We have in Van der Vaart-Asymptotic Statistics
The proof is given as follow.
Proof. (i) $\Rightarrow$ (ii). Assume first that the distribution function of $X$ is continuous. Then condition (i) implies that $\mathrm{P}\left(X_n \in I\right) \rightarrow \mathrm{P}(X \in I)$ for every rectangle $I$. Choose a sufficiently large, compact rectangle $I$ with $\mathrm{P}(X \notin I)<\varepsilon$. A continuous function $f$ is uniformly continuous on the compact set $I$. Thus there exists a partition $I=\cup_j I_j$ into finitely many rectangles $I_j$ such that $f$ varies at most $\varepsilon$ on every $I_j$. Take a point $x_j$ from each $I_j$ and define $f_{\varepsilon}=\sum_j f\left(x_j\right) 1_{I_j}$. Then $\left|f-f_{\varepsilon}\right|<\varepsilon$ on $I$, whence if $f$ takes its values in $[-1,1]$,
$$\begin{aligned} &\left|\mathrm{E} f\left(X_n\right)-\mathrm{E} f_{\varepsilon}\left(X_n\right)\right| \leq \varepsilon+\mathrm{P}\left(X_n \notin I\right), \\ &\left|\mathrm{E} f(X)-\mathrm{E} f_{\varepsilon}(X)\right| \leq \varepsilon+\mathrm{P}(X \notin I)<2 \varepsilon\end{aligned}$$
For sufficiently large $n$, the right side of the first equation is smaller than $2 \varepsilon$ as well. We combine this with $$ \left|\mathrm{E} f_{\varepsilon}\left(X_n\right)-\mathrm{E} f_{\varepsilon}(X)\right| \leq \sum_j\left|\mathrm{P}\left(X_n \in I_j\right)-\mathrm{P}\left(X \in I_j\right)\right|\left|f\left(x_j\right)\right| \rightarrow 0 $$ Together with the triangle inequality the three displays show that $\left|\mathrm{E} f\left(X_n\right)-\mathrm{E} f(X)\right|$ is bounded by $5 \varepsilon$ eventually. This being true for every $\varepsilon>0$ implies (ii).
My question regards the preservation of the rate of convergence. In my case I know that
$$\sup _\mathbf{x} |P(X_n \leq \mathbf{x}) - P(X \leq \mathbf{x})| = O(n^{-1})$$
Hence could I rewrite the proof in order to keep this rate? Basically by setting $\varepsilon = \frac{C}{n}$ for a positive constant $C$. The proof would be then:
Proof. (i) $\Rightarrow$ (ii). Assume first that the distribution function of $X$ is continuous. Then condition (i) implies that $\mathrm{P}\left(X_n \in I\right) \rightarrow \mathrm{P}(X \in I)$ for every rectangle $I$. Choose a sufficiently large, compact rectangle $I$ with $\mathrm{P}(X \notin I)< \varepsilon = C/n$ where $C$ is a positive constant. A continuous function $f$ is uniformly continuous on the compact set $I$. Thus there exists a partition $I=\cup_j I_j$ into finitely many rectangles $I_j$ such that $f$ varies at most $C/n$ on every $I_j$. Take a point $x_j$ from each $I_j$ and define $f_{\varepsilon}=\sum_j f\left(x_j\right) 1_{I_j}$. Then $\left|f-f_{\varepsilon}\right|<C/n$ on $I$, whence if $f$ takes its values in $[-1,1]$,
$$\begin{aligned} &\left|\mathrm{E} f\left(X_n\right)-\mathrm{E} f_{\varepsilon}\left(X_n\right)\right| \leq C/n+\mathrm{P}\left(X_n \notin I\right), \\ &\left|\mathrm{E} f(X)-\mathrm{E} f_{\varepsilon}(X)\right| \leq C/n+\mathrm{P}(X \notin I)<2 C/n\end{aligned}$$
For sufficiently large $n$, the right side of the first equation is smaller than $C/n + P(X \notin I) + O(n^{-1})$. We combine this with $$ \left|\mathrm{E} f_{\varepsilon}\left(X_n\right)-\mathrm{E} f_{\varepsilon}(X)\right| \leq \sum_j\left|\mathrm{P}\left(X_n \in I_j\right)-\mathrm{P}\left(X \in I_j\right)\right|\left|f\left(x_j\right)\right| \rightarrow 0 $$ Together with the triangle inequality the three displays show that $\left|\mathrm{E} f\left(X_n\right)-\mathrm{E} f(X)\right|$ is bounded by $5 C/n = O(n^{-1})$ eventually.