A property of measures on a Polish space

Question

Let $(\mathbf{Y}, \mathcal{Y})$ be a Polish space. Then, there exists a sequence of measurable functions $(\phi_{n})_{n\in\mathbb{N}}$ from $(\mathbf{Y}, \mathcal{Y})$ to $\mathbb{R}$ such that, for two probability measures $\nu_{1}$ and $\nu_{2}$ on $(\mathbf{Y}, \mathcal{Y})$, one has: $$ \left(\int_{\mathbf{Y}}\phi_{n}d\nu_{1}\right)_{n\in\mathbb{N}} = \left(\int_{\mathbf{Y}}\phi_{n}d\nu_{2}\right)_{n\in\mathbb{N}} \implies \nu_{1} = \nu_{2} $$ I think I must exploit the separability property of space $\mathbf{Y}$,but I can't see how to do so. Any hint/help would be appreciated.

Answer 1

This is how I see it. Clearly, $\nu_1(U) = \nu_2(U)$ for any open $U \subset Y$ is enough to conclude $\nu_1 = \nu_2$ (since open sets are stable by intersections and generate Borel $\sigma-$field).

Let $\{x_k\}_{k \in \mathbb N}$ be a countable dense subset of $Y$ (by separability). Define $\phi_{k,m} = 1_{B(x_k,\frac{1}{m})}$, that is an indicator function of open ball $B(x_k,\frac{1}{m}) = \{y \in Y : d(x_k,y) < \frac{1}{m}\}$.

Note that, since $Y$ is a separable metric space, any open set $U \subset Y$ can be written as $$U = \bigcup_{\substack{k,m \in \mathbb N \\ B(x_k,\frac{1}{m}) \subset U}} B(x_k,\frac{1}{m})$$ Indeed, inclusion $\supset$ is simple, since any $B(x_k,\frac{1}{m})$ is contained in $U$ (we're taking union of only those balls that are inside of $U$). For the other inclusion, take any $y \in U$. Then, we have some $B(y,\frac{1}{m}) \subset U$. By definition of a dense subset, we have some $x_k$ such that $d(x_k,y) < \frac{1}{2m}$. Then, by triangle inequality, $B(x_k,\frac{1}{2m}) \subset U$ and the result follows.

Let $(\phi_n)_n$ be an enumeration of $\phi_{k,m}$. Note that $\int_{Y}\phi_{k,m}d\nu_1 = \int_{Y}\phi_{k,m}d\nu_2$ means $\nu_1(B(x_k,\frac{1}{m})) = \nu_2(B(x_k,\frac{1}{m}))$. We want to prove $\nu_1(U) = \nu_2(U)$ for any open set $U$. Now, for any $n \in \mathbb N$ define $$ U_n = \bigcup_{ \substack{k,m \le n \\ B(x_k,\frac{1}{m}) \subset U}} B(x_k,\frac{1}{m})$$ (that is, similar as $U$ itself, but we take only a finite sum). Note that $(U_n)_n$ is an ascending sequence of sets and $U = \bigcup_{n} U_n$, so if we show $\nu_1(U_n) = \nu_2(U_n)$ we're done by continuity from below of probability.

But do we really need to show it? NO! For any $n$, there are finitelly many sets generated by $ B(x_k,\frac{1}{m})$ for $k,m \le n$ (What I mean by that, is for example for $n=2$ we have $B(x_1,1),B(x_2,1), B(x_1,\frac{1}{2}),B(x_2,\frac{1}{2}), B(x_1,1) \cup B(x_2,1), B(x_1,1) \cup B(x_2,\frac{1}{2})$, etc.) (If I'm counting correctly, there should be at most $2^{n^2}$ of such sets for given $n$). Now, your countable sequence of functions is just a sequence of indicators of all of those sets for any $n \in \mathbb N$ (since for given $n$ we have at most $2^{n^2}$ of them, we have countably many such sets in total)).