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 $M :\operatorname{Meas} \to \operatorname{Meas}$ ?

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    $\begingroup$ I think $\mathbb R$ should be a fixed object, since a measure on $\mathbb R$ is a countable sequence of real numbers satisfying certain inequalities, which we can easily encode as a single real number. $\endgroup$ – Will Sawin Dec 26 '12 at 20:24
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    $\begingroup$ This functor is a monad, assuming the measures involved are allowed to take the value $\infty$. The identity sends a point to the atomic measure with mass $1$ at that point. The multiplication sends a measure on the space of measures to its integral. The two coherence conditions arise because we can reverse the order of integration of nonnegative functions without issue, and because the integral of a function against the atomic matter with mass $1$ is just the evaluation of that function. I do not know if their is a natural monad structure without infinities involved. I presume not. $\endgroup$ – Will Sawin Dec 26 '12 at 20:33
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    $\begingroup$ @Will: I am perfectly happy to accept infinite measures; I have updated my question to say extended-real-valued measures. Unfortunately, I am not familiar enough with category theory (and monads in particular) to understand your response. Could you write out a longer answer describing the monad structure present? $\endgroup$ – Tom LaGatta Dec 26 '12 at 21:02
  • $\begingroup$ For others who are as unfamiliar with monads as I am, here is a nice video by the Catsters, featuring Eugenia Cheng presenting the basics of monads: youtube.com/… h/t @Joey Hirsh $\endgroup$ – Tom LaGatta Dec 27 '12 at 7:17

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.

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    $\begingroup$ Isn't this the Giry monad? $\endgroup$ – G. Rodrigues Dec 26 '12 at 22:13
  • $\begingroup$ @G, what is the Giry monad? $\endgroup$ – Tom LaGatta Dec 26 '12 at 22:44
  • $\begingroup$ Probably. What's the Giry monad? $\endgroup$ – Will Sawin Dec 26 '12 at 22:49
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    $\begingroup$ Reference: Michèle Giry, A categorical approach to probability theory. In B. Banaschewski, editor, Categorical Aspects of Topology and Analysis, Springer LNM 915, 1982. It's almost inconceivable that there could be two different reasonable monad structures on this endofunctor, so I bet it's the same. $\endgroup$ – Tom Leinster Dec 26 '12 at 23:51
  • $\begingroup$ Thanks @Tom. This sounds like an important structure, so I've posted another thread asking for possible applications of the Giry monad in probability and statistics: mathoverflow.net/questions/117294/… $\endgroup$ – Tom LaGatta Dec 27 '12 at 1:07

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.

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 $M(X_{\infty}) = \{0\} \ne X_{\infty}$.

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

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    $\begingroup$ One needs additional assumption for the projective limit to exist, usually of a quasi-topological kind. The classical counterexample can be found here: sdu.dk/media/bibpdf/Bind%2020-29%5CBind%5Cmfm-25-4.pdf Also, the projective limit seems to make only sense when we deal with probability measures. $\endgroup$ – Michael Greinecker Dec 27 '12 at 11:25
  • $\begingroup$ @Michael, your counterexample is sound but I am not certain it is relevant here. First, note that we are working with measurable spaces and not measure spaces. As a set, $X_{\infty}$ certainly exists, and it should be a measurable space when equipped with the minimal $\sigma$-algebra so that the projection maps are measurable. The example by Andersen & Jessen might be an obstruction to showing that $X_{\infty} \cong M(X_{\infty})$ in general; I do not readily see why this is the case. Why does the projective limit seem to make sense only when we deal with probability measures? $\endgroup$ – Tom LaGatta Dec 27 '12 at 17:04
  • $\begingroup$ @Tom Yes, the infinite product of measurable spaces clearly exists and the problem lies in identifying it with a measure on the product. The problem one has when working with non-probability measures that in general, all sets have infinte or zero measure (just think about $\mathbb{R}^\infty$ with Lebesgue measure on each factor). In the type space constructions used in game theory (and in the projective limit theorems of Daniell, Kolmogorov, Bochner etc.), only probability spaces are used. There is also a version of the Andersen-Jessen counterexample that is adapted to epistemic game theory... $\endgroup$ – Michael Greinecker Dec 27 '12 at 17:52
  • $\begingroup$ ...due to Heifetz and Samet: tau.ac.il/~samet/papers/coherent.pdf The problem does however not occur when one simply takes products of probability spaces. Every direct product of prob spaces can be extended to a prob measure on the product $\sigma$-algebra, something that kakutani has shown a long, long time ago. This result does hold only for probability measures. $\endgroup$ – Michael Greinecker Dec 27 '12 at 17:57
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    $\begingroup$ @Tom Can you explain how you construct your canonical section? My impression is that is quite different from the type space construction in epistemic game theory. For probability measures, the original problem seems to be rather easy. If $P(X)\subseteq M(X)$ consist of the probability measures on $X$, then $P$ is idempotent in that one can integrate with respect to a probability measure over probability measures to get a probability measures. As a matter of fact, measurable functions $f:X\to P(Y)$ correspond to transition probabilities or Markov kernels $k:X\times\mathcal{Y}\to[0,1]$... $\endgroup$ – Michael Greinecker Dec 28 '12 at 11:21

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