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. ∎