If $p_{k}\rightarrow x$ under the convergence of $(M,\tilde{\omega}_{t_{k}})$ to $(X,d_{X})$, we claim that a subsequence of $p_{k}$ converges to a point $p^{\prime}\in f^{-1}(\phi^{-1}(x))$ under the metric $\omega_{M}$ on $M$. By Lemma 5.1, there is a compact neighborhood $B\subset N_{0}$ of $\phi^{-1}(x)$ and a section $s:B\rightarrow f^{-1}(B)$ such that $s(\phi^{-1}(x))\rightarrow x$ under the Gromov-Hausdorff convergence of $(M,\tilde{\omega}_{t_{k}})$ to $(X,d_{X})$. Thus $d_{\tilde{\omega}_{t_{k}}}(p_{k},s(\phi^{-1}(x)))\rightarrow 0$ when $t_{k}\rightarrow 0$. By Lemma 4.1, there are curves $\gamma_{k}$ connecting $p_{k}$ and $s(\phi^{-1}(x))$ such that ${\rm length}_{\tilde{\omega}_{t_{k}}}(\gamma_{k})=d_{\tilde{\omega}_{t_{k}}}(p_{k},s(\phi^{-1}(x)))$, and

$${\rm length}_{\omega_{0}}(f(\gamma_{k})\cap B)={\rm length}_{f^{*}\omega_{0}}(\gamma_{k}\cap f^{-1}(B))\leqslant C^{\frac{1}{2}}{\rm length}_{\tilde{\omega}_{t_{k}}}(\gamma_{k})\rightarrow 0.$$

For a $k\gg 1$, if there is a $y_{k}\in f(\gamma_{k})\backslash B$, then

$${\rm length}_{\omega_{0}}(f(\gamma_{k})\cap B)\geqslant d_{\omega_{0}}(y_{k},\phi^{-1}(x))\geqslant\rho,$$

where $\rho>0$ such that $B_{\omega_{0}}(\phi^{-1}(x),\rho)\subset B$, which is a contradiction. Thus $f(\gamma_{k})\subset B$ for $k\gg 1$, ${\rm length}_{\omega_{0}}(f(\gamma_{k}))\rightarrow 0$ and $f(p_{k})$ converges to $\phi^{-1}(x)$ under the metric $\omega_{0}$. By passing to a subsequence, $p_{k}$ converges to a point $p^{\prime}$ under the metric $\omega_{M}$. Since $f^{*}\omega_{0}\leqslant C^{\prime}\omega_{M}$ for a constant $C^{\prime}>0$, $d_{\omega_{0}}(f(p_{k}),f(p^{\prime}))\leqslant C^{\prime\frac{1}{2}}d_{\omega_{M}}(p_{k},p^{\prime})\rightarrow 0.$ Hence $f(p^{\prime})=\phi^{-1}(x)$ and $p^{\prime}\in f^{-1}(\phi^{-1}(x))$.

Let $r$ satisfy $r\leqslant 1$, and $B_{\omega}(\phi^{-1}(x),2r)$ is a geodesically convex subset of $(N_{0},\omega)$. If $q\in f^{-1}(B_{\omega}(\phi^{-1}(x),2r))$, there is a curve $\bar{\gamma}$ connecting $p^{\prime}$ and $q$ such that $f(\bar{\gamma})$ is the unique minimal geodesic connecting $\phi^{-1}(x)$ and $f(q)$. Thanks to (4.18) we have

$$f^{*}\omega-\varepsilon(t_{k})\omega_{M}\leqslant\tilde{\omega}_{t_{k}}\leqslant f^{*}\omega+\varepsilon(t_{k})\omega_{M}$$

where $\varepsilon(t_{k})\rightarrow 0$ when $t_{k}\rightarrow 0$, on $f^{-1}(B_{\omega}(\phi^{-1}(x),2r))$. We obtain that

$$\begin{split}d_{\tilde{\omega}_{t_{k}}}(p^{\prime},q)&\leqslant{\rm length}_{\tilde{\omega}_{t_{k}}}(\bar{\gamma})\\ &\leqslant{\rm length}_{\omega}(f(\bar{\gamma}))+C\varepsilon(t_{k})^{\frac{1}{2}}\\ &=d_{\omega}(\phi^{-1}(x),f(q))+C\varepsilon(t_{k})^{\frac{1}{2}}.\end{split}$$

If $\bar{\gamma}_{k}$ is a minimal geodesic of $\tilde{\omega}_{t_{k}}$ connecting $p^{\prime}$ and $q$, then (4.19) gives

$$d_{\tilde{\omega}_{t_{k}}}(p^{\prime},q)={\rm length}_{\tilde{\omega}_{t_{k}}}(\bar{\gamma}_{k})\geqslant e^{-\frac{\varepsilon(t_{k})}{2}}{\rm length}_{\omega}(f(\bar{\gamma}_{k})\cap B_{\omega}(\phi^{-1}(x),2r)).$$

