How do we construct measures? In some simple cases, for example when considering discrete probability, we just specify a $\sigma$-algebra $\mathcal{A}$ and the value of $\mu$ on each set from $\mathcal{A}$. But in more complicated cases it is useful to specify $\mu$ on a smaller class of subsets. For example, recall how we constructed the Lebesgue measure: first, we defined it on intervals, then we extended it to the Lebesgue outer measure by covering an arbitrary subset by open intervals. The outer measure did not have the properties we wanted so we restricted it to the class of measurable sets. It turns out that this procedure works in a general case. In the next few lectures, we discuss the following procedure for defining a measure:
(4. We would also want that $\mathcal{A}$ contains the $\sigma$-algebra generated by $\mathcal{S}$ and agrees on it with the notion of measure we started with in point 1. Finally, we would want the resulting measure $\mu$ and $\sigma$-algebra to be unique with this property. This last point is more subtle and we return to it in the next lecture.)
The first step is somewhat involved so let us start with the second step, as it is directly analogous to earlier discussion of the Lebesgue outer measure.
Definition. Let X be a set. A function $\mu^*\colon P(X) \to [0,\infty]$ is an outer measure if
(Empty set) $\mu^*(\emptyset) = 0$
(Monotonicity) If $A\subset B$ then $\mu^(A) \leq \mu^(B)$
(Countably subadditivity) For any $A_1,A_2,\ldots \in P(X)$ we have
$\mu^(\bigcup_{n=1}^\infty A_n) \leq \sum_{n=1}^\infty \mu^(A_n)$
Rembember that we defined the Lebesgue outer measure by covering an arbitrary set with open intervals. This construction can be generalized as follows.
Proposition. Let $\mathcal{S} \subset P(X)$ be a set containing $\emptyset$ and $X$, and let $\rho \colon \mathcal{S} \to [0,\infty]$ be such that $\rho(\emptyset) = 0.$ The function $\mu^* \colon P(X) \to [0,\infty]$ defined by
$\mu^*(A) = \inf \{ \sum_{n=1}^\infty \rho(E_n) \ | \ E_n \in \mathcal{S} \text{ and } A \subset \bigcup_{n=1}^\infty E_n \}$
is an outer measure. (By convention, the infimum of the empty set is $\infty$, so the right-hand side is still well-defined even if $A$ cannot be covered by a countable collection of sets in $\mathcal{S}$)
The proof that $\mu^*$ is an outer measure is exactly the same as for the properties of the Lebesgue outer measure in Lecture 2. We leave it as an exercise verify the details.
Example. If $\mathcal{S}=\{\emptyset,X\}$, then $\mu^*$ takes the same value on all non-empty subsets of $X$.
Example. Let $X$ be a countable set (for simplicity), and let $\mathcal{S}$ consist of the empty set and all singletons, i.e. sets of the form $\{x\}$ for $x\in X$. A function $\rho \colon \mathcal{S} \to [0,\infty]$ satisfying $\rho(\emptyset) = 0$is simply a function on $X$ given by $x \mapsto \rho(\{x\})$ (by abuse of notation we will write simply $\rho(x)$ and consider $\rho$ as a function $\rho \colon X \to [0,\infty])$. The resulting outer measure $\mu^*$ is easily checked to be