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I am looking for examples of pairs ($(\Omega,\Sigma)$, ($\mathcal P(\Omega)$, $\tau$)), where $(\Omega,\Sigma)$ is a measurable space and ($\mathcal P(\Omega)$, $\tau$) is a space of probability measures, such that the latter fails to be a perfectly normal topological space.

Almost all results start with $\Omega$ separable metric to get nice topological properties on ($\mathcal P(\Omega)$, $\tau$). The answer to the following question seems to be closely related: Gaussian measures on non-separable spaces .

I suppose an easy example would be $\Omega$ normal but not perfectly normal and $\mathcal P(\Omega)$ the set of Dirac measures on $\Omega$, and $\tau$ being the topology induced from that on $\Omega$.

I guess some restriction on the $\tau$ to be "natural" is needed to make this question interesting.

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up vote 2 down vote accepted

One of the most natural topologies on spaces of probability measures on a compact space is the weak star topology. Suppose that $X$ is a compact Hausdorff space. Then a set of the form $f^{-1}[\{0\}]$ for some continuous $f:X\rightarrow\mathbb{R}$ is said to be a zero set. The Baire $\sigma$-algebra $\mathcal{M}$ on $X$ is the smallest $\sigma$-algebra on $X$ containing each zero set. Then $(X,\mathcal{M})$ is the smallest topology such that each continuous function is integrable. Let $\mathcal{P}(X)$ be the set of all probability measures on $(X,\mathcal{M})$. Then $\mathcal{P}(X)$ becomes a compact Hausdorff space as follows. Suppose $(\mu_{d})_{d\in D}$ is a net in $\mathcal{P}(X)$ and $\mu\in\mathcal{P}(X)$. Then $\mu_{d}\rightarrow\mu$ if and only if $\int_{X}fd\mu_{d}=\int_{X}fd\mu$ for each continuous $f:X\rightarrow\mathbb{R}$. This topology on $\mathcal{P}(X)$ is referred to as the weak star topology. Furthermore, the mapping $X\rightarrow\mathcal{P}(X),x\mapsto\delta_{x}$ that sends each element to it's Dirac measure is an injective continuous function, so $X$ can be embedded as a closed subspace of $\mathcal{P}(X)$.

Every subspace of a perfectly normal space is perfectly normal as well. Therefore, if $X$ is a compact Hausdorff space which is not perfectly normal, then $\mathcal{P}(X)$ is not perfectly normal either since $X$ is embedded as a closed subspace of $\mathcal{P}(X)$. I should mention that a compact Hausdorff space is perfectly normal if and only if every Borel set is in the Baire $\sigma$-algebra, so if $X$ is compact but not perfectly normal, then there are always more Borel sets than Baire sets.

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