Given $f\in\mathcal{D}(X)$, we have by definition that $\int f\MA(\varphi^{\langle t\rangle})\to\int f\MA(\varphi)$ as $t\to\infty$ and $\int f\MA(\varphi^{\langle t\rangle}_{j})\to\int f\MA(\varphi_{j})$ as $t\to\infty$ for every $j$. Moreover, Lemma 6.8 shows that the latter convergence is uniform in $j$. Since for each $t$ we have $\int f\MA(\varphi^{\langle t\rangle}_{j})\to\int f\MA(\varphi^{\langle t\rangle})$ as $j\to\infty$ by Theorem 3.1, the result follows. ∎