Let $E$ be a Borel subset of $X$. Since $Cap_{\omega}(E\cap{\mathcal{U}}_{\alpha}^{\delta})\leq Cap_{\omega}(E)\leq\sum_{\alpha}Cap_{\omega}(E\cap{\mathcal{U}}_{\alpha}^{\delta})$, it is sufficient to show that if $\Omega=\{x\in X\,/\,\varrho(x)<0\}$ is a smooth hyperconvex subset of $X$, then there exists $C\geq 1$ such that for all $E\subset\Omega_{\delta}$,

$$\frac{1}{C}Cap_{\omega}(E)\leq Cap_{BT}(E,\Omega)\leq C\cdot Cap_{\omega}(E),$$

where $\Omega_{\delta}=\{x\in X\,/\,\varrho(x)<-\delta\}$.

It is an easy and well known fact in the local theory that the capacities $Cap(\cdot,\Omega)$ and $Cap(\cdot,\Omega^{\prime})$ are comparable when $\Omega^{\prime}\subset\Omega$ (see e.g. theorem 6.5 in [12]). Therefore we can assume (passing to a finer covering if necessary) that $\omega=dd^{c}\psi$ near $\overline{\Omega}$. Fix $C_{1}>0$ such that $-C_{1}\leq\psi\leq C_{1}$ on $\Omega$. Fix $\varphi\in PSH(X,\omega)$ such that $0\leq\varphi\leq 1$ on $X$ and set $u=(2C_{1})^{-1}(\varphi+\psi+C_{1})$. Then $u\in PSH(\Omega)$ and $0\leq u\leq 1$, hence

$$\int_{E}\omega_{\varphi}^{n}=(2C_{1})^{n}\int_{E}(dd^{c}u)^{n}\leq(2C_{1})^{n}Cap_{BT}(E,\Omega),$$

which yields $Cap_{\omega}(E)\leq(2C_{1})^{n}Cap_{BT}(E,\Omega)$. Observe that we have not used here that $\omega$ is Kähler.

For the reverse inequality we consider $\chi\in{\mathcal{C}}^{\infty}(X)$ such that $\chi\equiv 0$ in $X\setminus\Omega$ and $\chi<0$ in $\Omega$. Replacing $\chi$ by $\varepsilon\chi$ if necessary, we can assume $\chi\in PSH(X,\omega)$. This is because $\omega$ is Kähler (and this is the only place where we shall use this crucial assumption). Fix $\varepsilon>0$ so small that $\chi\leq-\varepsilon$ on $\Omega_{\delta}$. Let now $u\in PSH(\Omega)$ be such that $0\leq u\leq 1$ on $\Omega$. Consider

$$\varphi(x)=\left\{\begin{array}[]{cl}\frac{u-\psi+C_{1}}{2+2C_{1}}&\text{ in }\Omega_{\delta}\\ \max\left(\frac{u-\psi+C_{1}}{2+2C_{1}},\frac{2}{\varepsilon}\chi(x)+1\right)&\text{ in }\Omega\setminus\Omega_{\delta}\\ 1&\text{ in }X\setminus\Omega\end{array}\right.$$

Observe that $0\leq u^{\prime}:=(u-\psi+C_{1})/(2+2C_{1})\leq(1+2C_{1})/(2+2C_{1})<1$ in $\Omega$. Therefore $\varphi\in PSH(X,\frac{2}{\varepsilon}\omega)$ since $\frac{2}{\varepsilon}\chi(x)+1\leq-1<u^{\prime}$ in $\Omega_{\delta}$, while $\frac{2}{\varepsilon}\chi(x)+1\equiv 1>u^{\prime}$ on $\partial\Omega$. Note also that $0\leq\varphi\leq 1$ thus for $E\subset\Omega_{\delta}$,

	$\displaystyle\frac{1}{(2+2C_{1})^{n}}\int_{E}(dd^{c}u)^{n}$	$\displaystyle=$	$\displaystyle\int_{E}\left(\frac{\omega}{2+2C_{1}}+dd^{c}\varphi\right)^{n}\leq\int_{E}\left(\frac{2}{\varepsilon}\omega+dd^{c}\varphi\right)^{n}$
		$\displaystyle\leq$	$\displaystyle Cap_{2\omega/\varepsilon}(E)\leq\left(\frac{2}{\varepsilon}\right)^{n}Cap_{\omega}(E)$

hence $Cap_{BT}(E,\Omega)\leq 4^{n}(1+C_{1})^{n}\varepsilon^{-n}Cap_{\omega}(E)$. ∎