# Is “conditioning to a sub-$\sigma$-algebra” a measurable operation?

Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $[0,1]$, and let $\mathcal{M}$ be the set of probability measures on $([0,1],\mathcal{B})$, equipped with the evaluation $\sigma$-algebra $\ \sigma(\rho \mapsto \rho(A):A \in \mathcal{B})$.

Let $\mathcal{M}_2$ be the set of probability measures on $([0,1] \times [0,1], \mathcal{B} \otimes \mathcal{B})$, again equipped with the associated evaluation $\sigma$-algebra $\ \sigma(\mu \mapsto \mu(A):A \in \mathcal{B} \otimes \mathcal{B})$.

Given a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$, for any $\mu \in \mathcal{M}_2$, define $\mathbb{E}_\mathcal{G}(\mu) \in \mathcal{M}_2$ to be the unique measure such that

• $\mathbb{E}_\mathcal{G}(\mu)$ agrees with $\mu$ on $\mathcal{G} \otimes \mathcal{B}$;
• there exists a $\mathcal{G}$-measurable function $\nu \colon [0,1] \to \mathcal{M}$ such that for all $A_1,A_2 \in \mathcal{B}$, $$\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ = \ \int_{A_1 \times [0,1]} \nu(x)(A_2) \, \mu(d(x,y)).$$

Is the map $\mathbb{E}_\mathcal{G} \colon \mathcal{M}_2 \to \mathcal{M}_2$ a measurable function? Or at the least, is $\mathbb{E}_\mathcal{G}^{-1}(\mathcal{A})$ a universally measurable subset of $\mathcal{M}_2$ for every measurable set $\mathcal{A} \subset \mathcal{M}_2$?

Remark: The existence and uniqueness of the measure $\mathbb{E}_\mathcal{G}(\mu)$ follows from the disintegration theorem applied to the measure $\mu|_{\mathcal{G} \otimes \mathcal{B}}$.

UPDATE: A simple "counting" argument as in the comments below yields that there must exist $\mathcal{G}$ such that $\mathbb{E}_\mathcal{G}$ is not measurable with respect to the evaluation $\sigma$-algebra. Nonetheless, the key question remains as to whether $\mathbb{E}_\mathcal{G}$ is necessarily universally measurable (in the sense that the pre-image of every member of the evaluation $\sigma$-algebra belongs to the universal completion of the evaluation $\sigma$-algebra).

[For an explicit example showing that $\mathbb{E}_\mathcal{G}$ is not necessarily continuous: Let $\mathcal{G}$ be the countable-cocountable algebra. Let $\rho_n$ be a sequence of atomless probability measures on $[0,1]$ converging weakly to a non-trivial purely atomic measure $\rho$. Writing $D \colon x \mapsto (x,x)$, we have that $\mathbb{E}_\mathcal{G}(D_\ast\rho_n)=\rho_n \otimes \rho_n$ but $\mathbb{E}_\mathcal{G}(D_\ast\rho)=D_\ast\rho$.]

A possible approach: Fix $A_1$ and $A_2$. Given a finite or countable partition $\mathcal{H}$ of $[0,1]$ by members of $\mathcal{G}$, we can "approximate" $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$ by $$\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ \approx \ \mathbb{E}_{\sigma(\mathcal{H})}(\mu)(A_1 \times A_2) \ = \ \sum_{G \in \mathcal{H}} \mu(G \times A_2)\mu((G \cap A_1) \times [0,1]).$$ Given an increasing sequence of such partitions $\mathcal{H}_n$ such that $\mathcal{G} \subset \sigma(\mathcal{N}_{\pi^1_\ast\mu} \cup \bigcup_{n=1}^\infty \mathcal{H}_n)$, Levy's upward theorem yields that $\mathbb{E}_{\sigma(\mathcal{H}_n)}(\mu)(A_1 \times A_2) \to \mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$.

[Here, $\mathcal{N}_{\pi^1_\ast\mu}$ is the set of $\pi^1_\ast\mu$-null sets, where $\pi^1_\ast\mu(A):=\mu(A \times [0,1])$.]

