[proofplan] We first obtain a weakly convergent subsequence of the pushed-forward Riemannian volume measures by combining the diameter bound, the Ricci lower bound, and Bishop volume comparison. We then transfer balls in $M_j$ to balls in $X$ using the $\varepsilon_j$-approximations and pass Bishop-Gromov monotonicity to the limit at continuity radii of the limiting measure. Finally, monotonicity at arbitrary radii follows by regularity and one-sided approximation, and the noncollapsed volume convergence theorem identifies the limit measure with a positive constant multiple of $\mathcal{H}^n$. [/proofplan]

[step:Extract a weak limit of the pushed-forward volume measures]For each $j\in\mathbb{N}$, let $d_j:=d_{g_j}$ and let $\nu_j:=\operatorname{Vol}_{g_j}$ denote the Riemannian volume measure on $M_j$. Since $\operatorname{Ric}_{g_j}\geq (n-1)K g_j$ and $\operatorname{diam}(M_j,g_j)\leq D$, the manifold version of the [Bishop-Gromov Volume Comparison Theorem](/page/Bishop-Gromov%20Volume%20Comparison%20Theorem) gives a uniform upper bound for total volume. Define \begin{align*} C_0:= \begin{cases} V_{K,n}(D), & K\leq 0,\\ V_{K,n}(\pi/\sqrt K), & K>0. \end{cases} \end{align*} Here $V_{K,n}(\rho)$ denotes the volume of a radius-$\rho$ ball in the simply connected $n$-dimensional space form of sectional curvature $K$. In the case $K>0$, the [Bonnet-Myers Theorem](/page/Bonnet-Myers%20Theorem) gives $\operatorname{diam}(M_j,g_j)\leq \pi/\sqrt K$, so the displayed value bounds the full model space volume. Hence \begin{align*} \nu_j(M_j)\leq C_0 \end{align*} for all $j\in\mathbb{N}$. Since $X$ is compact, the space of finite Radon measures on $X$ with total mass at most $C_0$ is weakly sequentially compact. Therefore the sequence \begin{align*} \mu_j := (\Phi_j)_*\nu_j \end{align*} has a weakly convergent subsequence. By hypothesis, after passing to that subsequence, there is a Radon measure $\mu$ on $X$ such that \begin{align*} \int_X f\,d\mu_j \longrightarrow \int_X f\,d\mu \end{align*} for every [continuous function](/page/Continuous%20Function) $f:X\to\mathbb{R}$. The lower volume bound gives \begin{align*} \mu(X)=\lim_{j\to\infty}\mu_j(X)=\lim_{j\to\infty}\nu_j(M_j)\geq v>0, \end{align*} so $\mu$ is nonzero.[/step]

[guided]For each $j\in\mathbb{N}$, write $d_j:=d_{g_j}$ for the Riemannian distance and $\nu_j:=\operatorname{Vol}_{g_j}$ for the Riemannian volume measure. We need a subsequential weak limit of the measures \begin{align*} \mu_j := (\Phi_j)_*\nu_j \end{align*} on the fixed [compact space](/page/Compact%20Space) $X$. The first point is uniform boundedness of the total masses. The Ricci lower bound and diameter bound imply, by the manifold [Bishop-Gromov Volume Comparison Theorem](/page/Bishop-Gromov%20Volume%20Comparison%20Theorem), that there is a finite constant depending only on $n,K,D$ such that \begin{align*} \nu_j(M_j)\leq C_0 \end{align*} for every $j$. More explicitly, set \begin{align*} C_0:= \begin{cases} V_{K,n}(D), & K\leq 0,\\ V_{K,n}(\pi/\sqrt K), & K>0, \end{cases} \end{align*} where $V_{K,n}(\rho)$ is the model $n$-volume of a radius-$\rho$ ball in the simply [connected space](/page/Connected%20Space) form