Synthetic measure theory with quasi-borel spaces I'm trying to understand this paper "A Convenient Category for Higher-Order Probability Theory" which is a new way to formulate probability theory by making random variables primary and deriving the rest (as far as I can tell).
They say a random variable is a quasi-Borel space, which consists of a sample space $\Omega$, which can be interpreted as a set of random seeds, a set of outcomes $X$, and a subset of functions $M\subseteq[\Omega\rightarrow X]$. This structure has to satisfy a few constraints. They say in practice $\Omega$ should be set to be the reals $\mathbb{R}$ since it is isomorphic to most structures that we care about in probability theory, so more specifically $M\subseteq[\mathbb{R}\rightarrow X]$.
The functions $M$ are the random elements of the space. I'm trying to see how this works exactly. Let's say I want to create a boolean random variable with $X={heads,tails}$ an underlying probability for each outcome of $P(heads)=0.25,P(tails)=0.75$, I think this means my quasi-Borel space would need to essentially partition $\mathbb{R}$ into two where one partition would represent random seeds that map onto "heads" and the other set would map onto "tails" such that the size (measure) of the heads subset of random seeds would be 0.25 proportionally. Is that right?
 A: Recall the classical measure-theoretic setup: Let $(\Omega, \Sigma_\Omega), (X,\Sigma_X)$ be measurable spaces, then we can consider measurable functions $Y : \Omega \to X$, capturing how elements of $X$ may depend measurably on random seeds $\omega \in \Omega$.
Now if $P$ is a probability measure on $\Omega$, the triple $(\Omega,\Sigma_\Omega,P)$ becomes a probability space. We can now consider $Y$ a random variable, as it defines probability measure $P_Y = Y_*P$ called the law of $Y$ on $X$ by pushing forward $P$ onto X
$$P_Y(A) = P(Y^{-1}(A))$$
Let's look at your example for $X=\lbrace H,T \rbrace$


*

*Let $P_1$ choose $0$ with probability .25 and $1$ with .75, and let $Y_1(x)$ be $H$ if $x=0$ and $T$ otherwise.

*Let $P_2$ be the uniform distribution on $[0,1]$ and let $Y_2(x)$ be $H$ if $x \leq 0.25$ and $T$ otherwise
Then the $Y_i : (\mathbb R, \Sigma_{\mathbb R}, P_i) \to X$ are both random variables on different probability spaces, which both encode your experiment and induce the same law on $X$.

You're correct in that quasi-Borel spaces take the notion of "random element" as primary, and that we restrict to $\Omega = \mathbb R$ for well-behavedness. A quasi-Borel space is a set $X$ together with a collection $M_X \subseteq [\mathbb R \to X]$ of dedicated maps called "random elements". These capture axiomatically how elements of $X$ may depend measurably on random seeds $\omega \in \mathbb R$.
But there are no probabilities involved so far! We need the analogue of a probability space. Given a probability measure $P$ on $\mathbb R$ and a random element $Y \in M_X$, we can think of $Y$ as an $X$-valued random variable. In fact, we can exactly copy the situation from above, as both $Y_i \in M_X$, and the $P_i$ are probability measures on $\mathbb R$.
The interesting question is: Do $Y_1,Y_2$ still induce the same law on $X$? We wish to pushforward $P_i$ onto $X$, but $X$ does not come equipped with a $\sigma$-algebra, so we can't a priori talk about what a measure on $X$ is. But it turns out we can endow every quasi-Borel space $X$ with a $\sigma$-algebra $\Sigma_X$ and define pushforwards on that algebra, and the laws such defined do indeed agree.
This is one of the selling points of quasi-Borel spaces: The $\sigma$-algebra on $X$ is a derived notion, and unlike for measurable spaces, does not determine which maps $\mathbb R \to X$ are admissible as random elements. This gives the theory the extra flexibility to deal with e.g. function spaces nicely. The $\sigma$-algebras are usually only needed at the end, to compare equality of laws.
