# 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

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

It is also possible to show that the category of measurable spaces, $$\newcommand{\Mble}{\mathbf{Mble}}\Mble$$, is not cartesian closed by using more category theory and less measure theory (though still some). We reason as follows. If $$\Mble$$ were cartesian closed, then for each measurable space $$Y$$, the functor $$\newcommand{\blank}{\mbox{-}}\blank \times Y : \Mble \rightarrow \Mble$$ would be a left adjoint, and therefore preserve coproducts. Therefore we can show that $$\Mble$$ is not cartesian closed by finding a coproduct that is not preserved.

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.

The category $$\Mble$$ also admits coproducts of arbitrary arity, which are preserved by the forgetful functor to $$\newcommand{\Set}{\mathbf{Set}}\Set$$, but all we need is the fact that for each set $$Y$$, the space $$\newcommand{\powerset}{\mathcal{P}}(Y,\powerset(Y))$$ is the coproduct of its elements (again, this comes down to proving that the set-theoretic coprojections $$\kappa_y : \{y\} \rightarrow Y$$ and universal map are measurable).

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.