Exercise 8.2.6

Proof. Let $=𝒫\left(S\right)$, which we know is the largest $\sigma$-algebra of subsets of $S$. We must show that $\mu$ as defined above satisfies the properties of $\sigma$-additive measure on $S$:

To show i), we clearly have that $a\notin \varnothing$ so that by definition $\mu (\varnothing )=0$. Also, clearly $a\in S$ so that $\mu (S)=1>0$.

Regarding ii), suppose that ${\left\{X_n\right\}}_{n=0}^{\infty }$ is a collection of mutually disjoint sets in $=𝒫\left(S\right)$.

Case: $a\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n$. Then by definition $\mu \left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right)=1$. There is also an $n\in 𝑵$ such that $a\in X_n$, and since the sets ${\left\{X_k\right\}}_{k=0}^{\infty }$ are mutually disjoint, it follows that $a\notin X_m$ for any natural $m\ne n$ (since otherwise $X_n$ and $X_m$ would not be disjoint). Thus we have $\mu (X_n)=1$ while $\mu (X_m)=0$ for all natural $m\ne n$. Hence

Thus clearly $\mu \left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }\mu (X_n)=1$.

Case: $a\notin {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n$. Then $\mu \left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n\right)=0$ by definition. It also follows that $a\notin X_n$ for every natural $X_n$ so that $\mu (X_n)=0$. Hence clearly

Thus ii) is shown in both cases so that $\mu$ is indeed a $\sigma$-additive measure on $S$ since we also showed i). □