Let $n \in \mathbb{N}$. Show that every symmetric function $f\colon E^n \rightarrow \mathbb{R}$ can be written in the form $f(x) = g\Bigl(\frac{1}{n}\sum_{i=1}^n \delta_{x_i} \Bigr)$, where $g$ has to be chosen appropriately (dependent on $f$).
$E$ is an arbitrary Polish space.
First I asked about the meaning of "$g$". Since we have concluded that it is a map $$g\colon \mathcal{M}_{\mathrm{fin}} (E) \rightarrow \mathbb{R} $$ where $\mathcal{M}_{\mathrm{fin}}(E)$ means the space of all finite measures on $(E,\mathcal{B}(E))$, it's now about solving the exercise.
We have to find a way to "extract" as much information about the vector $(x_1, \ldots, x_n)\in E^n$ from the measure $\mu := \frac{1}{n}\sum_{i=1}^n \delta_{x_i}$ as possible.
Since $E$ is a Polish space, there exists a countable, dense subset $Z= \{z_1, z_2, \ldots\}$ of $E$. Define the countable family of sets $$\mathcal{S}_\varepsilon := \{ B_\varepsilon(z_i)\in \mathcal{B}(E): \mu\bigl(B_\varepsilon(z_i)\bigr)>0 \text{ and } z_i \in Z\}\, ,$$ then $$ R := \bigcap_{m=1}^\infty\; \bigcup_{S\in\mathcal{S}_{1/m}} S \, .$$
Now $R = \{x_1, \ldots, x_n\}$, though we don't get any information about the original order of the elements. Assume $R$ consists of different $r_1, \ldots, r_k \in E$ with $k \leq n$. We construct a vector $\vec{v}_\mu\in E^n$ by $$ \vec{v}_\mu := (\underbrace{r_1, \ldots, r_1}_{n \mu(r_1)-\text{times}}, \ldots, \underbrace{r_k, \ldots, r_k}_{n \mu(r_k)-\text{times}})\, ,$$ so the number of occurrences of an $x_i$ is the same in $\vec{v}_\mu$ and $(x_1, \ldots, x_n)$.
Since $f$ is symmetric, which means $$ f(x_1, \ldots, x_n) = f(x_{\rho(1)}, \ldots, x_{\rho(n)}) \quad \forall \rho \in S_n \, ,$$ we also have $$f(x_1, \ldots, x_n) = f(\vec{v}_\mu)\, ,$$ this defines the map $g(\mu) := f(\vec{v}_\mu)$ (aside from the detail that we also have to arbitrarily define $g$ for all the measures we don't care about; only probability measures of the form $\mu = \frac{1}{n} \sum_{i=1}^n\delta_{x_i}$ are important). $\square$
Is that ok?