Consider a Polish metric space $(X,d)$ and the class of Borel probability measures on it, endowed with the usual topology of weak convergence of measures, say $(\mathcal{P},\tau_W)$. Then, it is well known that $(\mathcal{P},\tau_W)$ is separable and completely metrizable – e.g. via the Léevy-Prohorov metric, see
https://en.wikipedia.org/wiki/Lévy–Prokhorov_metric
For simplicity, let $X$ be a subset of an Euclidean space and denote by $\mathcal{P}_0\subset\mathcal{P}$ the subset of probability measures which are absolutely continuous with respect to the Lebesgue measure. Clearly, $\mathcal{P}_0$ is not closed, for a sequence of absolutely continuous measures can converge to a point mass. Is $\mathcal{P}_0$ a $G_\delta$ set in $(\mathcal{P},\tau_W)$?
ADDENDUM: Let $\nu$ be the Lebesgue measure on $X$ and define
$$
\Delta=\{f \in L_1(\nu):\, \int_Xf d\nu=1,\, f \geq 0 \,a.e.\}.
$$
Then, in a 1989 paper,
https://www.jstor.org/stable/pdf/4616132.pdf?refreqid=excelsior%3A4ad9120f5b9af9f4af6aa1c012f5b44a
Gaudard and Hadwin argued that $\Delta$ is a closed subset of $L_1(\nu)$ (in fact, a Standard Borel subspace) and that the map $\Phi:\Delta\mapsto\mathcal{P} $ defined via
$$
[\Phi(f)](D)=\int_Df d\nu, \quad \text{for all } D \text{ Borel subset of }X,
$$
is $1-\text{to}-1$ and continuous. In particular, $\Phi$ is an isomorphism of $\Delta$ into $\Phi(\Delta)=\mathcal{P}_0$. Is this of any help? Of course, the above arguments entail that $\mathcal{P}_0$ is a Borel subset of $(\mathcal{P},\tau_W)$, but this, per se, does not guarantee that $\mathcal{P}_0$ is also $G_\delta$.
Best Answer
I think it is not $G_\delta$.
Take $X = [0,1]$ for instance and let $m$ denote Lebesgue measure. First, note that $\mathcal{P}_0$ is dense in $\mathcal{P}$.
Now I claim it is possible to find a homeomorphism $F$ of $[0,1]$ and a Borel set $E \subset [0,1]$ such that $m(E) = 1$ and $m(F(E)) = 0$.
(For instance, let $\mu_0$ be an atomless probability measure of full support which is singular to Lebesgue measure, so that there exists a Borel set $E$ with $m(E) = 1$ and $\mu_0(E)=0$, and let $F(x) = \mu_0([0,x])$ be the cdf of $\mu_0$, which one can check is a homeomorphism. Then we have $m(F(A)) = \mu_0(A)$ for $A$ of the form $[0,x]$, hence for all Borel sets $A$, and taking $A=E$ we have what we want.)
Let $G = F^{-1}$ which is also a homeomorphism, and for $\mu \in \mathcal{P}$ let $\Gamma\mu = G_* \mu$ be the pushforward of $\mu$ under $G$, i.e. $(\Gamma\mu)(A) = \mu(G^{-1}(A)) = \mu(F(A))$. It is a standard exercise to check that $\Gamma$ is a homeomorphism of $\mathcal{P}$. Moreover, if $\mu \in \mathcal{P}_0$ then we have $\mu(E) = 1$ but $\Gamma \mu(E) = \mu(F(E)) = 0$, so that $\Gamma \mu$ is singular to $m$. In particular, $\Gamma \mathcal{P}_0 \cap \mathcal{P}_0 = \emptyset$. So if $\mathcal{P}_0$ were $G_\delta$, then $\mathcal{P}_0$ and $\Gamma \mathcal{P}_0$ would be two disjoint dense $G_\delta$ sets in $\mathcal{P}$, which contradicts the Baire category theorem.
Actually, we can soup this up to prove that the set is meager. On $[0,1]$, we may find an uncountable family $\{\mu_i : i \in I\}$ of probability measures which are atomless, full support, and mutually singular, and you may if you like take one of them to be Lebesgue measure. A standard example is to take $I = (0,1)$ and let $\mu_i$ be the law of a random variable $X_i = \sum_n 2^{-n} \xi_{i,n}$ where the $\xi_{i,n}, n = 1, 2, \dots$ are iid Bernoulli($i$). And by considering cdfs as above, for each $i,j$ there is a homeomorphism $F_{ij}$ of $[0,1]$ such that $\mu_j = \mu_i \circ F_{ij}^{-1}$. Hence the pushforward maps $\Phi_{ij} := (F_{ij})_*$ are homeomorphisms of $\mathcal{P}$.
Let $\mathcal{P}_i$ be the set of all probability measures absolutely continuous to $\mu_i$. Note the sets $\mathcal{P}_i$ are disjoint. I claim $\Phi_{ij} \mathcal{P}_i = \mathcal{P}_j$. For suppose $\nu_i \ll \mu_i$ and $\mu_j(E) = 0$. Then $\mu_i(F_{ij}^{-1}(E)) = \mu_j(E) = 0$ so we have $\Phi_{ij} \nu_i(E) = \nu_i(F_{ij}^{-1}(E)) = 0$, which is to say that $\Phi_{ij} \nu_i \ll \mu_j$. The reverse inclusion follows by symmetry.
So suppose one of the sets $\mathcal{P}_i$ is nonmeager. Then all of them are nonmeager. However, they are all Borel as you noted, and therefore they have the property of Baire: each $\mathcal{P}_i$ can be written as $\mathcal{P}_i = \mathcal{U}_i \triangle \mathcal{M}_i$ where $\mathcal{U}_i$ is open and $\mathcal{M}_i$ is meager. By assumption, each $\mathcal{U}_i$ must be nonempty. Now for $i \ne j$, since $\mathcal{P}_i \cap \mathcal{P}_j = \emptyset$, we have $\mathcal{U}_i \cap \mathcal{U}_j \subset \mathcal{M}_i \cup \mathcal{M}_j$. So $\mathcal{U}_i \cap \mathcal{U}_j$ is a meager open set, which by the Baire category theorem must be empty. Thus $\{\mathcal{U}_i\}$ is an uncountable collection of nonempty disjoint open sets, which is absurd since $\mathcal{P}$ is separable. We conclude, then, that all the sets $\mathcal{P}_i$ are meager.