# Every Polish Space is a Regular Conditional Probability Space

We say that topological space $$M$$ is polish if it is completely metrizable and separable.

We say that $$Q : \Omega \times \mathcal{B}M \to[0,1]$$ is a regular conditional probability measure for a random element $$X : (\Omega,\mathcal{F},\mathbb{P}) \to (M,\mathcal{B}M)$$ if for every $$\omega \in \Omega$$ map $$Q(\omega,\cdot)$$ is a probability measure and for every $$E \in \mathcal{B}M$$ equality $$Q(\omega,E) = \mathbb{P}( X \in E|\mathcal{A})(\omega)$$ holds almost surely $$\mathbb{P}(\omega)$$. Here $$(\Omega,\mathcal{F},\mathbb{P})$$ is a probability space , $$\mathcal{B}M$$ is a Borel $$\sigma$$-algebra over $$\mathcal{M}$$ and $$\mathcal{A}$$ is a $$\sigma$$-subalgebra of $$\mathcal{F}$$. $$\mathbb{P}(X \in E | \mathcal{A}) : \Omega \to [0,1]$$ is conditional probability with respect $$\mathcal{A}$$, defined as $$\mathbb{E}(I_E(X)|\mathcal{A})$$, where $$I_E$$ is indicator for $$E$$ .

We say that measurable space $$(M,\mathcal{BM})$$ is regular conditional probability space if for all random objects $$X$$ and subalgebras $$\mathcal{A}$$ exists regular conditional probability $$Q$$.

Wikipedia claims that every Polish space is regular conditional space and I'm trying to prove it.

I will treat space $$M$$ as a metric space with distance function $$d$$ providing complete metrization.

My idea of the proof were

Define RCPM on set on the countable set of rational balls $$\xrightarrow[]{(1)}$$ Increase to the set of all Balls of $$M$$ by computing limits on sets of measure 1 $$\xrightarrow[]{(2)}$$ extend it to all the open sets $$\xrightarrow{(3)}$$ extend to all Borel sets. Then, tho show that the connection with conditional probability is satisfied with monotone class theorem starting from rational balls as the basis for the monotone class.

It seems to me that transitions (1),(3) are pretty simple but I am not how to perform transition (2) in such a way that resulting $$Q(\omega,\cdot)$$ is still a measure. Is my sketch correct? Are where any insufficient steps? How to perform extension (2) correctly?

Here is my proof so far: Lets $$R$$ be countable dense subset of $$M$$. We say that ball $$B$$ is rational if it has center at $$R$$ and radius in $$\mathbb{Q}$$.

lets $$Z \subset \Omega$$ such that for all $$\omega \in Z$$ something of the following happens:

1. there are rational balls $$A,B$$ such that $$\mathbb{P}(X \in A|\mathcal{A})(\omega) < \mathbb{P}(X \in B|\mathcal{A})(\omega)$$ and in the same time $$B \subset A$$.

2. there are $$p \in R$$ and $$r \in Q$$ such $$\lim_{ n \to \infty} \mathbb{P}\Big( X \in\mathbb{B}(p,r - \gamma/n) \Big|\mathcal{A}\Big) \neq \mathbb{P}\Big(X \in \mathbb{B}(p,r)\Big| \mathcal{A} \Big)$$. Here we have ratioanl $$\gamma < r$$

3. there is $$p \in R$$ such that $$\lim_{n \to \infty} \mathbb{P}\Big(X \in \mathbb{B}(p,n) \Big| \mathcal{A} \Big) \neq 1$$

4. there is $$p \in R$$ such that $$\lim_{n \to \infty} \mathbb{P}\Big(X \in \mathbb{B}(p,1/n) \Big| \mathcal{A} \Big) \neq 0$$

It can be shown that $$\mathbb{P}(Z) = 0$$.

Assuming that $$M$$ is nonempty it is safe for $$\omega \in Z$$ to let $$Q(\omega,\cdot) = \delta_m$$ for some $$m \in M$$.

