I am learning measure theory this semester. The definition for sigma-algebra is "a collection of sets that is closed under complements and countable unions and intersections." I wonder what does it mean by "closed under complements and countable unions and intersections." Thank you so much for your help!

• Saying that a collection of objects is "closed under blah blah blah" means that if you perform the "blah blah blah" operation with any objects selected from the collection, the resulting object is again a member of the collection. – MPW Jan 18 '15 at 3:25

A family $$\mathcal{M}$$ is a sigma-algebra if it satisfies the following conditions:

It is closed under compliments: If $$E\in \mathcal{M}$$ then $$E^c\in\mathcal{M}$$

It is closed under countable unions: If $$E_1, E_2, E_3, \dots$$ are sets in $$\mathcal{M}$$ then $$\bigcup\limits_{i=1}^\infty E_i \in \mathcal{M}$$

It is closed under countable intersections: If $$E_1, E_2, E_3, \dots$$ are sets in $$\mathcal{M}$$ then $$\bigcap\limits_{i=1}^\infty E_i \in \mathcal{M}$$

In particular, being closed under countable unions implies also being closed under a finite number of unions since you can take $$E_{n> N} = \emptyset$$ to show $$\mathcal{M}\ni \bigcup\limits_{i=1}^\infty E_i = \bigcup\limits_{i=1}^N E_i$$.

It is also of interest to note that an equivalent definition asks instead of being closed under intersections that it instead be closed under set differences. It is a good exercise to try to prove that the two definitions are equivalent (i.e., given that it is closed under complement and countable unions, that it follows that it is closed under intersections if and only if it is also closed under differences as well).

For a concrete example, consider the sigma algebra $$(\mathscr{P}(X),X)$$ where $$X = \{1,2\}$$ and $$\mathscr{P}(X) =$$the power set of $$X$$. That is to say, $$\mathscr{P}(X)=\{\emptyset, \{1\},\{2\},\{1,2\}\}$$. The power set is always an example of a sigma-algebra.

It is closed under compliments since for any set (for example $$\{1\}$$) its complement is also in the algebra (in this case $$\{1\}^c = \{2\}\in\mathscr{P}(X)$$)

It is closed under union since for any two sets (for example $$\{1\},\{1,2\}$$) their union is also in the algebra (in this case $$\{1\}\cup\{1,2\} = \{1,2\}\in\mathscr{P}(X)$$)

It is similarly closed under intersection since for any two sets (for example $$\{1\},\emptyset$$) their intersection is also in the algebra (in this case $$\{1\}\cap\emptyset = \emptyset \in \mathscr{P}(X)$$)

• Thanks so much. I think to prove It is closed under countable unions implies closed under countable intersections, we can use Demorgan's law and the fact that the compliments are in the Family. Is that right? – Sissi Sue Jan 18 '15 at 14:42
• @sissisue yes, that is absolutely correct. That being said, really the only conditions necessary to be a sigma algebra are to be closed under compliment and closed under countable union (and be nonempty). Notice, are X and $\emptyset$ in the sigma algebra? One more thing to mention, algebras are defined the same way except they aren't required to be closed under countably many unions, just finitely many is good enough. – JMoravitz Jan 18 '15 at 15:50