of sectional curvature $K$. For $K>0$, the [Bonnet-Myers Theorem](/page/Bonnet-Myers%20Theorem) gives $\operatorname{diam}(M_j,g_j)\leq \pi/\sqrt K$, so the second line bounds the full total volume. Since $\mu_j(X)=\nu_j(M_j)$ by the definition of pushforward measure, this gives \begin{align*} \mu_j(X)\leq C_0. \end{align*} Because $X$ is compact, finite Radon measures on $X$ with uniformly bounded total mass are weakly sequentially compact. Thus, after passing to a subsequence, the measures $\mu_j$ converge weakly to a Radon measure $\mu$ on $X$, meaning that \begin{align*} \int_X f\,d\mu_j \longrightarrow \int_X f\,d\mu \end{align*} for every continuous function $f:X\to\mathbb{R}$. Finally, the noncollapse hypothesis prevents the limiting measure from being zero. Since each $\Phi_j$ maps $M_j$ into $X$, \begin{align*} \mu_j(X)=\nu_j(M_j)\geq v. \end{align*} [Weak convergence](/page/Weak%20Convergence) against the constant function $1:X\to\mathbb{R}$ therefore gives \begin{align*} \mu(X)=\lim_{j\to\infty}\mu_j(X)\geq v>0. \end{align*}[/guided]

[step:Compare pulled-back balls through the Gromov-Hausdorff approximations]For a [metric space](/page/Metric%20Space) $(Y,d_Y)$, a point $y\in Y$, and a radius $\rho>0$, define the open metric ball and metric sphere by \begin{align*} B_Y(y,\rho)&:=\{z\in Y:d_Y(z,y)<\rho\},\\ \partial B_Y(y,\rho)&:=\{z\in Y:d_Y(z,y)=\rho\}. \end{align*} When $Y=M_j$, the metric is $d_j=d_{g_j}$; when $Y=X$, the metric is $d_X$. Fix $x\in X$. Choose points $p_j\in M_j$ such that \begin{align*} d_X(\Phi_j(p_j),x)\leq \varepsilon_j. \end{align*} This is possible because $\Phi_j$ is an $\varepsilon_j$-approximation. For every $\rho>0$ and every $q\in M_j$, the distortion estimate for $\Phi_j$ gives \begin{align*} \left|d_X(\Phi_j(q),\Phi_j(p_j))-d_j(q,p_j)\right|\leq \varepsilon_j. \end{align*} Therefore \begin{align*} B_{M_j}(p_j,\rho-2\varepsilon_j) \subset \Phi_j^{-1}(B_X(x,\rho)) \subset B_{M_j}(p_j,\rho+2\varepsilon_j) \end{align*} whenever $\rho>2\varepsilon_j$. Applying $\nu_j$ to these inclusions yields \begin{align*} \nu_j(B_{M_j}(p_j,\rho-2\varepsilon_j)) \leq \mu_j(B_X(x,\rho)) \leq \nu_j(B_{M_j}(p_j,\rho+2\varepsilon_j)). \end{align*}[/step]

[guided]Fix $x\in X$. Because $\Phi_j:M_j\to X$ is an $\varepsilon_j$-approximation, its image is $\varepsilon_j$-dense in $X$. Hence we may choose a point $p_j\in M_j$ satisfying \begin{align*} d_X(\Phi_j(p_j),x)\leq \varepsilon_j. \end{align*} This point is the approximate preimage of the centre $x$ of the ball in $X$. Now let $\rho>0$ and $q\in M_j$. The distortion bound in the definition of an $\varepsilon_j$-approximation says that \begin{align*} \left|d_X(\Phi_j(q),\Phi_j(p_j))-d_j(q,p_j)\right|\leq \varepsilon_j. \end{align*} We prove the two inclusions separately. First suppose $q\in B_{M_j}(p_j,\rho-2\varepsilon_j)$, so $d_j(q,p_j)<\rho-2\varepsilon_j$. By the triangle inequality in $X$ and the distortion estimate, \begin{align*} d_X(\Phi_j(q),x) &\leq d_X(\Phi_j(q),\Phi_j(p_j))+d_X(\Phi_j(p_j),x)\\ &\leq d_j(q,p_j)+\varepsilon_j+\varepsilon_j\\ &<\rho. \end{align*} Thus $\Phi_j(q)\in B_X(x,\rho)$, proving \begin{align*} B_{M_j}(p_j,\rho-2\varepsilon_j) \subset \Phi_j^{-1}(B_X(x,\rho)). \end{align*} Conversely, suppose $q\in \Phi_j^{-1}(B_X(x,\rho))$, so $d_X(\Phi_j(q),x)<\rho$. Again using the triangle inequality and the distortion estimate, \begin{align*} d_j(q,p_j) &\leq d_X(\Phi_j(q),\Phi_j(p_j))+\varepsilon_j\\ &\leq d_X(\Phi_j(q),x)+d_X(x,\Phi_j(p_j))+\varepsilon_j\\ &<\rho+2\varepsilon_j. \end{align*} Hence $q\in B_{M_j}(p_j,\rho+2\varepsilon_j)$, and therefore \begin{align*} \Phi_j^{-1}(B_X(x,\rho)) \subset B_{M_j}(p_j,\rho+2\varepsilon_j). \end{align*} When $\rho>2\varepsilon_j$, the smaller radius $\rho-2\varepsilon_j$ is positive, so the first ball is an ordinary open metric ball. Applying the measure $\nu_j$ to the two inclusions and using the definition $\mu_j=(\Phi_j)_*\nu_j$, namely \begin{align*} \mu_j(B_X(x,\rho))=\nu_j(\Phi_j^{-1}(B_X(x,\rho))), \end{align*} gives \begin{align*} \nu_j(B_{M_j}(p_j,\rho-2\varepsilon_j)) \leq \mu_j(B_X(x,\rho)) \leq \nu_j(B_{M_j}(p_j,\rho+2\varepsilon_j)). \end{align*}[/guided]

[step:Pass Bishop-Gromov monotonicity to continuity radii]Let $0<r<R$, with $R<\pi/\sqrt K$ if $K>0$, and assume first that \begin{align*} \mu(\partial B_X(x,r))=\mu(\partial B_X(x,R))=0. \end{align*} For all sufficiently large $j$, the radii $r-2\varepsilon_j$ and $R+2\varepsilon_j$ are positive and satisfy $r-2\varepsilon_j<R+2\varepsilon_j$. If $K>0$, the additional assumption $R<\pi/\sqrt K$ and the convergence $\varepsilon_j\to 0$ imply that, for all sufficiently large $j$, \begin{align*} R+2\varepsilon_j<\pi/\sqrt K. \end{align*} Thus the radii lie in the admissible range for the positive-curvature [Bishop-Gromov Volume Comparison Theorem](/page/Bishop-Gromov%20Volume%20Comparison%20Theorem). The Bishop-Gromov inequality on $(M_j,g_j)$ gives \begin{align*} \frac{\nu_j(B_{M_j}(p_j,R+2\varepsilon_j))}{V_{K,n}(R+2\varepsilon_j)} \leq \frac{\nu_j(B_{M_j}(p_j,r-2\varepsilon_j))}{V_{K,n}(r-2\varepsilon_j)}. \end{align*} Using the ball inclusions from the previous step, we obtain \begin{align*} \frac{\mu_j(B_X(x,R))}{V_{K,n}(R+2\varepsilon_j)} \leq \frac{\mu_j(B_X(x,r))}{V_{K,n}(r-2\varepsilon_j)}. \end{align*} Since $\mu_j\rightharpoonup\mu$ weakly and the boundary measures at $r$ and $R$ vanish, the [Portmanteau Theorem](/page/Portmanteau%20Theorem) gives \begin{align*} \mu_j(B_X(x,r))\to\mu(B_X(x,r)), \qquad \mu_j(B_X(x,R))\to\mu(B_X(x,R)). \end{align*} The model volume function $V_{K,n}:(0,\pi/\sqrt K)\to(0,\infty)$ when $K>0$, and $V_{K,n}:(0,\infty)\to(0,\infty)$ when $K\leq 0$, is continuous. Passing to the limit in the preceding inequality gives \begin{align*} \frac{\mu(B_X(x,R))}{V_{K,n}(R)} \leq \frac{\mu(B_X(x,r))}{V_{K,n}(r)}. \end{align*}[/step]

