B. Proofs of Some Area Theorems

B.1   Theorem (Addition Theorem.) For any bounded sets $S$ and $T$ in $\mbox{{\bf R}}^2$

$$\alpha (S\cup T)=\alpha (S)+\alpha (T)-\alpha (S\cap T).$$ (B.2)

and consequently

$$\alpha (S\cup T)\leq\alpha (S)+\alpha (T).$$

Proof: We have

$$S\cup T=S\cup (T\setminus S) \mbox{ where } S\cap (T\setminus S)=\emptyset$$

and

$$T=(T\setminus S)\cup (T\cap S) \mbox{ where } (T\setminus S)\cap (T\cap S)=\emptyset .$$

Hence by the additivity of area

$$\alpha (S\cup T)=\alpha (S)+\alpha (T\setminus S)$$ (B.3)

and

$$\alpha (T)=\alpha (T\setminus S)+\alpha (T\cap S)$$ (B.4)

If we solve equation (B.4) for $\alpha (T\setminus S)$ and use this result in equation (B.3) we get the desired result. $\diamondsuit$

B.5   Corollary (Subadditivity of area.) Let $n \in \mbox{{\bf Z}}_{\geq 1}$, and let $A_1$, $A_2$, $\cdots$ $A_n$ be bounded sets in $\mbox{{\bf R}}^2$. Then

$$\alpha( \bigcup_{i=1}^n A_i) \leq \sum_{i=1}^n \alpha(A_i).$$ (B.6)

Proof: The proof is by induction. If $n=1$, then (B.6) says $\alpha(A_1) \leq \alpha(A_1)$, which is true. Suppose now that $k$ is a generic element of $\mbox{{\bf Z}}_{\geq 1}$, and that (B.6) is true when $n=k$. Let $A_1, \cdots, A_{k+1}$ be bounded sets in $\mbox{{\bf R}}^2$. Then

$$\begin{eqnarray*} \alpha(\bigcup_{i=1}^{k+1} A_i) &=& \alpha\left( \bigcup_{i=1}... ...^k \alpha(A_i) + \alpha(A_{k+1}) = \sum_{i=1}^{k+1} \alpha(A_i). \end{eqnarray*}$$

Hence (B.6) is true when $n = k+1$, and by induction the formula holds for all $n \in \mbox{{\bf Z}}_{n\geq k}.\mbox{ $\diamondsuit$}$

B.7   Theorem (Monotonicity of Area.) Let $S,T$ be bounded sets such that $S \subset T$. Then $\alpha (S)\leq\alpha (T)$.

Proof: If $S \subset T$ then $S\cap T=S$, and in this case equation (B.4) becomes

$$\alpha (T)=\alpha (T\setminus S)+\alpha (S).$$

Since $\alpha (T\setminus S)\geq 0$, it follows that $\alpha (T)\geq\alpha (S)$. $\diamondsuit$

B.8   Theorem (Additivity for almost disjoint sets.) Let $\{R_1,\cdots ,R_n\}$ be a finite set of bounded sets such that $R_i$ and $R_j$ are almost disjoint whenever $i\neq j$. Then

$$\alpha (\bigcup_{i=1}^n R_i)=\sum_{i=1}^n \alpha (R_i).$$ (B.9)

Proof: The proof is by induction on $n$. For $n=1$, equation (B.9) says that $\alpha (R_1)=\alpha (R_1)$, and this is true. Now suppose $\{ R_1\cdots R_{n+1}\}$ is a family of mutually almost-disjoint sets. Then

$$(R_1\cup\cdots\cup R_n)\cap R_{n+1}=(R_1\cap R_{n+1})\cup (R_2\cap R_{n+1})\cup\cdots\cup (R_n\cap R_{n+1})$$

and this is a finite union of zero-area sets, and hence is a zero-area set. Hence, by the addition theorem,

$$\alpha ((R_1\cup\cdots\cup R_n)\cup R_{n+1})=\alpha (R_1\cup\cdots\cup R_n)+\alpha (R_{n+1})$$

i.e.,

$$\alpha (\bigcup_{i=1}^{n+1}R_i)=\sum_{i=1}^n\alpha (R_i)+\alpha (R_{n+1})=\sum_{i=1}^{n+1}\alpha (R_i).$$

The theorem now follows from the induction principle. $\diamondsuit$