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

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    $\begingroup$ 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. $\endgroup$
    – Todd Trimble
    Jul 13, 2019 at 21:04
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    $\begingroup$ 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. $\endgroup$ Jul 13, 2019 at 22:40
  • 1
    $\begingroup$ This has also been mentioned here before: mathoverflow.net/a/28114/61785 $\endgroup$ Jul 14, 2019 at 9:07
  • 1
    $\begingroup$ And also here: mathoverflow.net/a/104656/61785 $\endgroup$ Jul 14, 2019 at 9:53
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    $\begingroup$ It could be interesting to ask for a maximal cartesian closed subcategory of ${\sf Meas}$, along the lines of Todd's suggestion above. $\endgroup$
    – Alec Rhea
    Jul 23, 2019 at 0:50

2 Answers 2


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.


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