03/10/24
Reference: Section 2.1 from lecture notes
If $I=[a,b]$ is an interval in $\R$ its length is $l(I) = b-a$, similarly for open intervals. We would like to extend this notion to all subsets of $\R$. For this, it is convenient to introduce the following.
Let $X$ be a set. The power set $P(X)$ is a set whose elements are subsets of $X$.
Example. $X = \{ 1, 2, 3 \}$, $P(X) = \{ \emptyset, \{1\}, \{2\}, \{3\}, \{1, 2 \}, \{2, 3\}, \{3,1\}, \{1,2,3\} \}$. In general, if $X$ has $n$ elements, then $P(X)$ has $2^n$ elements.
A set is countable if it is finite or in bijection with $\N$. Otherwise we say it is uncountable.
Example. $X=\N$, $P(X)$ is uncountable, in fact in bijection with $\R$ which can be seen by interpreting a subset of $\N$ as a sequence of $0$ and $1$’s and associating to it a real number in its binary expansion.
Let $[0,\infty] = [0,\infty)\cup\{\infty\}$. To extend length, we would like to have a function
$\mu \colon P(\R) \to [0,\infty]$
which satisfies the following natural properties
(Intervals) $\mu(I) = l(I)$ for all intervals.
(Translation invariance) $\mu(A+c) = \mu(A)$ for any $A \in P(\R)$ and $c \in \R$, where $A + c = \{ a + c : a \in A, \ c \in \R \}$.
(Countable additivity) If $A_1, A_2, \ldots, \in P(\R)$ are pairwise disjoint, i.e. $A_i \cap A_j = \emptyset$ for $i\neq j$, then
$\mu(\bigcup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \mu(A_i)$
(We interpret this as follows. If the sum on right-hand side converges, then the left-hand side is equal to its sum. If the sum diverges, then the left-hand side is $\infty$.)
Remark. We could ask for a similar function $\mu \colon P(\N) \to [0,\infty]$ measuring the size of subsets of $\N$. Of course, such a function exists, namely $\mu(A) = \#A$, the number of elements.
Unfortunately, this is too much to ask for. We will show that such a map $\mu$ does not exist, at least in the standard axioms of set theory. But we will learn something from understanding this failure.
Remark. We could relax the last condition to deal only with finite sums. This is an interesting question. It turns out that in that case such $\mu$ exists, but it is not unique and the construction is not very enlightening. However, we really want to deal with countable sums because in analysis we want to take limits of infinite sequences, so we will stick to the countably additivity.