Counting Measure

Define $n(\emptyset) = 0$ . When $1..n \sim A$ define $n(A) = n$ . This is a consistent definition since if $1..n \sim A$ and $1..m \sim A$ then $1..n \sim 1..m$ and so there is an isomorphism $\psi : 1..n \rightarrow 1..m$ . If $n > m$ then by the pigeonhole principle two numbers are assigned the same number, which is not possible and therefore $n \le m$ . Dually $m \le n$ . Hence $n = m$ . When $A$ is an infinite set, define $n(A) = \infty$ .

lemma: when $A_1,\ldots,A_n$ is a list of disjoint finite sets, then $n(\cup_{i=1..n} A_i) = \sum_{i = 1..n} n(A_i)$ .

The statement is true when $n = 1$ . Assume it to be true for a given $n$ and let $A_1,\ldots, A_n,A_{n+1}$ be a sequence of disjoint finite sets. Let $n(\cup_{i = 1..n} A_i) = m, n(A_{n+1}) = q$ . Then there are isomorphisms $\phi : 1..m \rightarrow \cup_{i = 1..n} A_i$ and $\psi : 1..q \rightarrow A_{n+1}$ . Define $\psi' : m+1,\ldots,m+q \rightarrow 1..q$ by $\psi'(m+i) = \psi(i)$ . Then $\phi \vee \psi' : 1..m+q \rightarrow \cup_{i = 1..n+1} A_i$ is an isomorphism and so $n(\cup_{i=1..n+1} A_i) = m + q = \sum_{i=1..n} n(A_i) + n(A_{n+1})$ .

lemma: when $A \subseteq B$ then $n(A) \le n(B)$ .

If $A$ is infinite, then $B$ is infinite and $n(A) = n(B) = \infty$ . If $B$ is infinite, then $n(A) \le \infty = n(B)$ . When both $A,B$ are finite then by prior result, $n(B) = n(A) + n(B-A) \ge n(A)$ .

lemma: If $\mathcal{S}$ is a collection of sets, then $n$ is a measure on $2^{\mathcal{S}}$ .

It is a non-negative function such that $n(\emptyset) = 0$ . To establish this as measure countable additivity has to be verified. Let $\left< A_i \right>$ be a sequence of disjoint subsets of $\mathbb{N}$ .

Either $n(A_i) < \infty$ for all $i$ or there is an $i$ such that $n(A_i) = \infty$ .

If there is an $i$ such that $n(A_i) = \infty$ then $n(\cup A_i) = \infty = \sum_i n(A_i)$ since a set that contains an infinite set is infinite.

Suppose $n(A_i) < \infty$ for all $i$ . Either $n(\cup_i A_i) = \infty$ or $n(\cup_i A_i) < \infty$ . In the first case, an infinite number of sets $A_i$ are nonempty and so $\sum_i n(A_i) =\infty$ . In the second case, all but finitely many sets are nonempty and $\sum_i n(A_i) = \sum_{i : A_i \neq \emptyset} n(A_i) = n(\cup_{i : A_i \neq \emptyset} A_i) = n(\cup_i A_i)$ .

Navigation

About

Raedwulf ….