of the unit sphere. The model volume is
\begin{align*}
V_{-K}^n(r)=\omega_{n-1}\int_0^r s_K(t)^{n-1}\,d\mathcal{L}^1(t).
\end{align*}
Define
\begin{align*}
A:(0,\infty)&\to (0,\infty)\\
r&\mapsto \frac{V_{-K}^n(r)}{r^n}.
\end{align*}
Since $s_K(t)/t \to 1$ as $t\to 0^+$, we have
\begin{align*}
\lim_{r\to 0^+} A(r)
=
\frac{\omega_{n-1}}{n}.
\end{align*}
Thus $A$ extends continuously to $[0,\infty)$ by setting $A(0):=\omega_{n-1}/n$.
For $0<R\leq R_0$,
\begin{align*}
\frac{V_{-K}^n(2R)}{V_{-K}^n(R)}
=
2^n\frac{A(2R)}{A(R)}.
\end{align*}
The function $A$ is continuous and positive on the compact intervals $[0,2R_0]$ and $[0,R_0]$. Hence
\begin{align*}
M&:=\max_{0\leq r\leq 2R_0} A(r),\\
m&:=\min_{0\leq r\leq R_0} A(r)
\end{align*}
are well-defined finite constants, and $m>0$. Therefore, for every $0<R\leq R_0$,
\begin{align*}
\frac{V_{-K}^n(2R)}{V_{-K}^n(R)}
\leq
2^n\frac{M}{m}.
\end{align*}
Set
\begin{align*}
C(n,K,R_0):=2^n\frac{M}{m}.
\end{align*}
Combining this bound with the estimate from the previous step gives
\begin{align*}
\operatorname{vol}_g(B_g(p,2R))
\leq
C(n,K,R_0)\operatorname{vol}_g(B_g(p,R))
\end{align*}
for every $p\in M$ and every $0<R\leq R_0$. This proves the local volume doubling conclusion.
[/step]
