# Exponential objects in the category of measurable spaces

Let $$\text{Meas}$$ be the category of measurable spaces with measurable functions as morphisms. Does $$\text{Meas}$$ have exponential objects?

• The question reads ambiguously to me. Almost surely $Meas$ is not cartesian closed although I'd have to think to give an example. However, the question of which objects are exponentiable in $Meas$ (see ncatlab.org/nlab/show/exponential+object#related_notions) is interesting, just as the corresponding question for $Top$ is interesting. Of course some objects will be exponentiable. Jul 13, 2019 at 21:04
• Reads ambiguously to me too. Indeed, the failure of Meas to be cartesian closed is discussed as Proposition 6 of arxiv.org/pdf/1701.02547.pdf, citing an old result of Aumann. Jul 13, 2019 at 22:40
• This has also been mentioned here before: mathoverflow.net/a/28114/61785 Jul 14, 2019 at 9:07
• And also here: mathoverflow.net/a/104656/61785 Jul 14, 2019 at 9:53
• It could be interesting to ask for a maximal cartesian closed subcategory of ${\sf Meas}$, along the lines of Todd's suggestion above. Jul 23, 2019 at 0:50

## 2 Answers

As mentioned in the comments, Meas of course has some exponential objects $$B^A$$, but not for all $$A$$ and $$B$$, i.e., it is not cartesian closed. This fact is discussed as Proposition 6 of A Convenient Category for Higher-Order Probability Theory by Heunen, Kammar, Staton, and Yang, citing an old result of Aumann:

• R. J. Aumann, "Borel structures for function spaces," Illinois Journal of Mathematics, vol. 5, pp. 614–630, 1961. project euclid


To follow this line of reasoning, we first need to go over how products and coproducts work in $$\Mble$$. The product $$(X,\Sigma_X) \times (Y,\Sigma_Y)$$ in $$\Mble$$ is given by $$(X \times Y, \Sigma_X \otimes \Sigma_Y)$$, where $$\Sigma_X \otimes \Sigma_Y$$ is the $$\sigma$$-algebra generated by rectangles (this is simple to prove and comes down to showing that the projections and universal mapping are measurable). We will also use the fact that for a singleton $$\{x\}$$ with its unique $$\sigma$$-algebra, $$\{x\} \times Y \cong Y$$ (measurably). You can deduce this from singletons being terminal objects, if that's your bag.


As discussed in this question and its answers, if $$|Y| > 2^{\aleph_0}$$, the diagonal $$\Delta = \{ (y,y) \mid y \in Y \} \subseteq Y \times Y$$ is an element of $$\powerset(Y \times Y)$$ but not of $$\powerset(Y) \otimes \powerset(Y)$$. A summary of the proof is:

1. Because of how $$\sigma$$-algebras are generated, there must exist a countable family of rectangles $$(S_i\times T_i)_{i \in \omega}$$ such that $$\Delta$$ is in the $$\sigma$$-algebra $$\Sigma$$ generated by $$(S_i \times T_i)_{i \in \omega}$$.
2. The "distinguishability relation" defined by $$(T_i)_{i \in \omega}$$ on $$Y$$ has at most $$2^{\aleph_0}$$ equivalence classes, so $$|Y| > 2^{\aleph_0}$$ implies that there exist distinct elements $$y_1,y_2 \in Y$$ such that for all $$i \in \omega$$, $$y_1 \in T_i$$ iff $$y_2 \in T_i$$. (We could also have done this for $$(S_i)_{i \in \omega}$$, but this is not needed.)
3. Therefore $$(y_1,y_2) \in S_i \times T_i$$ iff $$(y_1,y_1) \in S_i \times T_i$$ for all $$i \in \omega$$. Proceeding inductively, this holds for all elements of the $$\sigma$$-algebra $$\Sigma$$ generated by $$(S_i \times T_i)_{i \in \omega}$$.
4. As $$\Delta \in \Sigma$$, we have $$(y_1,y_2) \in \Delta$$, which contradicts $$y_1 \neq y_2$$.

If $$\blank \times Y$$ preserved coproducts, then we would have $$\coprod_{y \in Y}\{y\} \times Y \cong \left(\coprod_{y \in Y}\{y\}\right) \times Y$$ in $$\Mble$$. The $$\sigma$$-algebra on the left is $$\powerset(Y \times Y)$$ because $$\coprod_{y \in Y}\{y\} \times Y \cong \coprod_{y \in Y} Y \cong \coprod_{y_1 \in Y} \coprod_{y_2 \in Y} \{y_2\}$$ (this is where we use the fact about singletons). But on the right we have $$\left(\coprod_{y \in Y}\{y\}\right) \times Y \cong Y \times Y$$, so the $$\sigma$$-algebra is $$\powerset(Y) \otimes \powerset(Y)$$, and so there is no isomorphism if $$|Y| > 2^{\aleph_0}$$.

Of course, Aumann's result, proved using the Baire hierarchy, is stronger and shows that the spaces $$[0,1]$$, $$2^\omega$$, $$\mathbb{R}$$ etc., which we are often more interested in, are not exponentiable.