Showing finite additivity of a given measure

Question

Let $(X,\mathcal{A},\mu)$ be a measure space. For $A\in\mathcal{A}$, define $S(A)=\{B\in \mathcal{A}:B\subset A,\mu(B)<\infty\}$ Define $\nu:\mathcal{A}\to[0,\infty]$ by $\nu(A)=\sup\limits_{B\in S(A)}\{\mu(B)\}$. I have shown $\nu$ is $\sigma$-subadditive, and I want to show that $\nu$ is finitely additive. So I let $A_1,A_2\in\mathcal{A}$, disjoint. Then by the $\sigma$-subadditivity, I know that $\nu(A_1\cup A_2)\leq\nu(A_1)+\nu(A_2)$. Next I want to prove the reverse inequality. If I fix $B\in S(A_1\cup A_2)$, I see that $$\nu(A_1\cup A_2)\geq\mu(B)=\mu((B\cap A_1)\cup(B\cap A_2))=\mu(B\cap A_1)+\mu(B\cap A_2).$$ It follows that $$ \nu(A_1\cup A_2)\geq\sup\limits_{B\in S(A_1\cup A_2)}\{\mu(B\cap A_1)+\mu(B\cap A_2)\}\underbrace{=}_{?}$$$$\sup\limits_{B\in S(A_1\cup A_2)}\{\mu(B\cap A_1)\}+\sup\limits_{B\in S(A_1\cup A_2)}\mu(B\cap A_2)\underbrace{=}_{?}\nu(A_1)+\nu(A_2). $$ Are the equality signs indicated by a question mark correct?

Answer 1

The equality is correct, but not because of some property of the supremum (in general the supremum of the sum is less than the sum of suprema). The key is that for every choice of argument of the supremum on the left there is a corresponding choice on the right and vice versa.

Choose $\epsilon > 0$, and pick sets $B_i \subset A_i$ of finite measure such that $\mu(B_i) \geq \nu(A_i) - \epsilon/2$. Set $A = \cup_i A_i$ and $B = \cup_i B_i$. Then $B \in S(A)$ and $\mu(B) \geq \nu(A_1) + \nu(A_2) - \epsilon$, implying that $\nu(A) \geq \nu(A_1) + \nu(A_2) - \epsilon$. Since $\epsilon$ was arbitrary you get $\nu(A) \geq \nu(A_1) + \nu(A_2)$.

In the other direction, for any $B \in S(A)$ set $B_i = B \cap A_i$. Then $$\mu(B) = \mu(B_1) + \mu(B_2) \leq \nu(A_1) + \nu(A_2).$$ Taking a supremum on the left gives $\nu(A) \leq \nu(A_1) + \nu(A_2)$.

You can extend this proof to countable additivity if $\mu$ is countably additive. Just replace $\epsilon/2$ in the first part with $\epsilon/2^i$.