[guided]Fix $0<r<R$, and in the case $K>0$ assume $R<\pi/\sqrt K$. We first impose the additional continuity-radius hypotheses \begin{align*} \mu(\partial B_X(x,r))=\mu(\partial B_X(x,R))=0. \end{align*} These hypotheses are exactly what allows weak convergence of measures to identify the limiting masses of the open balls. For all sufficiently large $j$, the radius $r-2\varepsilon_j$ is positive and $r-2\varepsilon_j<R+2\varepsilon_j$. If $K>0$, the strict inequality $R<\pi/\sqrt K$ and the convergence $\varepsilon_j\to 0$ give \begin{align*} R+2\varepsilon_j<\pi/\sqrt K \end{align*} for all sufficiently large $j$. Thus the pair of radii $r-2\varepsilon_j$ and $R+2\varepsilon_j$ lies in the admissible range of the manifold [Bishop-Gromov Volume Comparison Theorem](/page/Bishop-Gromov%20Volume%20Comparison%20Theorem) on $(M_j,g_j)$. Applying that theorem at the centre $p_j\in M_j$ gives \begin{align*} \frac{\nu_j(B_{M_j}(p_j,R+2\varepsilon_j))}{V_{K,n}(R+2\varepsilon_j)} \leq \frac{\nu_j(B_{M_j}(p_j,r-2\varepsilon_j))}{V_{K,n}(r-2\varepsilon_j)}. \end{align*} The ball-comparison step gives two inequalities in the needed directions: \begin{align*} \mu_j(B_X(x,R)) &\leq \nu_j(B_{M_j}(p_j,R+2\varepsilon_j)),\\ \nu_j(B_{M_j}(p_j,r-2\varepsilon_j)) &\leq \mu_j(B_X(x,r)). \end{align*} Since the model volumes are positive at positive admissible radii, substituting these bounds into Bishop-Gromov yields \begin{align*} \frac{\mu_j(B_X(x,R))}{V_{K,n}(R+2\varepsilon_j)} \leq \frac{\mu_j(B_X(x,r))}{V_{K,n}(r-2\varepsilon_j)}. \end{align*} We now pass to the limit. The [Portmanteau Theorem](/page/Portmanteau%20Theorem) says that if $\mu_j\rightharpoonup\mu$ weakly and a Borel set $A\subset X$ satisfies $\mu(\partial A)=0$, then $\mu_j(A)\to\mu(A)$. We apply this with $A=B_X(x,r)$ and $A=B_X(x,R)$. The boundary hypotheses above give \begin{align*} \mu_j(B_X(x,r))\to\mu(B_X(x,r)), \qquad \mu_j(B_X(x,R))\to\mu(B_X(x,R)). \end{align*} The model volume function $V_{K,n}$ is continuous and positive on the admissible interval, so \begin{align*} V_{K,n}(R+2\varepsilon_j)\to V_{K,n}(R), \qquad V_{K,n}(r-2\varepsilon_j)\to V_{K,n}(r). \end{align*} Taking the limit in the preceding inequality therefore gives \begin{align*} \frac{\mu(B_X(x,R))}{V_{K,n}(R)} \leq \frac{\mu(B_X(x,r))}{V_{K,n}(r)}. \end{align*} This is the Bishop-Gromov inequality at continuity radii in the limit space.[/guided]

[step:Remove the continuity-radius restriction]Now let $0<r<R$ be arbitrary, again with $R<\pi/\sqrt K$ if $K>0$. Since $\mu$ is a finite Radon measure on the compact metric space $X$, the spheres $\partial B_X(x,\rho)$ are pairwise disjoint as $\rho$ varies. Therefore the set of radii $\rho>0$ for which \begin{align*} \mu(\partial B_X(x,\rho))>0 \end{align*} is at most countable. Choose sequences $(r_i)_{i=1}^{\infty}$ and $(R_i)_{i=1}^{\infty}$ of positive continuity radii such that \begin{align*} r_i\uparrow r,\qquad R_i\uparrow R,\qquad r_i<R_i<R \end{align*} for every $i\in\mathbb{N}$. If $K>0$, this also ensures $R_i<\pi/\sqrt K$. Applying the continuity-radius result to $r_i$ and $R_i$ gives \begin{align*} \frac{\mu(B_X(x,R_i))}{V_{K,n}(R_i)} \leq \frac{\mu(B_X(x,r_i))}{V_{K,n}(r_i)}. \end{align*} Because $B_X(x,R_i)\uparrow B_X(x,R)$ and $B_X(x,r_i)\uparrow B_X(x,r)$, continuity from below for the finite measure $\mu$ gives \begin{align*} \mu(B_X(x,R_i))\uparrow\mu(B_X(x,R)), \qquad \mu(B_X(x,r_i))\uparrow\mu(B_X(x,r)). \end{align*} The model volume function $V_{K,n}$ is continuous and positive on the admissible interval, so \begin{align*} V_{K,n}(R_i)\to V_{K,n}(R), \qquad V_{K,n}(r_i)\to V_{K,n}(r). \end{align*} Letting $i\to\infty$ in the displayed inequality yields \begin{align*} \frac{\mu(B_X(x,R))}{V_{K,n}(R)} \leq \frac{\mu(B_X(x,r))}{V_{K,n}(r)}. \end{align*} This proves Bishop-Gromov monotonicity for all admissible radii.[/step]

