Loeb measure

Remarks

The map $\operatorname{st}\mu$ is finitely additive on $\mathcal{A}$. By countable saturation, if a countable union of sets of $\mathcal{A}$ lies in $\mathcal{A}$, then it is already a finite union, so $\operatorname{st}\mu$ is countably additive on $\mathcal{A}$. Hence Carathéodory's extension theorem gives the extension, and it is unique because $\operatorname{st}\mu(\Omega)$ is finite. Many authors call the completion of this measure the Loeb measure.

Example: let $N$ be an infinite hypernatural number, let $\Omega = \set{1, \ldots, N}$, and let $\mathcal{A}$ be the internal power set of $\Omega$. Let $\mu(A) = \abs{A}/N$, where $\abs{A}$ is the internal number of elements of $A$. Then $\mu(\Omega) = 1$ is finite, and $L_\mu$ is a probability measure. The standard part map $k \mapsto \operatorname{st}(k/N)$ pushes $L_\mu$ forward to Lebesgue measure on $[0, 1]$.

Non-example: on the same $\Omega$, the internal counting measure $\mu(A) = \abs{A}$ has $\mu(\Omega) = N$, which is infinite. It fails the finiteness clause, and the construction above does not apply directly. For such $\mu$ one can still define a Loeb measure, through an outer measure, but it need not be the unique extension.