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?

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 $\Sigma^{\infty}(X)$?