Exercise 8.2.3

Proof. Assume that $A\cap B=\varnothing$ and define $A_0=A$, $A_1=B$, and $A_n=\varnothing$ for all natural $n>1$. Then clearly each of the sets in ${\left\{A_n\right\}}_{n=0}^{\infty }$ are mutually disjoint. It is also trivial to show that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{A}_n=A\cup B$. We then have by property ii) that

noting that we have also used property i) according to which $\mu (\varnothing )=0$. This shows the desired result. □

Proof. Clearly $A$ and $B-A$ are disjoint sets such that $A\cup (B-A)=B$ so that $\mu (A)+\mu (B-A)=\mu (B)$ by property iv). The result then clearly follows by subtracting $\mu (A)$ from both sides. □

Proof. Suppose that $A\subseteq B$ so that $\mu (A)+\mu (B-A)=\mu (B)$ by Lemma 1. Since $\mu$ is a function into $[0,\infty )\cup \left\{\infty \right\}$, it follows that

as desired. □

Proof. Let $C=A\cap B$, $A^{\prime }=A-C$, and $B^{\prime }=B-C$. It is then trivial to show that $A^{\prime }$, $B^{\prime }$, and $C$ are mutually disjoint sets such that $A^{\prime }\cup B^{\prime }\cup C=A\cup B$. We then have by a straightforward extension of property iv) that

as desired. Note that we have also used Lemma 1 since clearly $C\subseteq A$ and $C\subseteq B$ so that $\mu (A-C)=\mu (A)-\mu (C)$ and $\mu (B-C)=\mu (B)-\mu (C)$. □

Proof. Supposing that we have a system of sets ${\left\{A_n\right\}}_{n=0}^{\infty }$, first we define a sequence of corresponding sets recursively:

Note that it is clear that $A_n^{\prime }\subseteq A_n$ for any $n\in 𝑵$ so that $\mu (A_n^{\prime })\le \mu (A_n)$ by property v).

We now show that each of these sets are mutually disjoint. So consider any natural $m$ and $n$ where $m\ne n$. Without loss of generality, we can then assume that $m<n$. Suppose that $A_m^{\prime }$ and $A_n^{\prime }$ are not disjoint so that there is an $x\in A_m^{\prime }\cap A_n^{\prime }$. Thus $x\in A_n^{\prime }=A_n-{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{n-1}{A}_k^{\prime }$ so that $x\notin {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{n-1}{A}_k^{\prime }$. However since also $x\in A_m^{\prime }$ and $0\le m\le n-1$, we also can conclude that $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{n-1}{A}_k^{\prime }$. Since this is a contradiction, it must be that $A_m^{\prime }$ and $A_n^{\prime }$ are in fact disjoint, which shows mutual disjointedness since $m$ and $n$ were arbitrary.

Next we show that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{A}_n^{\prime }={\bigcup }_{n=0}^{\infty }{A}_n$. The $\subseteq$ direction is clear since, for any $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{A}_n^{\prime }$, there a natural $n$ where $x\in A_n^{\prime }$. Since $A_n^{\prime }\subseteq A_n$ it follows that $x\in A_n$ so that clearly $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{A}_n$. Now consider any $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{A}_n$ so that there is a natural $n$ where $x\in A_n$. Clearly if $n=0$ then $x\in A_0=A_0^{\prime }$. So assume that $n>0$ so that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{n-1}{A}_k^{\prime }$ is defined. If $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{n-1}{A}_k^{\prime }$ then there is a $0\le k\le n-1$ where $x\in A_k^{\prime }$. On the other hand, if $x\notin {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{n-1}{A}_k^{\prime }$ then clearly $x\in A_n-{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{n-1}{A}_k^{\prime }=A_n^{\prime }$. Thus in all cases there is a natural $k$ such that $x\in A_k^{\prime }$ so that $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{A}_n^{\prime }$ as desired.

We therefore have

as desired. □