Let
- $(\Omega,\mathcal A)$ be a measurable space
- $\mu$ be a real-valued measure on $(\Omega,\mathcal A)$
Are we able to show that $$\mathcal M:=\left\{A\in\mathcal A:\mu(A)\ge0\right\}$$ is a monotone class?
Let $(A_n)_{n\in\mathbb N}\subseteq\mathcal M$ and $A:=\bigcup_{n\in\mathbb N}A_n$. We need to show that $A\in\mathcal M$. Let $$B_n:=\bigcup_{i=1}^nA_i$$$$C_1:=B_1\;,\;\;\;C_{n+1}:=B_{n+1}\setminus B_n=A_{n+1}\setminus B_n$$ $(B_n)_{n\in\mathbb N}$ is nondecreasing and $$A=\bigcup_{n\in\mathbb N}B_n=\biguplus_{n\in\mathbb N}C_n\;.$$ We've clear got $$\mu(C_{n+1})=\mu(B_{n+1})-\mu(B_n)\;.$$ So, it seems like we need to show $$0\le\mu(B_n)\le\mu(B_{n+1})\;.$$
Can we do that?
Oops: I said
No, the set where $\mu\ge0$ need not even be closed under finite unions. Say $X=\{1,2,3\}$ and $\mu(\{j\})=c_j$, $c_1=-2$, $c_2=3$, $c_3=-2$. Let $A=\{1,2\}$ and $B=\{2,3\}$; then $\mu(A),\mu(B)>0$, $\mu(A\cup B)<0$.
The positive sets for $\mu$ do form a monotone class; here $A$ is a positive set for $\mu$ if $\mu(E)\ge0$ for every $E\subset A$.
That's wrong. It's true that the sets where $\mu\ge0$ are not closed under finite unions, but I was forgetting the definition of "monotone class". The sets where $\mu\ge0$ are closed under countable increasing unions by "continuity from below". And the magical fact that a signed measure has finite total variation show that it is also continuous from above, hence the sets where $\mu\ge0$ are also closed under decreasing unions.