Is the category of measurable spaces MONOIDAL closed? (fake-proof)

Question

As discussed on MathOverflow, there is a preprint on ArXiv that claims the category of measurable spaces is monoidal closed (in particular that exponential objects $Y^X$ always exist and have measurable evaluation maps $X \times Y^X \to Y$).

Also as discussed in the same post, such a claim appears to contradict results by Robert Aumann that exponential objects do not always exist in the category of measurable spaces (i.e. there exists no $\sigma$-algebra on $Y^X$ for which the evaluation map $X \times Y^X \to Y$ is measurable). So the claim is likely to be false.

I have been scrutinizing the preprint and I think I may have identified the error in the argument. Am I correct? Or is the claim actually true?

High-level overview of my argument: Because $\sigma$-algebras are only closed under countable operations, rather than arbitrary ones, it is easier for $\sigma$-algebras that make all members of a certain class of functions measurable to not exist? (And therefore easier for there to not be any initial or final $\sigma$-algebras generated by those functions?) Cf. this comment on the MathOverflow post.

I.e. in contrast to the situation with topologies, which are closed under arbitrary unions, which makes it easier to guarantee that there always exists some topology for which a collection of functions is measurable? (Thus allowing the existence of initial or final topologies.)

Details of attempted disproof: The main issue seems to be that the preprint uses a non-standard definition for the $\sigma$-algebra on the Cartesian product of two measurable spaces $X \times Y$ when defining the monoidal product $X \otimes Y$:

The coinduced (final) $\sigma$-algebra [on $X \times Y$] such that all graph functions $$\begin{array}{rccl} \Gamma_f: & X &\to & X \times Y \\ & x & \mapsto & (x, f(x)) \end{array}$$ for $f: X \to Y$ measurable, as well as all the graph functions $$\begin{array}{rccl} \Gamma_g: & Y &\to & X \times Y \\ & y & \mapsto & (g(y), y) \end{array}$$ for $g: Y \to X$ measurable, are measurable.

Main question: The above $\sigma$-algebra isn't even guaranteed to exist in general, correct?

Asides/extra questions/comments: Normally the product $\sigma$-algebra would be defined as the induced (initial) $\sigma$-algebra such that the projections $(x,y) \mapsto x$ and $(x,y) \mapsto y$ are measurable, right?

In particular the definition from the preprint, and the standard definition, are not always equivalent, correct? (As an aside, are they ever equivalent?)

I can go into the details of how the preprint then defines the initial topology on $Y^X$ generated by the pointwise projections/evaluations, and from there the argument for why the evaluation map $X \otimes Y^X = X \times Y^X \to Y$ is measurable, but if the above topology isn't even guaranteed to exist, then the argument definitely seems to fail. (With Robert Aumann providing a counterexample in the case that $X = Y = [0,1]$.)

Terminology: By "initial $\sigma$-algebra" on $X$ generated by functions $f_{\alpha}: X \to Y$, with $Y$ a measurable space, I mean the "coarsest" or "smallest" possible $\sigma$-algebra on $X$ such that all of the $\{f_{\alpha}\}$ are measurable.

By "final $\sigma$-algebra on $Y$ generated by functions $f_{\alpha}: X \to Y$, with $X$ a measurable space, I mean the "finest" or "largest" possible $\sigma$-algebra on $Y$ such that all of the $\{f_{\alpha}\}$ are measurable. (I guess technically one could also require that $\{f_{\beta}\}$ for $f_{\beta}: X_2 \to Y$ are measurable for $X_2$ a measurable space, which is what the preprint appears to do.)

Answer 1

The coinduced/final $\sigma$-algebra here is in fact guaranteed to exist in general. It is constructed analogously to the final topology.

In particular, suppose we have some set $V$ on which we wish to define the final $\sigma$-algebra. Let $\mathcal{F}$ denote a collection of functions $f : U_f \to V$, where $U_f$ is part of a measurable space $(U_f, \Sigma_{U_f})$, which we emphasise may vary depending on $f$. The final $\sigma$-algebra $\Sigma_V$ is the largest $\sigma$-algebra that makes each $f \in \mathcal{F}$ measurable. Explicitly, $B \in \Sigma_V$ iff $f^{-1}(B) \in \Sigma_{U_f}$ for all $f \in \mathcal{F}$.

To see that this is a $\sigma$-algebra, for each $f \in \mathcal{F}$, if we denote $$ \Delta_{f} := \{B \subseteq V : f^{-1}(B) \in \Sigma_{U_f} \}, $$ then we can write $$ \Sigma_V = \bigcap_{f \in \mathcal{F}} \Delta_f. $$ But now it is easily checked that each $\Delta_f$ is a $\sigma$-algebra, and hence $\Sigma_V$ is too.

In the case that interests you, take $V = X \times Y$, and let $\mathcal{F}$ consist of all graph functions $X \to X \times Y$ and $Y \to X \times Y$. I'm not sure if the final $\sigma$-algebra on $X \times Y$ defined here is different to the usual product $\sigma$-algebra (which you are right is usually taken to be initial with respect to the projection maps), but I suspect this is the case. If so, then there is not necessarily a contradiction between Sturtz's work and Aumann's, since these would be talking about different monoidal products.

Answer 2

Sturtz's claim does contradict Aumann's, although not for the reason you described.

As noted in @12qu's answer, the final $\sigma$-algebra is perfectly well-defined. Under the assumption that $\otimes$ is a symmetric monoidal structure on measurable spaces, we can conclude that it is the cartesian monoidal structure. To see that $\otimes = \times$, note that the unit for $\otimes$ is still the one-point space. This means that $\otimes$ is a semicartesian monoidal structure on measurable spaces. Further, the diagonal transformation $\Delta \colon X \to X \otimes X$ is measurable because $\Delta$ is the graph of the identity function. As described on nLab, any semicartesian monoidal category with diagonals is cartesian monoidal. Thus, $\otimes$ is the cartesian product, and so Aumann's contradiction applies. There are contradictions even before considering a closed monoidal structure. If $\otimes$ is monoidal, then it is the cartesian product, but graphs are not generally measurable into the cartesian product, as in this question. I have not checked which axiom of a monoidal category is broken.

I suspect Sturtz knows about this issue since in a more recent preprint of Sturtz's, the $\sigma$-algebra is instead generated by the constant graphs. In fact, the paper you cited only uses constant graphs to show that the evaluation map is measurable. This is an error in the paper; the proof that the evaluation map is measurable is faulty. To show that the evaluation is measurable, it must be measurable after precomposing with any graph, not just the constant graphs.

Of course, if the $\sigma$-algebra for $\otimes$ is generated only by constant graphs, then the evaluation map is measurable by Sturtz's argument. In this situation, there is no reason for the diagonal map of an arbitrary space to be measurable, so my argument that $\otimes$ is the cartesian monoidal structure does not apply.