[guided]The only obstruction in the previous step was that weak convergence of measures gives direct convergence of ball masses only when the boundary of the ball has zero limiting measure. We remove that assumption without ever passing through closed-ball mass. For a fixed point $x\in X$, the spheres $\partial B_X(x,\rho)$ are pairwise disjoint as $\rho$ ranges over positive radii. Since $\mu$ is finite on the compact space $X$, only countably many of these pairwise disjoint spheres can have positive $\mu$-measure. Hence the continuity radii, namely those $\rho>0$ satisfying \begin{align*} \mu(\partial B_X(x,\rho))=0, \end{align*} are dense in the admissible interval. Fix arbitrary radii $0<r<R$, with $R<\pi/\sqrt K$ if $K>0$. Choose continuity radii $r_i$ and $R_i$ such that \begin{align*} r_i\uparrow r,\qquad R_i\uparrow R,\qquad r_i<R_i<R \end{align*} for every $i\in\mathbb{N}$. The choice from below is the key point: increasing open balls have the correct open ball as their union, so the limiting mass will be $\mu(B_X(x,r))$ rather than the mass of a closed ball. If $K>0$, the condition $R_i<R<\pi/\sqrt K$ keeps every $R_i$ in the positive-curvature comparison range. For each $i$, both $r_i$ and $R_i$ are continuity radii. The result already proved at continuity radii therefore gives \begin{align*} \frac{\mu(B_X(x,R_i))}{V_{K,n}(R_i)} \leq \frac{\mu(B_X(x,r_i))}{V_{K,n}(r_i)}. \end{align*} Now pass to the limit. Since $R_i\uparrow R$, the open balls $B_X(x,R_i)$ increase to $B_X(x,R)$, and continuity from below gives \begin{align*} \mu(B_X(x,R_i))\uparrow\mu(B_X(x,R)). \end{align*} Similarly, since $r_i\uparrow r$, the open balls $B_X(x,r_i)$ increase to $B_X(x,r)$, so \begin{align*} \mu(B_X(x,r_i))\uparrow\mu(B_X(x,r)). \end{align*} The model volume function $V_{K,n}$ is continuous and positive on the admissible interval, hence \begin{align*} V_{K,n}(R_i)\to V_{K,n}(R), \qquad V_{K,n}(r_i)\to V_{K,n}(r). \end{align*} Letting $i\to\infty$ in the continuity-radius inequality gives \begin{align*} \frac{\mu(B_X(x,R))}{V_{K,n}(R)} \leq \frac{\mu(B_X(x,r))}{V_{K,n}(r)}. \end{align*} This proves Bishop-Gromov monotonicity for every pair of admissible radii.[/guided]

[step:Identify the noncollapsed limit measure with Hausdorff measure]The hypotheses \begin{align*} \operatorname{Ric}_{g_j}\geq (n-1)K g_j,\qquad \operatorname{diam}(M_j,g_j)\leq D,\qquad \operatorname{Vol}_{g_j}(M_j)\geq v>0 \end{align*} place the sequence in the noncollapsed Ricci-limit setting: the dimension is fixed, the Ricci lower bound is uniform, the diameter bound gives compactness of the underlying spaces, and the lower total-volume bound is precisely the noncollapse condition. The maps $\Phi_j:M_j\to X$ are the same Borel $\varepsilon_j$-approximations used to define the pushed-forward measures \begin{align*} \mu_j=(\Phi_j)_*\operatorname{Vol}_{g_j}, \end{align*} and the subsequence chosen above satisfies $\mu_j\rightharpoonup\mu$. Thus $\mu$ is the measured Gromov-Hausdorff volume limit associated to these approximations. We apply the [Cheeger-Colding Volume Convergence Theorem](/page/Cheeger-Colding%20Volume%20Convergence%20Theorem) for noncollapsed Ricci limits, also called the Noncollapsed Ricci Limit Volume Convergence Theorem. Its hypotheses are the fixed dimension $n$, the uniform lower Ricci bound $\operatorname{Ric}_{g_j}\geq (n-1)K g_j$, Gromov-Hausdorff convergence to $X$, and the noncollapse condition $\operatorname{Vol}_{g_j}(M_j)\geq v>0$. The measured convergence is formulated using Borel $\varepsilon_j$-approximations whose distortion tends to zero and whose images are $\varepsilon_j$-dense; these are precisely the maps $\Phi_j$ in the statement. Therefore the theorem applies to the pushed-forward Riemannian volume measures \begin{align*} \mu_j=(\Phi_j)_*\operatorname{Vol}_{g_j}\rightharpoonup\mu. \end{align*} With the standard metric normalization of [Hausdorff measure](/page/Hausdorff%20Measure) used in the theorem, the limiting Radon measure is a positive constant multiple of the $n$-dimensional Hausdorff measure on $X$. Hence there exists $c>0$ such that \begin{align*} \mu = c\,\mathcal{H}^n \end{align*} as Radon measures on $X$. Combining this identification with the limiting Bishop-Gromov inequality proved above completes the proof.[/step]

[guided]We now identify which Radon measure the weak limit $\mu$ actually is. The hypotheses put the sequence in the noncollapsed Ricci-limit framework: the manifolds have the same dimension $n$, the Ricci lower bound \begin{align*} \operatorname{Ric}_{g_j}\geq (n-1)K g_j \end{align*} is uniform, the spaces converge to $X$ in Gromov-Hausdorff distance, and the lower total-volume bound \begin{align*} \operatorname{Vol}_{g_j}(M_j)\geq v>0 \end{align*} prevents collapse of $n$-dimensional volume. The measured Gromov-Hausdorff normalization also matches the theorem being invoked. The maps $\Phi_j:M_j\to X$ are Borel $\varepsilon_j$-approximations: their distortion tends to zero and their images are $\varepsilon_j$-dense in $X$. The measures used in the convergence statement are exactly the pushed-forward Riemannian volume measures \begin{align*} \mu_j=(\Phi_j)_*\operatorname{Vol}_{g_j}. \end{align*} After passing to the subsequence fixed at the beginning of the proof, these measures converge weakly to $\mu$ as Radon measures on $X$. We may therefore apply the [Cheeger-Colding Volume Convergence Theorem](/page/Cheeger-Colding%20Volume%20Convergence%20Theorem) for noncollapsed Ricci limits. Under the standard metric normalization of $\mathcal H^n$, that theorem identifies any such measured Gromov-Hausdorff volume limit with a positive constant multiple of the $n$-dimensional Hausdorff measure on the limit space. Consequently there exists a constant $c>0$ such that \begin{align*} \mu=c\,\mathcal H^n \end{align*} as Radon measures on $X$. The positivity of $c$ is forced by the already proved lower mass bound $\mu(X)\geq v>0$. Combining this Hausdorff-measure identification with the Bishop-Gromov monotonicity inequality established for $\mu$ proves both conclusions of the theorem.[/guided]