The Giry monad $G : \textbf{Meas} \to \textbf{Meas}$ maps a measurable space $(X, \mathcal{F})$ to its set of probability measures. The $\sigma$-algebra of $G(X, \mathcal{F})$ is the smallest algebra such that the evaluation map $p \mapsto p(E)$ is measurable for all $E \in \mathcal{F}$. I am aware that this condition is needed to define the integral in $\mu\_X : \mathcal{G}^2X \to \mathcal{G}X$, but what is the intuitive meaning of the $\sigma$-algebra? Is there an intuitive example of a subset $A \subseteq \mathcal{G}(X, \mathcal{F})$ that is **not** in the $\sigma$-algebra?



Intuitive meaning of Giry monad's $\sigma$-algebra
(Disclaimer: I’m not at all seriously experienced with the Giry monad; I’ve not read much beyond the original Giry paper and what I’ve picked up on the street.)


Here are two kinds of sets which typically aren’t in the Giry $\sigma$-algebra $G(X)$:


* $\{ p \mid p(E) \in A \}$, for some measurable $E \subseteq X$ and *non-measurable* $A \subseteq [0,1]$. (For specific spaces $X$, it’s not hard to check this is non-measurable; I guess that should be true in fairly wide generality, but I don’t quickly see how to prove that.)
* $\{ p \mid p(E\_i) \in A\_i,\, \text{for all}\ i \in I \}$, for some arbitrary *uncountable family* of measurable sets $E\_i \subseteq X$, $A\_i \subseteq [0,1]$. For instance, take $X$ to be an uncountable set with the $\sigma$-algebra of just its countable and co-countable subsets. Then it’s not hard to check that for every measurable set $S$ of $G(X)$, the set $\{ p(\{x\}) \mid p \in S \}$ is an interval $[0,a]$ for all but countably many $x \in X$.


My overall intuition is that **very crudely, the Giry $\sigma$-algebra is a bit like the (infinite-indexed) product topology:** on the one hand there’s a “componentwise” constraint, as exemplified by the former counterexample above; on the other hand theres an “across-components” constraint, as seen with the latter.


