# Definition of sigma algebra - beyond pi and lambda system

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?

• You'd like the whole space to be measurable, I suppose, and this and axiom (2) then makes the empty set measurable. – kimchi lover Feb 3 '19 at 22:17
• Sorry I missed out some conditions - I included them just now – Extra Learn Feb 3 '19 at 22:25

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.