Fixed objects of the M endofunctor on category Meas

Question

Consider the category $\operatorname{Meas}$ of measurable spaces: its objects are sets equipped with $\sigma$-algebras, and its morphisms are measurable functions between spaces.

As Gerald Edgar & Michael Greinecker pointed out in this thread, there is the natural endofunctor $M : \operatorname{Meas} \to \operatorname{Meas}$ which sends a space $X$ to the collection $M(X)$ of (extended-real-valued) measures on $X$. This collection $M(X)$ is a measurable space, equipped with the minimal $\sigma$-algebra so that the evaluation functions $\mu \mapsto \mu(A)$ are measurable. A morphism $f : X \to Y$ is mapped to the push-forwarding map $f_* : \mu \mapsto \mu \circ f^{-1}$.

We may naturally iterate this endofunctor. Thinking of a measure on $X$ as a statistical ensemble, the space $M^2(X) = M(M(X))$ consists of ensembles-of-ensembles. Such hierarchical spaces are important in probability theory and dynamical systems. We may go further, defining $M^3(X) = M(M(M(X)))$ and so forth.

This simply generates a dynamical system on the category of measurable spaces, where the initial condition $X \in \operatorname{Meas}^{\operatorname{ob}}$ gets mapped to its successors $M(X)$, $M^2(X)$, $M^3(X)$, etc. Understanding these categorical dynamics is a hard problem, to say the least. Understanding the fixed ``points'', on the other hand, might actually be tractable. Hence the question:

What are the fixed objects of the endofunctor $M :\operatorname{Meas} \to \operatorname{Meas}$ ?

Answer 1

This answer is incomplete, in that it's missing all the hard parts. I'll finish it soon.

You have an endofunctor that "looks", in the vaguest possible sense, like the archetypical monad - the functor $F$ taking a set to the set of elements in the free group generated by that set. Whenever that happens, it's natural to ask if they are similar in a specific sense - that is, if your functor is a monad. I just did this by looking up the formal definition of a monad on wikipedia.

To make it a monad, we need to choose two natural transformations.

The first one "looks like" the map from a set $S$ to $F(S)$ that sends each element to the corresponding generator. We need a map from a set to the to $M(S)$. The obvious choice is a measure that assigns mass $1$ to that point and mass $0$ to every other point. (Using the fact that it is a natural transformation, it is easy to see that it assigns mass $0$ to the rest of the set. $1$ is the obvious choice for the mass on that point.)

The second one "looks like" the map from $F(F(S))$ to $F(S)$ that sends each a word of words of elements of $S$ to the corresponding word, through concatenation. The map that "combines" a measure of measures $\nu$ on $M(S)$ to a single measure by integration:

$\int_{\mu \in M(S)} \mu d\nu$

seems very analogous. This integration can be infinite, which is why we need extended real measures.

We must check that both these functions are actually measurable, but this is trivial - we just take a basis for measurable sets, pull them back, and check that they are measurable

We must check that these natural transformations satisfy two commutative diagrams. This is simply a calculation - we take a single element of the appropriate measurable set and follow it through the diagram both ways to check that it ends up in the same place.

I think one can describe the algebras of this monad as being some form of cone.

Answer 2

Here is a candidate class of examples. I have made this community wiki so please feel free to edit it if you can answer it. Or, copy the text and make a new answer so we can give you reputation points.

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Let $X_0 := X$ be any non-empty measurable space, and for each $n \in \mathbb N$, define the product $X_{n+1} := M(X_n) \times X_n$. This is a measurable space when equipped with the tensor product of $\sigma$-algebras. Define $X_{\infty}$ to be the projective limit of the sequence $X_0 \leftarrow X_1 \leftarrow X_2 \leftarrow \cdots$, where the arrows denote the projections onto the second components of the products.

The existence of the limit gives a canonical section $X_{\infty} \to M(X_{\infty}) \times X_{\infty}$. By iterating this map with the projection onto the first component, we have defined a natural measurable function $\varphi : X_{\infty} \to M(X_{\infty})$.

Consequently, a point $x$ in $X_{\infty}$ contains the data of a measure $\varphi(x)$ on the space. It may be the case that this is all the data possessed by the point, in which case $X_{\infty} \cong M(X_{\infty})$. Is this the case? That is,

is the function $\varphi : X_{\infty} \to M(X_{\infty})$ one-to-one and onto?

Note that the requirement that $X_0$ be non-empty is necessary. If $X_0 = \varnothing$, then $M(X_0) = \{0\}$ consists of the zero measure, but $X_1 = \{0\} \times \varnothing = \varnothing$. Consequently $X_{\infty} = \varnothing$ and $M(X_{\infty}) = \{0\} \ne X_{\infty}$.

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This construction is based on the concept of an epistemic type space, which encodes the belief hierarchies of players in epistemic game theory.