For any two probability measures $\mu,\nu$ over $\mathbb{R}^d$, the Prokhorov distance is defined to be $$d_P(\mu,\nu)=\inf\{\epsilon:\mu(A)\le \nu(A^{\epsilon})+\epsilon\text{ and }\nu(A)\le\mu(A^{\epsilon})+\epsilon~\text{for all Borel set }A\subset\mathbb{R}^d\},$$ where $A^{\epsilon}=\{x\in \mathbb{R}^d:\forall y\in A,~||x-y||_2\le\epsilon\}$.
Suppose we have $i.i.d.$ samples $X_1,X_2,\cdots,X_n$ from some probability measure $\mu$. The empirical measure is defined to be $$\hat{\mu}_n=\frac{1}{n}\sum_{i=1}^n\delta_{X_i},$$ where $\delta_{X_i}$ is Dirac measure on $X_i$.
My question is: for any $\epsilon>0$ do we have concentration inequality $$\mathrm{Pr}[d_P(\hat{\mu}_n,\mu)\ge \epsilon]\le C(\epsilon, n)$$ where $C(\epsilon, n)\rightarrow 0$ as $n\rightarrow\infty$ and depends only on $\epsilon$ and $n$.
Note that, similar concentration ineqaulity holds for the Komogrov-Smirnov distance by the DKW-inequality. But KS-distance is not consistent with weak convergence. I'm wondering whether similar result hold for the Prokhorov distance.
By the law of large numbers, \begin{equation*} \frac{1}{n} \sum_{i = 1}^{n} f(X_{i}) \to \int_{\mathbb{R}^{d}} f(x) \, \mu(dx) \quad \text{a.s. if} \, \, f \in BC(\mathbb{R}^{d}) \end{equation*} Thus, since $BC(\mathbb{R}^{d})$ is separable, $\hat{\mu}_{n} \to \mu$ almost surely.
This implies, in particular, that $\lim_{n \to \infty} \mathbb{P}\{d_{P}(\hat{\mu}_{n},\mu) \geq \epsilon\} = 0$ for each $\epsilon > 0$. Indeed, we know that $d_{P}(\hat{\mu}_{n},\mu) \to 0$ almost surely, and almost sure convergence implies convergence in probability.
You were looking for a $C(\epsilon,n)$ depending only on $\epsilon$ and $n$ with $\lim_{n \to \infty} C(\epsilon,n) = 0$. I found a $C(\epsilon,n,\mu)$. My best guess is this is the best one can do without imposing additional assumptions on $\mu$.