For all the other $$\omega \in Z^\complement$$, for rational ball $$B$$ define $$Q(\omega,B) = \mathbb{P}\Big( X \in B \Big| \mathcal{A} \Big) .$$

I think we may also define it on the set of finite intersections of rational balls $$A = \bigcap^n_{i=1} B_{i}$$ in the same way $$Q(\omega,A) = \mathbb{P}\Big(X \in A \Big| \mathcal{A} \Big)$$. Set of such intersections is still countable, so it might be possible to extend $$Z$$ so it covers such intersections too.

With this we can extend to finite unions with

$$Q\left(\omega, \bigcup^n_{i=1} A_i\right) = \sum^n_{k = 1} (-1)^{n- k} \sum_{I \in 2^n : |I| = k} Q\left(\omega, \bigcap_{i \in I} A_i\right)$$ and differences with

$$Q(\omega, A \setminus C) = Q(\omega,A) - Q(\omega,A \cap C)$$

Now I don't think that step (1) is required untill we have $$Q(\omega,\emptyset) = 0$$ on $$Z^\complement$$

{(1) to define $$Q$$ at $$\mathbb{B}(p,r)$$ for $$(p,r) \in M \times \mathbb{R}_{+}$$ select sequences $$a \to p$$ and $$s \uparrow r$$ such that $$d(a,p) \downarrow 0$$ and for each $$n \in \mathbb{N}$$ we have $$q_n \le r + d(a_n,p)$$ and resulting balls are rational. Then $$Q(\omega,\mathbb{B}(a,s))$$ converges to some number as bounded and monotonic sequence. Define $$Q(\omega,\mathbb{B}(p,r))$$ as a limit of this sequence. $$Q\Big(\omega,\mathbb{B}(p,r)\Big) = \mathbb{P}\Big(X \in \mathbb{B}(p,r) |\mathcal{A}\Big)(\omega)$$ almost surely by dominated convergence theorem for conditional expectation and corresponding indicator functions. At this stage we may define} $$Q(\omega, \emptyset) = 0 \quad Q(\omega,M) = 1$$ (2) We may represent open sets as $$U = \bigcup^\infty_{n=1}\bigcap^{N_n}_{m=1} B_{m,n}$$, where $$B_{m,n}$$ are rational balls with $$N_n \in \mathbb{N}$$.

This means that me may define

$$Q(\omega,U) = \lim_{n \to \infty} Q\left(\omega, \bigcup^n_{i =1} \bigcap^{N_i}_{j} B_{i,j} \right).$$ This limit must exist as it is a bounded above non decreasing sequence (requires additional sub-proof). For countable union of disjoint balls we get $$Q\left(\omega, \bigcup^\infty_{n=1} B_n\right) = \sum^\infty_{n=1}Q(\omega,B_n)$$ by union formula. I think we can extend it arbitrary open sets now.

$$\ldots$$

But how to define it for arbitrary open sets or unions/intersection of not disjoint balls?

$$\ldots$$

(3) for arbitrary Borel set $$B$$ selecting $$Q(\omega,B) = \inf\{ Q(\omega,U) | U : \textrm{Open},B \subset U\}$$ must hold for any probability measure by outer regularity in Radon space $$M$$. In this case we can say that $$Q(\omega,B) = \lim_{n \to \infty} Q(\omega,U_n)$$ with $$U$$ being a decreasing sequence of open sets.

$$\ldots$$

• "for arbitrary Borel set $B$ there must be representation $B = \bigcap^\infty_{n=1} U_n$ with $U_n$ being open." This claim is false; such intersections are only a tiny fraction of all possible Borel sets.
– user940
Sep 3, 2017 at 21:11
• 1. Every Polish space is Borel isomorphic to a subset of $\mathbb R$. Hence the result. 2. Without 1, it is not easy to proceed. In order to construct a measure, you need first to define it on a semiring (I believe, we can take differences of two finite unions of rational balls). But how can one guarantee $\sigma$-(sub)additivity? Sep 4, 2017 at 7:31
• @ByronSchmuland, I have edited the post to correct this. What, I probably wanted to say that this sets have "equal measure" but wrote it ambiguously. Sep 4, 2017 at 12:06
• @zhoraster , I know the fact that every space Borel isomorphic to subset Borel subset of $\mathbb{R}$ is conditional regular, however I don't know how to build such measurable isomorphism for arbitrary Polish space. I would be grateful If you suggest a source. Sep 4, 2017 at 12:12