Prove that the empirical measure is a measurable fucntion

This problem came from Schervish, Theory of Statistics, Sec. 1.4 Prob. 24.

Suppose that $X_1, \ldots, X_n$ are exchangeable and take values in the Borel space $(\mathcal{X}, \mathcal{B})$. Prove that the empirical probability measure $P_n$ is a measurable function from the $n$-fold product space $(\mathcal{X}^n,\mathcal{B}^n)$ to $(\mathcal{P}, \mathcal{C_p})$, where $\mathcal{P}$ is the sets of all probability measures on $(\mathcal{X}, \mathcal{B})$.

The question is how does the exchangeable condition on $X_1, \ldots, X_n$ work?

Here is what I have been tried:

Let $S \in \mathcal{C_p}$, we need to prove $P_n^{-1}(S) \in \mathcal{B}^n$. By the definition of the empirical probability measure, we have:

P_n^{-1}(S) = \Big\{(\omega_1,\ldots, \omega_n) \in \mathcal{X}^n | \frac{1}{n}\sum_{i=1}^n\delta_{X_i(\omega_i)} \in S\Big\}

Since $S$ contains probablity measures on $\mathcal{X}$, for any points in $\mathcal{B}^n$, we can get:

\frac{1}{n}\sum_{i=1}^n\delta_{X_i(\omega_i)}(B_i) \leq t \text{ for each } B_i \in \mathcal{B} \text{ and } t \in [0, 1]

Then we have $P_n^{-1}(S) \in \mathcal{B}^n$ since each $X_i$ is defined on $\mathcal{X}$.

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