# Atoms of a sigma algebra

I've been reading Schilling's Measures, Integrals, and Martingales, and I ran across a remark (on page 21) that I don't understand.

Here is the setup: let $$A_{1},\ldots,A_{n}$$ be non-empty, disjoint subsets of X with $$\cup A_{n}=X$$. Schilling says that "a set A in a $$\sigma$$-algebra $$\mathscr{A}$$ is called an atom, if there is no proper subset $$\emptyset\neq B \subsetneq A$$ such that $$B \in \mathscr{A}$$. In this sense all $$A_{n}$$ are atoms [of the $$\sigma$$-algebra generated by $$A_{1},\ldots,A_{n}]$$."

Why are the $$A_{n}$$ atoms? This seems intuitive, but I don't see how to prove it.

Since $$A_i$$ are disjoint and their union is the entire space $$\mathcal{X}$$, clearly you cannot construct a set $$B$$ such that $$B$$ is a strict subset of any $$A_i$$ using complimentation or union.
Since no such set can be constructed, it immediately follows that the $$\sigma$$-algebra generated by them, defined as containing any set that can be expressed as a countable union or compliment of these $$A_i$$, can contain a strict subset of any $$A_i$$. If it was true, it would contradict the prior observation. Hence all $$A_i$$ are atoms.
To see why no strict subset can be constructed, draw a diagram of a space and partition it into disjoint parts. It should be clear you cannot construct a strict subset of any the parts using the others (though you can construct any $$A_i$$ itself by complimenting the union of all other $$A_j, j \neq i$$.
If such a strict subset could be constructed, it would imply that some $$x \in A_i$$ also belongs to another $$A_j$$, contradicting the disjoint property of $$A_i$$'s.
• Thanks, Xiaomi. That's along the lines of what I was thinking. Another approach that I thought of is to first show that any element of the $\sigma$-algebra must either be the empty set or a union of some of the $A_i$. (This is mostly just definition-chasing, but also requires, I think, an appeal to the fact that the $\sigma$-algebra generated by the $A_i$ is the minimal $\sigma$-algebra containing the $A_i$.) It then follows immediately that the $A_i$'s are atoms. Oct 18, 2018 at 22:32
• This argument would extend to an arbitrary partition of $X$, not just a finite partition $A_1,\ldots,A_n$. Right? Dec 13, 2022 at 16:59