Let $A$ be a countable dense subset of $N_{0}$, and $K\subset N_{0}$ be a compact subset with the interior $\operatorname{int}K$ non-empty. Let $\{B_{i}\}$ be a finite covering of $K$ with small Euclidean balls such that each the concentric balls $B_{i}^{\prime}$ of half radius still cover $K$. Let $s_{i}:B_{i}\rightarrow f^{-1}(B_{i})$ be sections on $B_{i}$, i.e., holomorphic maps with $f\circ s_{i}={\rm id}$.

Now, we define a map $\phi$ from $A\cap K=\{a_{1},a_{2},\cdots\}$ to $X$. Suppose that the point $a_{1}$ lies inside the ball $B_{i}^{\prime}$, and consider the points $s_{i}(a_{1})$ inside $M$. Under the Gromov-Hausdorff convergence of $(M,\tilde{\omega}_{t_{k}})$ to $(X,d_{X})$, a subsequence of these points converges to a point $b_{1}$ in $X$, because the diameter of $(M,\tilde{\omega}_{t_{k}})$ is uniformly bounded. If $a_{1}$ also lies inside another ball $B_{j}^{\prime}$, then (1.3) (or also [38, (2.10)]) shows that $d_{\tilde{\omega}_{t_{k}}}(s_{i}(a_{1}),s_{j}(a_{1}))\rightarrow 0$ when $t_{k}\rightarrow 0$. Thus, by passing to subsequences, both $s_{i}(a_{1})$ and $s_{j}(a_{1})$ converge to the same point $b_{1}\in X$ under the Gromov-Hausdorff convergence of $(M,\tilde{\omega}_{t_{k}})$ to $(X,d_{X})$. We then define $\phi(a_{1})=b_{1}$. For $a_{2}$, by repeating the above procedure, we obtain that a subsequence $s_{i_{j}}(a_{j})$, $j=1,2$, converges to $b_{j}\in X$, $j=1,2$, respectively. Define $\phi(a_{2})=b_{2}$. By repeating this procedure and with a diagonal argument, we can find a subsequence of $(M,\tilde{\omega}_{t_{k}})$, denoted by $(M,\tilde{\omega}_{t_{k}})$ also, such that $s_{i_{j}}(a_{j})$ converges to $b_{j}\in X$ along the Gromov-Hausdorff convergence. For any $a_{j}\in A\cap K$, define $\phi(a_{j})=b_{j}$.

Now, we prove that $\phi:A\cap\operatorname{int}K\rightarrow X$ is injective. If it is not true, there are $y_{1}$, $y_{2}\in A\cap\operatorname{int}K$ such that $y_{1}\neq y_{2}$, and $\phi(y_{1})=\phi(y_{2})$, which implies $d_{\tilde{\omega}_{t_{k}}}(s_{i_{1}}(y_{1}),s_{i_{2}}(y_{2}))\rightarrow 0$. If $\gamma_{k}$ is a minimal geodesic in $(M,\tilde{\omega}_{t_{k}})$ connecting $s_{i_{1}}(y_{1})$ and $s_{i_{2}}(y_{2})$, then

$$C^{-1}{\rm length}_{\omega_{N}}(f(\gamma_{k})\cap K)\leqslant{\rm length}_{\tilde{\omega}_{t_{k}}}(\gamma_{k}\cap f^{-1}(K))\leqslant d_{\tilde{\omega}_{t_{k}}}(s_{i_{1}}(y_{1}),s_{i_{2}}(y_{2})),$$

by (4.1) for a constant $C>0$ independent of $k$. Thus, if $f(\gamma_{k})\subset K$ for $t_{k}\ll 1$,

$$d_{\omega_{N}}(y_{1},y_{2})\leqslant C{\rm length}_{\omega_{N}}(f(\gamma_{k}))\longrightarrow 0,$$

or, if $f(\gamma_{k})\cap N\backslash K$ are not empty by passing to a subsequence,

$$d_{\omega_{N}}(y_{1},\partial K)+d_{\omega_{N}}(\partial K,y_{2})\leqslant C{\rm length}_{\omega_{N}}(f(\gamma_{k})\cap K)\longrightarrow 0.$$

In both cases, we obtain contradictions. Thus $\phi:A\cap\operatorname{int}K\rightarrow X$ is injective.

Note that there is a $r>0$ such that, for any $y\in\operatorname{int}K$, the metric ball $B_{\omega}(y,r)$ is a geodesically convex set, i.e. for any $y_{1}$ and $y_{2}\in B_{\omega}(y,r)$, there is a minimal geodesic $\gamma\subset B_{\omega}(y,r)$ connecting $y_{1}$ and $y_{2}$, which implies

$$d_{\omega}(y_{1},y_{2})={\rm length}_{\omega}(\gamma)\leqslant 2r.$$

We take $r\ll 1$ such that there is a $B_{i}^{\prime}$ with $B_{\omega}(y,2r)\subset B_{i}^{\prime}$. If $y_{1},y_{2}\in A$, by Proposition 4.6,

$$\begin{split}d_{X}(\phi(y_{1}),\phi(y_{2}))&=\lim_{t_{k}\rightarrow 0}d_{\tilde{\omega}_{t_{k}}}(s_{i}(y_{1}),s_{i}(y_{2}))\\ &\leqslant\lim_{t_{k}\rightarrow 0}{\rm length}_{\tilde{\omega}_{t_{k}}}(s_{i}(\gamma))\\ &={\rm length}_{\omega}(\gamma)\\ &=d_{\omega}(y_{1},y_{2}).\end{split}$$

