Suppose $\kappa$ is a measurable cardinal. Then $\kappa$ also admits a $[0,1]$-valued measure.

Now let's construct a hyperreal field of cardinality $\kappa$, and look at the closed hyperreal interval $[0,1]^*$. Then the $[0,1]$ valued measure on $\kappa$ can be extended to this hyperreal interval.

Is there a way to do this which is translation-invariant, and furthermore which resembles the Lebesgue measure on the ordinary real unit interval, but with the property that every subset is measurable?

I'm leaving the notion of "resembles" deliberately vague here, as I feel there might be more than one satisfactory way to do this that's in the spirit of my question.

I'm also interested in the answer in which a hyperreal $[0,1]^*$-valued measure is permitted, if that makes anything easier.

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:

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

Now, here are the questions:

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