If $f(\bar{\gamma}_{k})\subset B_{\omega}(\phi^{-1}(x),2r)$, then

$${\rm length}_{\omega}(f(\bar{\gamma}_{k})\cap B_{\omega}(\phi^{-1}(x),2r))\geqslant{\rm length}_{\omega}(f(\bar{\gamma}))=d_{\omega}(\phi^{-1}(x),f(q)),$$

and, otherwise,

$${\rm length}_{\omega}(f(\bar{\gamma}_{k})\cap B_{\omega}(\phi^{-1}(x),2r))\geqslant 2r\geqslant{\rm length}_{\omega}(f(\bar{\gamma}))=d_{\omega}(\phi^{-1}(x),f(q)),$$

by the same argument as in the proof of Lemma 5.1. Thus

$$e^{-\frac{\varepsilon(t_{k})}{2}}d_{\omega}(\phi^{-1}(x),f(q))\leqslant d_{\tilde{\omega}_{t_{k}}}(p^{\prime},q)\leqslant d_{\omega}(\phi^{-1}(x),f(q))+C\varepsilon(t_{k})^{\frac{1}{2}}$$

where $C$ is a constant independent of $t_{k}$, $p^{\prime}$ and $q$. Of course if $k$ is large we will have that

$$d_{\omega}(\phi^{-1}(x),f(q))-C\varepsilon(t_{k})^{\frac{1}{2}}\leqslant e^{-\frac{\varepsilon(t_{k})}{2}}d_{\omega}(\phi^{-1}(x),f(q)).$$

Thanks to (4.1), there is constant $C>0$ independent of $t_{k}$ such that $\tilde{\omega}_{t_{k}}\leqslant C\omega_{M}$ on $f^{-1}(B_{\omega}(\phi^{-1}(x),2r))$. Let $\gamma^{\prime}_{k}$ be minimal geodesics of $\omega_{M}$ connecting $p_{k}$ and $p^{\prime}$, which satisfy $\gamma^{\prime}_{k}\subset f^{-1}(B_{\omega}(\phi^{-1}(x),2r))$ for $k\gg 1$. Thus

$$d_{\tilde{\omega}_{t_{k}}}(p^{\prime},p_{k})\leqslant{\rm length}_{\tilde{\omega}_{t_{k}}}(\gamma^{\prime}_{k})\leqslant C^{\frac{1}{2}}{\rm length}_{\omega_{M}}(\gamma^{\prime}_{k})=C^{\frac{1}{2}}d_{\omega_{M}}(p^{\prime},p_{k})\rightarrow 0.$$

The triangle inequality shows that

$$|d_{\tilde{\omega}_{t_{k}}}(p_{k},q)-d_{\omega}(\phi^{-1}(x),f(q))|\leqslant C\varepsilon(t_{k})^{\frac{1}{2}}+C^{\frac{1}{2}}d_{\omega_{M}}(p^{\prime},p_{k}).$$

Hence there is a function $\rho(t_{k})$ of $t_{k}$ such that $\rho(t_{k})\rightarrow 0$ when $t_{k}\rightarrow 0$, and

$$f^{-1}(B_{\omega}(\phi^{-1}(x),r-\rho(t_{k})))\subset B_{\tilde{\omega}_{t_{k}}}(p_{k},r)\subset f^{-1}(B_{\omega}(\phi^{-1}(x),r+\rho(t_{k}))).$$

We obtain that

$$\lim_{t_{k}\rightarrow 0}\int_{B_{\tilde{\omega}_{t_{k}}}(p_{k},r)}\omega_{M}^{n}=\int_{f^{-1}(B_{\omega}(\phi^{-1}(x),r))}\omega_{M}^{n}.$$

Note that

$$\tilde{\omega}_{t_{k}}^{n}=c_{t_{k}}t_{k}^{n-m}\omega_{M}^{n}.$$

Hence

$$\begin{split}\underline{V}_{k}(p_{k},r)&=\frac{{\rm Vol}_{\tilde{\omega}_{t_{k}}}(B_{\tilde{\omega}_{t_{k}}}(p_{k},r))}{{\rm Vol}_{\tilde{\omega}_{t_{k}}}(B_{\tilde{\omega}_{t_{k}}}(\bar{p}_{k},1))}\\ &=\frac{\int_{B_{\tilde{\omega}_{t_{k}}}(p_{k},r)}c_{t_{k}}t_{k}^{n-m}\omega_{M}^{n}}{\int_{B_{\tilde{\omega}_{t_{k}}}(\bar{p}_{k},1)}c_{t_{k}}t_{k}^{n-m}\omega_{M}^{n}}\to\frac{\int_{f^{-1}(B_{\omega}(\phi^{-1}(x),r))}\omega_{M}^{n}}{\int_{f^{-1}(B_{\omega}(\phi^{-1}(\bar{x}),1))}\omega_{M}^{n}},\end{split}$$

when $t_{k}\rightarrow 0$. By (5.1),

$$\underline{V}_{0}(x,r)=\upsilon\int_{f^{-1}(B_{\omega}(\phi^{-1}(x),r))}\omega_{M}^{n},\ \ {\rm where}\ \ \upsilon=\left(\int_{f^{-1}(B_{\omega}(\phi^{-1}(\bar{x}),1))}\omega_{M}^{n}\right)^{-1}.$$

Recall the diameter bound (1.4)

$${\rm diam}_{\tilde{\omega}_{t_{k}}}(M)\leqslant D$$

for a constant $D>0$. Using (5.1), we have

$$\nu(X)=\underline{V}_{0}(x,D)=\lim_{t_{k}\rightarrow 0}\underline{V}_{k}(p_{k},D)=\upsilon\int_{M}\omega_{M}^{n}.$$

∎