I understand the motivation behind measurable cardinals is to ask the question: "is there any set large enough to admit a non-trivial measure on all of its subsets?"

Hence, it's also worthwhile to ask the question "is there any real-closed field large enough to admit a non-trivial 'Lebesgue' measure on all of its subsets?" Here's one way to formulate this question precisely:

First, let $\Bbb R^*$ be our real-closed field, noting that it can be non-Archimedean.

Second, let's generalize our notion of "measure." Given some $\Bbb R^*$, let $\Bbb R^{**}$ be a strictly larger real-closed field. Then we will allow our measure on $\Bbb R^*$ to take values in $\Bbb R^{**}$.

Now, we want to find real-closed fields $\Bbb R^*$ and $\Bbb R^{**}$, and a function $\mu: 2^{\Bbb R^*} \to \Bbb R^{**}$ which obeys the following:

- For all $S \subseteq \Bbb R^*$, $\mu(S) \geq 0$.
- $\mu(\emptyset) = 0$.
- Countable additivity of pairwise disjoint sets.
- $\mu([0,1]^*) = 1^{**}$.
- $\mu$ is a complete measure.
- $\mu$ is translation-invariant.

Now, here are the questions:

**Can this measure space exist?**- If so, then what relationship does this measure space have with the concept of a measurable cardinal?