Show that the family $\left\{A\in\mathcal{S}:\mu(A)=0 \text{ or } \mu(A^c)=0\right\}$ is the $\mu$-trivial $\sigma$-algebra.

Question

Show that the family $\left\{A\in\mathcal{S}:\mu(A)=0 \text{ or } \mu(A^c)=0\right\}$ is a $\sigma$-algebra (the so-called $\mu$-trivial $\sigma$-algebra).

• Clearly $\phi \in \mathcal{S}$ since $\mu(\phi)=0$.
• If $A\in\mathcal{S}$ then $\mu(A)=0$ or $\mu(A^c)=0$ which means either $\mu((A^c )^c )=\mu(A)=0$ or $\mu(A^c )=0$. Hence $A^c\in \mathcal{S}$
• Here I want to show that $A_n\in \mathcal{S} \implies \cup_n A_n\in \mathcal{S}$ and here lies my problem

Let $A_1,A_2\in\mathcal{S}$ such that $\mu(A_1)=0$ and $\mu({A_2}^c)=0$ but $\mu(A_2)\neq0$ and $\mu({A_1}^c)\neq0$ then neither $\mu(A_!\cup A_2)=0$ nor $\mu({A_1}^c\cup{A_2}^c)=0$ hence $A_!\cup A_2 \notin\mathcal{S}$.

Where am I wrong here and how do I show $A_n\in \mathcal{S} \implies \cup_n A_n\in \mathcal{S}$?

Also, how do I show this is the trivial $\sigma$ -algebra?

Answer 1

In the case of your two sets $A_1$ and $A_2$, what you want to show is that either $\mu(A_1\cup A_2) = 0$ or $\mu((A_1\cup A_2)^c) = 0$. The latter is true, since $(A_1\cup A_2)^c = A_1^c\cap A_2^c \subseteq A_2^c$, so $\mu((A_1\cup A_2)^c) \leq \mu(A_2^c) = 0$.

For a countable union $\bigcup_n A_n$, the same sort of argument will show that if one of the $A_i$ satisfies $\mu(A_i^c) = 0$, then $\mu((\bigcup_n A_n)^c) = 0$. And on the other hand if $\mu(A_i) = 0$ for all $i$, then countable additivity implies that $\mu(\bigcup_n A_n) = 0$.