Pursuant to this question and the comments therein, it seems natural to ask:
Is there a maximal cartesian closed subcategory of ${\sf Meas}$, the category of measurable spaces and measurable functions?
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$\begingroup$ OMG! Awful category name. $\endgroup$– Fernando MuroCommented Jul 25, 2019 at 10:55
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$\begingroup$ What exactly counts as a "cartesian-closed subcategory"? Is it a full subcategory that happens to be cartesian-closed w.r.t. its own finite products (not necessarily the same as those of $\mathbf{Mble}$), or a cartesian-closed subcategory such that the inclusion functor preserves finite products and internal homs? I ask because $\mathbf{Set}$ forms a cartesian-closed subcategory of $\mathbf{Mble}$ (using the full power set as a $\sigma$-algebra), but the inclusion functor does not preserve binary products or map solely to spaces that are exponentiable in $\mathbf{Mble}$. $\endgroup$– Robert FurberCommented Jul 28, 2019 at 15:35
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$\begingroup$ Contrariwise, it seems that although we can exponentiate $\mathbb{N}$ in $\mathbf{Mble}$, the internal hom $\mathbb{N} \multimap \mathbb{N} \cong \prod_{n \in \mathbb{N}} \mathbb{N}$ is measurably isomorphic to $[0,1]$, and is therefore a space to which Aumann's theorem applies and so is not exponentiable. $\endgroup$– Robert FurberCommented Jul 28, 2019 at 15:38
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1 Answer
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It's a symmetric monoidal closed category under the tensor product (so that the constant graph maps are all measurable.) I am not so sure if there exists a maximal cartesian closed subcategory.
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$\begingroup$ I believe, assuming you want Cartesian closed and ``not just'' monoidal closed, then that will occur when the monoidal closed structure coincides with the Cartesian closed structure - so that your evaluation maps are measurable. But that is only going to occur for countable spaces (in which case the tensor product sigma-algebra coincides with the product sigma algebra structure). $\endgroup$ Commented Jul 25, 2019 at 11:56