5.1 Basic Assumptions about Area

5.3   Definition (Bounded Sets.) A subset $S$ of $\mbox{{\bf R}}^2$ is bounded if
$S\subset B(a,b\colon c,d)$ for some box $B(a,b\colon c,d)$. A subset of $\mbox{{\bf R}}^2$ that is not bounded is said to be unbounded.

It is clear that every subset of a bounded set is bounded. It is not difficult to show that if $B$ is a bounded set then $\mbox{{\bf a}}+ B$ is bounded for every $\mbox{{\bf a}}\in\mbox{{\bf R}}^2$, and $S(B)$ is bounded for every symmetry of the square, $S$.

5.4   Example. The set

$$\{ (n,{1\over n} ): n \in \mbox{{\bf Z}}^+ \}$$

is an unbounded subset of $\mbox{{\bf R}}^2$. I cannot draw a picture of an unbounded set, because the sheet of paper on which I make my drawing will represent a box containing any picture I draw.

5.5   Definition (Bounded Function.) Let $S$ be a set. A function $f:S \to \mbox{{\bf R}}$ is called a bounded function if there is a number $M$ such that $\vert f(x)\vert \leq M$ for all $x \in S$. It is clear that if $f$ is a bounded function on an interval $[a,b]$, then graph$(f)$ is a bounded subset of $\mbox{{\bf R}}^2$, since graph $(f) \subset B(a,b:-M,M).$ If $f$ is bounded on $S$, then any number $M$ satisfying

$$\vert f(x)\vert \leq M \mbox{ for all }x \in S$$

is called a bound for $f$ on $S$.

We are now ready to state our official assumptions about area. At this point you should officially forget everything you know about area. Unofficially, however, you remember everything you know so that you can evaluate whether the theorems we prove are reasonable. Our aim is not simply to calculate areas, but to see how our calculations are justified by our assumptions.

We will assume that there is a function $\alpha$ from the set of bounded subsets of $\mbox{{\bf R}}^2$ to the real numbers that satisfies the conditions of positivity, additivity, normalization, translation invariance and symmetry invariance described below. Any function $\alpha$ that satisfies these conditions will be called an area function.

5.6   Assumption (Positivity of area.)

$$\alpha (S)\geq 0 \mbox{ for every bounded subset } S \mbox{ of } \mbox{{\bf R}}^2 .$$

5.7   Definition (Disjoint sets.) We say two sets $S,T$ are disjoint if and only if $S\cap T=\emptyset$.

5.8   Assumption (Additivity of area.) If $S,T$ are disjoint bounded subsets of $\mbox{{\bf R}}^2$, then

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

5.9   Assumption (Normalization property of area.) For every box
$B(a,b\colon c,d)$ we have

$$\alpha \Big(B(a,b\colon c,d)\Big)=(b-a)(d-c),$$

i.e., the area of a box is the product of the length and the width of the box.

5.10   Assumption (Translation invariance of area.) Let $S$ be a bounded set in $\mbox{{\bf R}}^2$, and let $\mbox{{\bf a}}\in\mbox{{\bf R}}^2$, then

$$\alpha (S)=\alpha (\mbox{{\bf a}}+ S).$$

5.11   Assumption (Invariance under symmetry.) Let $S$ be a bounded subset of $\mbox{{\bf R}}^2$. Then if $F$ is any symmetry of the square

$$\alpha (F(S))=\alpha (S).$$

(See definition 4.12 for the definition of symmetry of the square.)

Remark: I would like to replace the assumptions 5.10 and 5.11 by the single stronger assumption:

If $A$ and $B$ are bounded subsets of $\mbox{{\bf R}}^2$, and $A$ is congruent to $B$, then $\alpha(A) = \alpha(B)$.

However the problem of defining what congruent means is rather complicated, and I do not want to consider it at this point.

5.12   Entertainment (Congruence problem.) Formulate a definition of what it means for two subsets of $\mbox{{\bf R}}^2$ to be congruent.

5.13   Example. Let

$$\begin{eqnarray*} S &=& B(0,1:0,1) \cap \{(x,y) \in \mbox{{\bf R}}^2: x \in \mb... ...) \cap \{(x,y) \in \mbox{{\bf R}}^2: x \not\in \mbox{{\bf Q}}\}. \end{eqnarray*}$$

I do not know how to make any reasonable drawing of $S$ or $T$. Any picture I draw of $S$ would look just like a picture of $T$, even though the two sets are disjoint. By additivity and the normalization property for area

$$\alpha(S) + \alpha(T) = \alpha(S \cup T) = \alpha(B(0,1:0,1)) = 1.$$

Since areas are non-negative, it follows that

$$0 \leq \alpha(S) \leq 1 \mbox{ and }0 \leq \alpha(T) \leq 1.$$

The problem of calculating $\alpha(S)$ exactly cannot be answered on the basis of the assumptions we have made.

Remarks: The assumptions we have just made are supposed to be intuitively plausible. When we choose to make a particular set of assumptions, we hope that the assumptions are consistent, i.e., that no contradictions follow from them. If we were to add a new assumption:

$$\mbox{ The area of a circle with radius $1$ is $3.14159$,}$$

then we would have an inconsistent set of assumptions, because it follows from the assumptions we have already made that the area of a circle of radius $1$ is greater than $3.141592$.

In 1923 Stefan Banach(1892-1945) [5] showed that area functions exist, i.e., that the assumptions we have made about area are consistent. Unfortunately Banach showed that there is more than one area function, and different area functions assign different values to the set $S$ described in the previous example.

A remarkable result of Felix Hausdorff(1868-1942) [24, pp469-472] shows that the analogous assumptions for volume in three dimensional space (if we include the assumption that any two congruent sets in 3 dimensional space $\mbox{{\bf R}}^3$ have the same volume) are inconsistent. If one wants to discuss volume in $\mbox{{\bf R}}^3$ then one cannot consider volumes of arbitrary sets. One must considerably restrict the class of sets that have volumes. A discussion of Hausdorff's result can be found in [20].