In Royden's Real Analysis book, he defines, for a collection $S$ of subsets of a set $X$ a premeasure to be a set function $\mu:S\rightarrow[0,\infty]$ that is both finitely additive and countably monotone and if $\emptyset\in S$ then $\mu(\emptyset)=0$. A problem in the book asks to show that a premeasure on a sigma algebra is a measure (where a measure is defined to satisfy countable additivity).
So, I'd like to show ultimately that given a set function $\mu:S\rightarrow[0,\infty]$ on a sigma algebra $S$, that countably monotone and finitely additive imply countable additive.
I am having trouble figuring this out. It seems to amount to showing that $\mu(\bigcup_{n=1}^\infty E_k)\geq \sum_{n=1}^\infty\mu(E_k)$, as the reverse inequality holds due to countable monotone. How to go about proving this inequality?
Best Answer
Let $E_j$ be a countable disjoint sequence of sets in $S$. Let $E = \bigcup_1^\infty E_j$. Let $F_i = \bigcup_1^i E_j$. Let $G_i = E - F_i$. Then $\mu(E) = \mu(G_i \cup F_i) = \mu(G_i) + \sum_{j=1}^n \mu(E_j)$. Then if $\mu(E) = \infty$, we have $\mu(E) \leq \sum_j \mu(E_j)$, so $\sum_j \mu(E_j) = \infty = \mu(E)$. If $\mu(E) < \infty$, then $\mu(G_i)$ is a decreasing sequence bounded below by $0$, so has a limit $\mu(G_i) \to C$. Then $\mu(E) = C + \sum_1^\infty \mu(E_j)$ by taking limits. But $\mu(E) \leq \sum_j \mu(E_j) = \mu(E) - C$, so $C =0$, and $\sum_j \mu(E_j) = \mu(E)$.