Is there a natural measurable structure on the $\sigma$-algebra of a measurable space?

Question

Let $(X, \Sigma)$ denote a measurable space. Is there a non-trivial $\sigma$-algebra $\Sigma^1$ of subsets of $\Sigma$ so that $(\Sigma, \Sigma^1)$ is also a measurable space?

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Here is one natural candidate. I'm not certain, but based on answers to related questions, I think this might be the Effros Borel structure that Gerald Edgar has mentioned here and here.

The $\sigma$-algebra $\Sigma$ is an ordered set under the canonical relation given by subset inclusion $\subseteq$, and is therefore naturally equipped with a specialization topology. The closed sets are generated by downward-closed sets, and the closure of a singleton is its down-set:$$\overline{\{A\}} = \{ B \in \Sigma : B \subseteq A \}.$$ Even though this topology is highly non-Hausdorff, it's still pretty nice. For example, it's an Alexandroff space: arbitrary unions of closed sets are closed.

Being a topological space, $\Sigma$ now has a natural measurable structure, namely, the one generated by the Borel $\sigma$-algebra $\Sigma^1 := \mathcal B_{\subseteq}(\Sigma)$.

• Is this space $(\Sigma, \Sigma^1)$ a reasonable one on which to do measure theory and probability?

Whether it is or not, there's some non-trivial structure present. For example, we can iterate this procedure. Set $\Sigma^0 = \Sigma$, and define $\Sigma^n := \mathcal B_{\subseteq}(\Sigma^{n-1}).$ Then each one of these spaces $\Sigma^n(X) := (\Sigma^{n}, \Sigma^{n+1})$ is measurable.

• Is $\Sigma : \mathrm{Meas} \to \mathrm{Meas}$ an endofunctor on the category of measurable spaces?

• Under what conditions does the sequence of measurable spaces $\Sigma^n(X)$ have a limit $\Sigma^{\infty}(X)$?

Answer 1

If $(X,\Sigma)$ is a measurable space, I think you are asking for a $\sigma$-algebra structure on $|\Sigma|$, the underlying set of $\Sigma$. We can identify this set with the set of measurable functions $$|\Sigma|\cong \text{Hom}_{\text{Meas}}\;(X,2),$$ where $2$ is a two-point space with discrete $\sigma$-algebra.

Thus it suffices to prove a more general result: that $\text{Meas}$ is a closed monoidal category. In other words, we would like to know that for any two measurable spaces, $X,Y$, there is a $\sigma$-algebra on the set of maps $\text{Hom}_{\text{Meas}}\;(X,Y)$, which has good formal properties (functoriality, right adjointness to $\otimes$).

The fact that $\text{Meas}$ is a symmetric monoidal closed category was proven by Kirk Sturtz in the paper Categorical Probability Theory. See Section 2.3.

Answer 2

$\newcommand\sdiff{\mathbin\triangle}$Starting from literally just the measurable space $(X,\Sigma)$, I wouldn't think there's any particularly nice and natural $\sigma$-algebra on $\Sigma$; but with just a tiny bit more structure, there is! Namely:

For any given equivalence class $\mathfrak{p}$ of probability measures on $(X,\Sigma)$ (where two probability measures being equivalent means that they agree on which sets have zero measure), we can define the $\sigma$-algebra $\Sigma_\mathfrak{p}^1$ on $\Sigma$ by $$ \Sigma_\mathfrak{p}^1 \ = \ \sigma( \, \{ A \in \Sigma : \mathbb{P}(A \cap E) \leq r \} \, : \, E \in \Sigma, \, r \in [0,1] \, ), $$ i.e. the smallest $\sigma$-algebra for which the map $A \mapsto \mathbb{P}(A \cap E)$ is measurable for each $E \in \Sigma$, where $\mathbb{P}$ may be any element of $\mathfrak{p}$. (This does not depend on the choice of $\mathbb{P}$, as I will justify shortly.)

This can equivalently be expressed as $$ \Sigma_\mathfrak{p}^1 \ = \ \sigma( \, \{ A \in \Sigma : \mathbb{P}(A \sdiff E) \leq r \} \, : \, E \in \Sigma, \, r \in [0,1] \, ), $$ since \begin{align*} \mathbb{P}(A \sdiff E) &= \mathbb{P}(A \cap (X \setminus E)) + \mathbb{P}(E) - \mathbb{P}(A \cap E) \\ \mathbb{P}(A \cap E) &= \tfrac{1}{2}\!\bigl( \mathbb{P}(E) + 1 - \mathbb{P}(A \sdiff X) - \mathbb{P}(A \sdiff E) \bigr). \end{align*}

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A slightly cleaner way of thinking about essentially the same thing: Let $\tilde{\Sigma}_\mathfrak{p}$ be the quotient of $\Sigma$ under the equivalence relation $$ A \sim B \ \Longleftrightarrow \ A \sdiff B \,\text{ is a $\mathfrak{p}$-null set.} $$ Given any $\mathbb{P} \in \mathfrak{p}$, we can define a metric $d_\mathbb{P}$ on $\tilde{\Sigma}_\mathfrak{p}$ by $$ d_\mathbb{P}([A],[B]) = \mathbb{P}(A \sdiff B). $$ The topology induced by this metric is independent of the choice of $\mathbb{P}$. [Proof: given $A \in \Sigma$ and a sequence $(A_n)$ in $\Sigma$, if $\mathbb{P}$ and $\mathbb{P}'$ are probability measures with $\mathbb{P}(A_n \sdiff A) \to 0$ and $\mathbb{P}'(A_n \sdiff A) \not\to 0$, then we can find a subsequence $(A_{k_n})$ with $\inf_{n \geq 1} \mathbb{P}'(A_{k_n} \sdiff A) > 0$ and $\mathbb{P}(A_{k_n} \sdiff A) \leq \frac{1}{2^n}$, and so the set $\,\bigcap_{n \geq 1} \bigcup_{m \geq n} (A_{k_m} \sdiff A)\,$ is $\mathbb{P}$-null but not $\mathbb{P}'$-null, and so $\mathbb{P}$ is not equivalent to $\mathbb{P}'$.]

This topology is quite nice; e.g. it is completely metrisable (with $d_\mathbb{P}$ being a complete metric), and if $\Sigma$ is countably generated then this topology is also separable (and hence Polish). The Borel $\sigma$-algebra $\smash{\tilde{\Sigma}}_\mathfrak{p}^1$ of this topology is precisely $$ \smash{\tilde{\Sigma}}_\mathfrak{p}^1 \ = \ \sigma( \, \{ [A] \in \tilde{\Sigma}_\mathfrak{p} : \mathbb{P}(A \sdiff E) \leq r \} \, : \, E \in \Sigma, \, r \in [0,1] \, ). $$ Hence there is a trivial identification between $\smash{\tilde{\Sigma}}_\mathfrak{p}^1$ and $\Sigma_\mathfrak{p}^1$.

Answer 3

One way to approach this would be to ask the same question inside a suitable topos in which "everything is measurable" and such that each object is naturally equipped with the structure of a $\sigma$-algebra. In effect you would be expanding the notion of measure space to accommodate better structure, as such toposes typically contain the "classical" measure spaces.

For example, Matthew Jackson's Ph.D. dissertation "A sheaf theoretic approach to measure theory" might be a starting point.