When I am reviewing the definition of sigma-algebra: (suppose M is a non-empty collection of subsets of a set X) (1) empty set and X belong to M (2) M closed under complement (3) M closed under countable unions. M is also closed under countable intersections.

Now if I just have M closed under complements and the fact that M is closed under countable unions, why does it not suffice to say that M is a sigma algebra? Is there a counter example?

  • $\begingroup$ You'd like the whole space to be measurable, I suppose, and this and axiom (2) then makes the empty set measurable. $\endgroup$ Commented Feb 3, 2019 at 22:17
  • $\begingroup$ Sorry I missed out some conditions - I included them just now $\endgroup$ Commented Feb 3, 2019 at 22:25

1 Answer 1


For $\mathcal{A} \subseteq \mathscr{P}(X)$ to be a $\sigma$-algebra, three axioms are usually stated:

  1. $\emptyset \in \mathcal{A}$.
  2. $\mathcal{A}$ is closed under complements.
  3. $\mathcal{A}$ is closed under countable unions.

These will imply that also $X \in \mathcal{A}$ by combining 1 and 2, and that $\mathcal{A}$ is closed under countable intersections (by de Morgan, combining 2 and 3); note that we could have replaced 3 by the axiom that $\mathcal{A}$ be closed under countable intersections (and unions would also have followed). Also closedness under finite unions follows from 3 too, and then using 2 we have finite intersections too, and then differences too are preserved as $A \setminus B=A \cap B^\complement$ etc. So we get an algebra that is also closed under countable set operations. These 3 axioms (or their variants, we could also have demanded axiom 1 to be $X \in \mathcal{A}$ (and derive the empytyset using 2) are pretty minimal. We can easily come up with examples that only obey 2 out of 3 of them and are not $\sigma$-algebras.

The only example of a family obeying 2 and 3 and not 1 is $\mathcal{A} = \emptyset$, boringly enough. Because if we have some $A \in \mathcal{A}$ (so when $\mathcal{A} $ is non-empty), we get $A^\complement \in \mathcal{A} $ by 2, $A \cup A^\complement = X \in \mathcal{A}$ by 3 and then $\emptyset \in \mathcal{A}$ by 2 again. Voidly, $\mathcal{A} = \emptyset$ does obey 2 and 3, but not 1.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .