Proof of Basic Properties of a Measure

Throughout, $(X,\mathcal{F},\mu)$ and the $[0,\infty]$ conventions are as in the statement. We use repeatedly that $\mathcal{F}$ is closed under differences: for $A,B\in\mathcal{F}$ we have $B\setminus A=X\setminus\bigl((X\setminus B)\cup A\bigr)$, which lies in $\mathcal{F}$ by the closure properties recorded in Sigma-Algebra and Measurable Space.

Claim 1. Define a sequence $(D_m)_{m\in\mathbb{N}}$ in $\mathcal{F}$ by $D_m=A_m$ for $m\le r$ and $D_m=\varnothing$ for $m>r$. Its members are pairwise disjoint and $\bigcup_{m\in\mathbb{N}}D_m=\bigcup_{i=1}^{r}A_i$, so countable additivity in Measure, Measure Space, and Probability Measure gives

$$\mu\Bigl(\bigcup_{i=1}^{r}A_i\Bigr)=\sum_{m\in\mathbb{N}}\mu(D_m).$$

Since $\mu(\varnothing)=0$, the $k$-th partial sum of $(\mu(D_m))_{m\in\mathbb{N}}$ equals $\sum_{i=1}^{\min\{k,r\}}\mu(A_i)$, and in particular equals $\sum_{i=1}^{r}\mu(A_i)$ for every $k\ge r$.

If every $\mu(A_i)$ with $i\le r$ is real, then every $\mu(D_m)$ is real and the partial sums form a nondecreasing sequence that is constant with value $\sum_{i=1}^{r}\mu(A_i)$ from index $r$ on; that value is therefore an upper bound for the partial sums and is itself a partial sum, hence it is their least upper bound. By the definition of the sum of a sequence in $[0,\infty]$, the displayed sum equals $\sum_{i=1}^{r}\mu(A_i)$.

If $\mu(A_i)=\infty$ for some $i\le r$, then not every term of $(\mu(D_m))$ is real, so by that same definition $\sum_{m}\mu(D_m)=\infty$; and the finite sum $\sum_{i=1}^{r}\mu(A_i)$, formed in $[0,\infty]$ with the convention $a+\infty=\infty+a=\infty$, is also $\infty$. In either case the two sides agree, and the finite sum is $\infty$ precisely when some term is $\infty$, since a finite sum of nonnegative real numbers is real.

Claim 2. The sets $A$ and $B\setminus A$ are disjoint members of $\mathcal{F}$ whose union is $B$, so claim 1 with $r=2$ gives

$$\mu(B)=\mu(A)+\mu(B\setminus A).$$

If $\mu(A)=\infty$ or $\mu(B\setminus A)=\infty$, then $\mu(B)=\infty$ and $\mu(A)\le\mu(B)$ holds, because every element of $[0,\infty]$ is at most $\infty$ by the convention $a<\infty$ for real $a\ge0$. Otherwise both are nonnegative real numbers and $\mu(A)\le\mu(A)+\mu(B\setminus A)=\mu(B)$ by the order of the ordered field of real numbers.

Claim 3. By claim 2, $\mu(A)\le\mu(B)<\infty$ and $\mu(B\setminus A)\le\mu(B)<\infty$, so both $\mu(A)$ and $\mu(B\setminus A)$ are real. The identity $\mu(B)=\mu(A)+\mu(B\setminus A)$ of claim 2 is then an identity between real numbers, and subtracting $\mu(A)$ gives $\mu(B\setminus A)=\mu(B)-\mu(A)$. If $\mu$ is finite, apply this with $B=X$, which is permitted since $\mu(X)<\infty$ and $A\subseteq X$.

Claim 4. Set $B_1=A_1$ and, for $m\ge2$,

$$B_m=A_m\setminus\bigcup_{i=1}^{m-1}A_i .$$

Each $B_m$ lies in $\mathcal{F}$, and $B_m\subseteq A_m$. They are pairwise disjoint: if $m<m'$ then $B_m\subseteq A_m\subseteq\bigcup_{i=1}^{m'-1}A_i$, while $B_{m'}$ is disjoint from that union. An induction on $k$ shows $\bigcup_{m=1}^{k}B_m=\bigcup_{m=1}^{k}A_m$: this is clear for $k=1$, and if it holds for $k$ then

$$\bigcup_{m=1}^{k+1}B_m=\Bigl(\bigcup_{m=1}^{k}A_m\Bigr)\cup\Bigl(A_{k+1}\setminus\bigcup_{m=1}^{k}A_m\Bigr)=\bigcup_{m=1}^{k+1}A_m .$$

Consequently $\bigcup_{m\in\mathbb{N}}B_m=\bigcup_{m\in\mathbb{N}}A_m$, and countable additivity gives

$$\mu\Bigl(\bigcup_{m\in\mathbb{N}}A_m\Bigr)=\sum_{m\in\mathbb{N}}\mu(B_m),$$

while $\mu(B_m)\le\mu(A_m)$ for every $m$ by claim 2.

If some $\mu(A_m)=\infty$, or if every $\mu(A_m)$ is real but the partial sums of $(\mu(A_m))$ are not bounded above, then $\sum_{m}\mu(A_m)=\infty$ and the asserted inequality holds. Otherwise every $\mu(A_m)$ is real with partial sums bounded above; then every $\mu(B_m)$ is real, and comparing the sums term by term, the $k$-th partial sum of $(\mu(B_m))$ is at most the $k$-th partial sum of $(\mu(A_m))$, hence at most $\sum_{m}\mu(A_m)$. Thus $\sum_{m}\mu(A_m)$ is an upper bound for the partial sums of $(\mu(B_m))$, and their least upper bound $\sum_{m}\mu(B_m)$ is at most $\sum_{m}\mu(A_m)$.

Claim 5. Set $C_1=A_1$ and $C_m=A_m\setminus A_{m-1}$ for $m\ge2$; these lie in $\mathcal{F}$. They are pairwise disjoint: if $m<m'$ then $C_m\subseteq A_m\subseteq A_{m'-1}$ by the nesting hypothesis, while $C_{m'}$ is disjoint from $A_{m'-1}$. An induction on $k$ shows $\bigcup_{m=1}^{k}C_m=A_k$: this is clear for $k=1$, and if it holds for $k$ then $\bigcup_{m=1}^{k+1}C_m=A_k\cup(A_{k+1}\setminus A_k)=A_{k+1}$, using $A_k\subseteq A_{k+1}$. Hence $\bigcup_{m\in\mathbb{N}}C_m=A$, and countable additivity gives $\mu(A)=\sum_{m\in\mathbb{N}}\mu(C_m)$, while by claim 1 the $k$-th partial sum of $(\mu(C_m))$ is

$$\sum_{m=1}^{k}\mu(C_m)=\mu\Bigl(\bigcup_{m=1}^{k}C_m\Bigr)=\mu(A_k).$$

Suppose first that every $\mu(A_m)$ is real and that $S=\{\mu(A_m):m\in\mathbb{N}\}$ is bounded above. Each $\mu(C_m)$ is then real, since $\mu(C_m)\le\mu(A_m)<\infty$ by claim 2, and the partial sums of $(\mu(C_m))$ are exactly the members of $S$, which are bounded above. By the definition of the sum of a sequence in $[0,\infty]$, $\mu(A)$ is the least upper bound $\sigma$ of $S$. The sequence $(\mu(A_k))_{k\in\mathbb{N}}$ is nondecreasing by claim 2 and the nesting hypothesis. Given a real $\varepsilon>0$, the number $\sigma-\varepsilon$ is not an upper bound of $S$, so there is $K\in\mathbb{N}$ with $\sigma-\varepsilon<\mu(A_K)$; for every $k\ge K$ we then have $\sigma-\varepsilon<\mu(A_K)\le\mu(A_k)\le\sigma$, whence $|\mu(A_k)-\sigma|<\varepsilon$. Thus $(\mu(A_k))$ converges to $\sigma=\mu(A)$ in the sense of Limit of a Sequence of Real Numbers.

Suppose instead that some $\mu(A_{m_0})=\infty$. Then $A_{m_0}\subseteq A$ and claim 2 give $\mu(A)=\infty$. Finally, suppose every $\mu(A_m)$ is real but $S$ is not bounded above. The partial sums of $(\mu(C_m))$ are the members of $S$ and are therefore not bounded above, so by the definition of the sum of a sequence in $[0,\infty]$ we again get $\mu(A)=\sum_{m}\mu(C_m)=\infty$.