Exercise 8.2.8

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 $S$:

For i) we have that $\mu (\varnothing )=\left|\text{∅}\right|=0$ since $\text{∅}$ is finite. If $S$ is finite then $\mu (S)=\left|S\right|>0$ since $S$ is nonempty. If $S$ is infinite then $\mu (S)=\infty >0$ as well so that in either case $\mu (S)>0$ as desired.

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

Case: There is an $m\in 𝑵$ where $X_m$ is infinite. Then obviously $\mu (X_m)=\infty$ by definition. We also clearly have that $$ is infinite since $X_m\subseteq$, and hence $\mu ()=\infty$. Then

since $\mu (X_n)\in 𝑵\cup \left\{\infty \right\}$ for every $n\ne m$. Therefore $\mu ()={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }\mu (X_n)=\infty$ as desired.

Case: $X_n$ is finite for every $n\in 𝑵$. Clearly then $\mu (X_n)=\left|X_n\right|$ for every natural $n$. First, if there is a natural $N$ such that $X_n=\varnothing$ for all $n>N$, then clearly $={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^N{X}_n$, i.e. the union is finite. It then follows that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^N{X}_n$ is finite by Theorem 4.2.7 so that $\mu \left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^N{X}_n\right)=\left|{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^N{X}_n\right|$. Then, since the sets ${\left\{X_n\right\}}_{n=0}^N$ are mutually disjoint, we have

by the definition of cardinal addition, noting that clearly the last step follows from the fact that $\mu (X_n)=\left|\text{∅}\right|=0$ for all $n>N$.

On the other hand, if there is no such $N$, then it follows that, for every $N\in 𝑵$, there is a natural $n>N$ such that $X_n\ne \varnothing$. At this point we need two facts from real analysis, supposing that ${⟨a_n⟩}_{n=0}^{\infty }$ is a real sequence:

1.

By definition, the sequence converges to a real $a$ if, for every real $𝜀>0$, there is an $N\in 𝑵$ such that $\left|a_n-a\right|<𝜀$ for every natural $n\ge N$.

2.

If the infinite series ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{a}_n$ converges (to a finite value) then the sequence itself must converge to zero.

We shall show that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }\mu (X_n)$ diverges by showing that the sequence ${⟨\mu (X_n)⟩}_{n=0}^{\infty }$ does not converge to zero (i.e. the contrapositive of 2). So let $𝜀=1∕2$, noting that clearly $𝜀=1∕2>0$. Then consider any natural $N$ so that there is a natural $n>N$ such that $X_n\ne \varnothing$. It then follows that, since $X_n$ is finite but nonempty, $\left|\mu (X_n)-0\right|=\left|\mu (X_n)\right|=\left|\left|X_n\right|\right|=\left|X_n\right|\ge 1\ge 1∕2=𝜀$. Therefore we have shown

which shows by definition that ${⟨\mu (X_n)⟩}_{n=0}^{\infty }$ does not converge to zero. Hence ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }\mu (X_n)$ diverges so that by convention ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }\mu (X_n)=\infty$.

Lastly, we show that $$ must be infinite. Suppose to the contrary that $$ is finite. We then construct a function $f:\to 𝑵$ as follows: for each $x\in$ there is a unique $m\in 𝑵$ where $x\in X_m$. Clearly such an $m$ exists since $x\in$, and it is unique because the sets ${\left\{X_n\right\}}_{n=0}^{\infty }$ are mutually disjoint (if it was not unique then there would be distinct $n$ and $m$ where $x\in X_n$ and $x\in X_m$ so that $X_n\cap X_m\ne \varnothing$). We then simply set $f(x)=m$.

It then follows from Theorem 4.2.5 that $\mathrm{ran}(f)$ is finite since $\mathrm{dom}(f)=$ is. Since $\mathrm{ran}(f)$ is then a finite set of natural numbers, it has greatest natural number $N$. But we know that there is an $m>N$ such that $X_m\ne \varnothing$ so that there is an $x\in X_m$. It then follows that clearly $x\in$ and that $f(x)=m$. However, then $m$ would be in $\mathrm{ran}(f)$ so that $m\le N$ since $N$ is the greatest element of $\mathrm{ran}(f)$. But we already know that $m>N$, which is a contradiction. So it has to be that $$ is in fact infinite as desired.

We therefore have $\mu \left(\right)=\infty ={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }\mu (X_n)$ and hence ii) has been shown in every case and sub-case so that $\mu$ is indeed a $\sigma$-additive measure on $S$ by definition. □