Now such a sequence $\mathcal{H}_n$ always exists---based on the proof of the equivalence between "separable" and "countably generated mod 0" given here, I think the following construction works: Letting $\mathcal{U}$ be a countable algebra generating $\mathcal{B}$, let $\{G_i\}_{i=1}^\infty \subset \mathcal{G}$ be a countable set such that for every $U \in \mathcal{U}$ and $r \in \mathbb{N}$, if $\mathcal{G}_{U,r}:=\{G \in \mathcal{G} : \mu(G \triangle U) < \frac{1}{r} \}$ is non-empty then $\exists \, i$ s.t. $G_i \in \mathcal{G}_{U,r}$. Then take $\mathcal{H}_n$ to be the partition generated by $\{G_1,\ldots,G_n\}$.

So then, to prove the desired result, it is sufficient to show that there is a sequence of "partition-valued" functions $$\mathcal{H}_n \colon \mathcal{M}_2 \to \ \{\textrm{finite or countable partitions contained in } \mathcal{G}\}$$ such that $\mu \mapsto \sum_{G \in \mathcal{H}_n(\mu)} \mu(G \times A_2)\mu((G \cap A_1) \times [0,1])$ is universally measurable for each $n$, and $\mathcal{G} \subset \sigma(\mathcal{N}_{\pi^1_\ast\mu} \cup \bigcup_{n=1}^\infty \mathcal{H}_n(\mu))$ for every $\mu$.

[Remark: One might hope that it could even be possible to take $\mathcal{H}_n$ to be independent of $\mu$. However, this is not possible in general. This follows from the answer to https://math.stackexchange.com/questions/2156998/is-every-sub-sigma-algebra-of-mathcalb-mathbbr-universally-countabl; but also, if it were possible to take $\mathcal{H}_n$ to be independent of $\mu$, then $\mathbb{E}_\mathcal{G}$ would be measurable with respect to the evaluation $\sigma$-algebra, which we have established not always to be the case.]

Another characterisation of $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$:

In the above discussion "A possible approach", I gave a sort-of-explicit construction of $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$, but the difficulty is that it relies on "making a choice" - which could then lead to measurability issues when we allow $\mu$ to vary.

Now one of the comments suggested looking directly at the proof of the disintegration theorem and hoping that the desired measurability might become more clear from there. Disintegration relies fundamentally on the Radon-Nikodym theorem; on the basis of the proof of the Radon-Nikodym theorem, $\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2)$ can be characterised as follows:

Let $\mathcal{C}$ be the set of $(\mathcal{G},\mathcal{B})$-measurable functions $g \colon [0,1] \to [0,1]$ such that for every $G \in \mathcal{G}$, $$\int_{G \times [0,1]} g(x) \, \mu(d(x,y)) \ \leq \ \mu(G \times A_2) \, ;$$ then $$\mathbb{E}_\mathcal{G}(\mu)(A_1 \times A_2) \ = \ \sup_{g \in \mathcal{C}} \int_{A_1 \times [0,1]} g(x) \, \mu(d(x,y)).$$

Again, it is difficult to see any measurability from this.

Increasingly I am starting to suspect that (assuming the axiom of choice) there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ such that $\mathbb{E}_\mathcal{G}$ is not universally measurable!

