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Fix an infinite number $\omega$, and a finite number $n$. Let $$ \Omega = \left\{\frac{k}{\omega}: k=1,2,\ldots,n\right\}, $$ then $\Omega$ is a subset of the infinitesimal in $[0,1]$.
Is it possible to construct the Loeb measure associated to $$ X = (\Omega,\sigma(\Omega),\text{ counting measure}), $$ where $\sigma(\Omega)=2^\Omega$ is the power set of $\Omega$.
That is $$ \text{$X$ is a uniform random variable taking value in $\Omega$}. $$

From this question, I should look for a first order description of $\sigma(\Omega)$.
But I don't see such description.

Community
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    Perhaps also say what you mean by $\sigma(\Omega)$. – GEdgar Apr 22 '13 at 14:40
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    Any finite object you write down is "first order" by just stringing together all its elements. – Mikhail Katz Apr 22 '13 at 15:21
  • @user72694 How can I string finite numbers to get say $A={\frac{1}{\omega},\frac{2}{\omega}$}? My thinking is $\mathrm{st}(A)={0,0}$, so I don't see what to string. –  Apr 22 '13 at 15:33
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    Remember that $\Omega$ is still an infinite set. Hyper reals can make the infinite seem finite, but it requires, tweaking, in this instant. (You may need to construct some sort of extension of the counting measure to hyper-reals.) – Christopher King May 22 '13 at 00:35

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The answer to the question in the title is no. A Loeb measure is an extension of the standard part of an internal measure defined on an internal measure space. This means that you can't construct any sort of Loeb measure on the set of positive infinitesimals because that set is external.

The answer to the question in the body is yes. Since $n \in \mathbb{N}$, the set $\Omega = \{1/\omega, 2/\omega, \ldots, n/\omega\}$ is finite. An explicit bijection with a finite set is $k\mapsto k/\omega$ for $k = 1,\ldots, n$. Since $2^\Omega$ is already a $\sigma$-algebra (since it's a finite algebra) and the counting measure already takes values in $\mathbb{R}$, the Loeb construction leaves the probability space $(\Omega, 2^\Omega, \text{counting measure})$ unchanged.

Markus Vasquez
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