If $\gamma_{k}$ is a minimal geodesic in $(M,\tilde{\omega}_{t_{k}})$ connecting $s_{i}(y_{1})$ and $s_{i}(y_{2})$, then (4.19) implies that

$$e^{-\frac{\varepsilon(t_{k})}{2}}{\rm length}_{\omega}(f(\gamma_{k})\cap B_{\omega}(y,2r))\leqslant{\rm length}_{\tilde{\omega}_{t_{k}}}(\gamma_{k})\longrightarrow d_{X}(\phi(y_{1}),\phi(y_{2})),$$

for some function $\varepsilon(t)\to 0$ as $t\to 0$. If $f(\gamma_{k})\subset B_{\omega}(y,2r)$ for $t_{k}\ll 1$ by passing to a subsequence,

$${\rm length}_{\omega}(f(\gamma_{k}))\geqslant{\rm length}_{\omega}(\gamma),$$

since $\gamma$ is a minimal geodesic in $(N_{0},\omega)$. If $f(\gamma_{k})\cap N_{0}\backslash B_{\omega}(y,2r)$ is not empty for $t_{k}\ll 1$, then there is a $\bar{y}\in f(\gamma_{k})\cap N_{0}\backslash B_{\omega}(y,2r)$. Since $y_{1}$, $y_{2}\in B_{\omega}(y,r)$ and $f(\gamma_{k})$ connects $y_{1}$ and $y_{2}$,

$${\rm length}_{\omega}(f(\gamma_{k})\cap B_{\omega}(y,2r))\geqslant d_{\omega}(y_{1},\bar{y})+d_{\omega}(y_{2},\bar{y})\geqslant 2r\geqslant{\rm length}_{\omega}(\gamma).$$

In both cases,

$$\begin{split}d_{\omega}(y_{1},y_{2})&={\rm length}_{\omega}(\gamma)\\ &\leqslant\lim_{t_{k}\rightarrow 0}{\rm length}_{\omega}(f(\gamma_{k})\cap B_{\omega}(y,2r))\\ &\leqslant d_{X}(\phi(y_{1}),\phi(y_{2})).\end{split}$$

Thus

$$d_{\omega}(y_{1},y_{2})=d_{X}(\phi(y_{1}),\phi(y_{2})),$$

i.e. $\phi:(A\cap\operatorname{int}K,d_{\omega})\longrightarrow(X,d_{X})$ is a local isometric embedding. If $\{y_{1,j}\}$ and $\{y_{2,j}\}$ are two sequences in $A\cap\operatorname{int}K$ such that $\lim\limits_{j\rightarrow\infty}d_{\omega}(y_{i,j},y)=0$ for $i=1,2$, then $\lim\limits_{j\rightarrow\infty}d_{\omega}(y_{1,j},y_{2,j})=0$ and $\{y_{1,j},y_{2,j}\}\subset B_{\omega}(y,r)$ for $j\gg 1$. Hence $d_{\omega}(y_{1,j},y_{2,j})=d_{X}(\phi(y_{1,j}),\phi(y_{2,j}))$ and $d_{\omega}(y_{i,j},y_{i,j+\ell})=d_{X}(\phi(y_{i,j}),\phi(y_{i,j+\ell}))$ for $j\gg 1$ and any $\ell\geqslant 0$, which implies that $\{\phi(y_{1,j})\}$ and $\{\phi(y_{2,j})\}$ are two Cauchy sequences, and converge to a unique point $x\in X$. By defining $\phi(y)=x$, $\phi$ extends to a unique map, denoted still by $\phi$, from $\operatorname{int}K$ to $X$ which is also a local isometric embedding.

Now we prove that $\phi(\operatorname{int}K)$ is an open subset of $X$. Let $x\in\phi(\operatorname{int}K)$, i.e. there is a $y\in\operatorname{int}K$ such that $\phi(y)=x$, and let $x^{\prime}\in X$ with $d_{X}(x,x^{\prime})<\rho$ for a constant $\rho<\frac{1}{8}d_{\omega}(y,\partial K)$. From the above construction, $y\in B_{i}^{\prime}$ for a $B_{i}^{\prime}$, and $s_{i}(y)\rightarrow x$ under Gromov-Hausdorff convergence. There is a sequence of points $p_{k}\in(M,\tilde{\omega}_{t_{k}})$ such that $p_{k}\rightarrow x^{\prime}$ under the Gromov-Hausdorff convergence. If $\gamma_{k}^{\prime}$ is a minimal geodesic connecting $s_{i}(y)$ and $p_{k}$ in $(M,\tilde{\omega}_{t_{k}})$, then

$$d_{\tilde{\omega}_{t_{k}}}(s_{i}(y),p_{k})={\rm length}_{\tilde{\omega}_{t_{k}}}(\gamma_{k}^{\prime})\longrightarrow d_{X}(x,x^{\prime}).$$