• In the integral, $\nu$ seems to be a function of both coordinates. – Michael Greinecker Feb 21 '17 at 22:37
• For each $x \in [0,1]$, $\nu(x)$ is a measure on $[0,1]$. An alternative way to write the integral is $\ \int_{A_1} \nu(x)(A_2) \, \pi_{1\ast}\mu(dx)\,$, where $\pi_1 \colon [0,1] \times [0,1] \to [0,1]$ is the projection $\pi_1(x,y)=x$. – Julian Newman Feb 21 '17 at 22:44
• We probably just need to look at a constructive proof of the disintegration theorem, and the measurability may become obvious. By the way, isn't the evaluation $\sigma$-algebra equal to the Borel $\sigma$-algebra of the weak topology? That could be helpful; the conditioning map may turn out to be (semi)continuous or otherwise nice with respect to that topology. – Nate Eldredge Feb 21 '17 at 23:46
• Yes, the evaluation $\sigma$-algebra is precisely the Borel $\sigma$-algebra of the topology of weak convergence. I thought about the possibility of trying to use some kind of continuity property, but it felt unlikely to me that we would have such continuity, since $\mathcal{G}$ is an arbitrary sub-$\sigma$-algebra. Some kind of continuity with respect to total-variation distance seems more likely, but this generates the wrong $\sigma$-algebra. – Julian Newman Feb 22 '17 at 1:53
• One thing that's potentially alarming is that there are $2^{\mathfrak{c}}$ sub-$\sigma$-algebras of $\mathcal{B}$, whereas there are only $\mathfrak{c}$ many Borel maps from $\mathcal{M}_2$ to itself. (For instance, given any $A \subset [0,1]$, let $\mathcal{G}_A$ be generated by all the countable subsets of $A$.) Is the map $\mathcal{G} \mapsto \mathbb{E}_{\mathcal{G}}$ 1-1? – Nate Eldredge Mar 7 '17 at 23:44

Okay, I think I've worked out that the answer is no, i.e. there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ such that $\mathbb{E}_\mathcal{G}$ is not universally measurable.
(We will write $\lambda$ for the Lebesgue measure on $([0,1],\mathcal{B})$.)
It is well-known that the evaluation $\sigma$-algebra is precisely the Borel $\sigma$-algebra of the topology of weak convergence, which is a Polish topology. So then, to show that (for some given $\mathcal{G}$) $\mathbb{E}_\mathcal{G}$ is not universally measurable, it is sufficient (by Lusin's theorem) to find a probability measure $Q$ on $\mathcal{M}_2$ such that for every measurable set $E \subset \mathcal{M}_2$ with $Q(E)>0$, the restriction of $\mathbb{E}_\mathcal{G}$ to $E$ is not a continuous function (with respect to the topology of weak convergence).
Let $A \subset [0,1]$ be a set such that $A$ and $[0,1] \setminus A$ intersect every Lebesgue-positive measure set. (Assuming the axiom of choice, such a set $A$ exists, as shown here.) Let $\mathcal{G}$ be the $\sigma$-algebra consisting of all countable subsets of $A$ and their complements.
Let $Q$ be the image measure of $\lambda \otimes \lambda$ under the map $D : (x,y) \mapsto \frac{1}{2}(\delta_{x,x}+\delta_{y,y})$. Let $E \subset \mathcal{M}_2$ be any measurable set with $Q(E)>0$, and let $F=D^{-1}(E)$. (So $\lambda \otimes \lambda(F)>0$.) Let $F'=\{x \in [0,1] : \lambda(y:(x,y)\in F)>0\}$. Obviously $\lambda(F')>0$, so fix a point $x \in F' \setminus A$. For any $y \in [0,1]$, we have that $$\mathbb{E}_\mathcal{G}(D(x,y)) \ = \ \left\{ \begin{array}{c l} \frac{1}{2}(\delta_{x,x}+\delta_{y,y}) & y \in A \\ \frac{1}{4}(\delta_{x,x} + \delta_{x,y} + \delta_{y,x} + \delta_{y,y}) & y \not\in A. \end{array} \right.$$ So since $Y:=\{y \in [0,1] : (x,y) \in F\}$ has positive Lebesgue measure, it is clear that $\mathbb{E}_\mathcal{G}(D(x,\cdot))$ is not continuous on $Y$, so $\mathbb{E}_\mathcal{G} \circ D$ is not continuous on $F$, so (since $D$ is obviously continuous) $\mathbb{E}_\mathcal{G}$ is not continuous on $E$.
• Just out of curiosity, can we say the operation of conditioning on a sub $\sigma$ algebra is measurable as long as it is non-atomic w.r.t. any element in $\mathcal{M}$? Or is there any sufficient condition? – Henry.L Mar 14 '17 at 0:11