The intersection of all generators of a sigma-algebra never has measurable subsets

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Let $\mathcal B = \sigma (D_1,...)$ be a countably generated $\sigma$-algebra. Let $D = \cap D_i \in \mathcal B$.

It seems intuitive that generating a $\sigma$-algebra from $\{D_1,...\}$ cannot produce anything smaller than $\cap D_i$. To put it in precise terms:

Let $B \subseteq D$ be a nonempty set, and assume $B \in \mathcal B$. Must we have $B=D$?

In other words, we ask if $D$ is a $\mathcal B$-atom.

In the finitely generated case, the sigma algebra is just all sets created by the Venn diagram so we see graphically that the claim holds. For non-countably generated algebras the intersection may no longer be an event so we may run into problems.

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We call the ground set $X$.

We say two elements $x$ and $y$ in X are related if $x\in D_i \implies y\in D_i$ and visceversa for all $i$.

Let $C_\alpha$ be the equivalence classes of our relation.

It is not hard to see that the set $\mathcal C$ of subsets of $X$ that are unions of some $C_\alpha$ is a sigma algebra that contains each $D_i$.

Notice that the intersections of all the $D_i$ is one of the $C_i$ and so clearly no element of $\mathcal C$ is a proper subset of the intersection of all the $D_i$.