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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Oct 18, 2018 at 22:32
  • $\begingroup$ This argument would extend to an arbitrary partition of $X$, not just a finite partition $A_1,\ldots,A_n$. Right? $\endgroup$
    – James
    Dec 13, 2022 at 16:59

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