We may assume $0\leq h\leq 1$, $-M\leq\varphi\leq 0$ and $-M\leq\varphi_{j}\leq 0$ for all $j$, where $M\geq 1$. Given $\varepsilon>0$, let $G$ be an open set such that $\Capa_{\omega}(G)<\varepsilon$ and $h$ is continuous on $G^{c}$, see Definition 4.4. Using the Tietze extension theorem, we extend $h|_{G^{c}}$ to a continuous function $\tilde{h}$ on all of $X$ such that $0\leq\tilde{h}\leq 1$. We then have

	$\displaystyle\int h\MA(\varphi_{j})-\int h\MA(\varphi)$	$\displaystyle=\int\tilde{h}\MA(\varphi_{j})-\int\tilde{h}\MA(\varphi)$
		$\displaystyle+\int\limits_{G}(h-\tilde{h})\MA(\varphi_{j})-\int\limits_{G}(h-\tilde{h})\MA(\varphi).$

It follows from Lemma 4.6 that

$$\left|\int h\MA(\varphi_{j})-\int h\MA(\varphi)\right|\leq\left|\int\tilde{h}\MA(\varphi_{j})-\int\tilde{h}\MA(\varphi)\right|+2\sup|h-\tilde{h}|M^{n}\,\Capa_{\omega}(G).$$

Since $\tilde{h}$ is continuous, $\int\tilde{h}\MA(\varphi_{j})\to\int\tilde{h}\MA(\varphi)$ as $j\to\infty$, thus

$$\limsup_{j}\left|\int h\MA(\varphi_{j})-\int h\MA(\varphi)\right|\leq 4\varepsilon.$$

Letting $\varepsilon$ tend to zero completes the proof. ∎