Equation (4.19) implies that, for $k\gg 1$,

$$\begin{split}\frac{1}{2}{\rm length}_{\omega}(f(\gamma_{k}^{\prime})\cap K)&\leqslant e^{-\frac{\varepsilon(t_{k})}{2}}{\rm length}_{\omega}(f(\gamma_{k}^{\prime})\cap K)\\ &\leqslant{\rm length}_{\tilde{\omega}_{t_{k}}}(\gamma_{k}^{\prime})\\ &<2\rho<\frac{1}{4}d_{\omega}(y,\partial K).\end{split}$$

Thus $f(p_{k})\in K^{\prime}\subset\operatorname{int}K$ where $K^{\prime}$ is a compact subset of $\operatorname{int}K$. By passing to a subsequence, $f(p_{k})\rightarrow y^{\prime}$ in $(K^{\prime},\omega)$. By Proposition 4.6, $d_{\tilde{\omega}_{t_{k}}}(p_{k},s_{i_{k}}(f(p_{k})))\rightarrow 0$ when $t_{k}\rightarrow 0$, and, thus, $s_{i_{k}}(f(p_{k}))\rightarrow x^{\prime}$ under the Gromov-Hausdorff convergence. The above construction shows that $\phi(y^{\prime})=x^{\prime}$, which implies that $\{x^{\prime}|d_{X}(x,x^{\prime})<\rho\}\subset\phi(\operatorname{int}K)$. Hence $\phi(\operatorname{int}K)$ is open, and $\phi:\operatorname{int}K\longrightarrow\phi(\operatorname{int}K)$ is a homeomorphism.

Let $K_{0}\subset\cdots\subset K_{j}\subset K_{j+1}\subset\cdots\subset N_{0}$ be a family of compact subsets with $N_{0}=\bigcup\limits_{j}\operatorname{int}K_{j}$. Given each $K_{j}$, the above argument constructs a local isometric embedding $\phi_{j}:(\operatorname{int}K_{j},\omega)\longrightarrow(X,d_{X})$, which is a homeomorphism onto the image $\phi_{j}(\operatorname{int}K_{j})$. By the same argument as above, $\phi_{j}$ extends to a local isometric embedding $\phi_{j+1}:(\operatorname{int}K_{j+1},\omega)\longrightarrow(X,d_{X})$, i.e. $\phi_{j+1}|_{\operatorname{int}K_{j}}=\phi_{j}$, which is a homeomorphism onto the image $\phi_{j+1}(\operatorname{int}K_{j+1})$. By a diagonal argument, we obtain a local isometry $\phi:(N_{0},\omega)\longrightarrow(\phi(N_{0}),d_{X})\subset(X,d_{X})$. ∎

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

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

∎

We prove that $X_{0}\subset X$ is dense. If this is not true, there is a metric ball $B_{d_{X}}(x^{\prime},\rho)\subset X\backslash X_{0}$. Note that

$${\rm diam}_{d_{X}}(X)=\lim_{t_{k}\rightarrow 0}{\rm diam}_{\tilde{\omega}_{t_{k}}}(M)\leqslant D.$$

Because of (5.2), we have

$$\nu(B_{d_{X}}(x^{\prime},\rho))\geqslant\mu(\rho,D)\nu(X)=\varpi>0.$$

For any compact subset $K\subset X_{0}$,

$$\nu(K)\leqslant\nu(X)-\varpi=\upsilon\int_{M}\omega_{M}^{n}-\varpi$$

by Lemma 5.2. If $B_{d_{X}}(x_{i},r_{i})$ is a family of metric balls in $(X,d_{X})$ such that $r_{i}<\delta\ll 1$, $B_{d_{X}}(x_{i},2r_{i})$ is a geodesically convex subset of $X_{0}$, and $\bigcup\limits_{i}B_{d_{X}}(x_{i},r_{i})\supset K$, then

$$\sum_{i}\underline{V}_{0}(x_{i},r_{i})=\sum_{i}\upsilon\int_{f^{-1}(\phi^{-1}(B_{d_{X}}(x_{i},r_{i})))}\omega_{M}^{n}\geqslant\upsilon\int_{f^{-1}(\phi^{-1}(K))}\omega_{M}^{n}$$

by Lemma 5.2. Thus

$$\upsilon\int_{f^{-1}(\phi^{-1}(K))}\omega_{M}^{n}\leqslant\lim_{\delta\rightarrow 0}\nu_{\delta}(K)=\lim_{\delta\rightarrow 0}\inf\left\{\sum_{i}\underline{V}_{0}(x_{i},r_{i})|r_{i}<\delta\right\}=\nu(K).$$

By taking $K$ large enough such that

$$\nu(K)\geqslant\upsilon\int_{f^{-1}(N_{0})}\omega_{M}^{n}-\frac{\varpi}{2}=\upsilon\int_{M}\omega_{M}^{n}-\frac{\varpi}{2},$$

we obtain a contradiction. ∎