6
$\begingroup$

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?
Improve this question
$\endgroup$
5
  • $\begingroup$ Why the votes to close? I certainly have not come across any discussion of hyperreal numbers when reading about measurable cardinals, and I don't understand why this would be off-topic. $\endgroup$
    – Mike Battaglia
    Commented Sep 12, 2015 at 23:14
  • 1
    $\begingroup$ Regarding the new question, the countably additivity requirement is problematic for $\mathbb{R}^{**}$-valued measures, since the field itself does not have the LUB property, and so one cannot just take the limit of the finite partial sums. Thus, it will be difficult in this context to assign a value to many infinite sums. $\endgroup$
    – Joel David Hamkins
    Commented Sep 13, 2015 at 8:18
  • $\begingroup$ I thought that using the larger field would fix that problem. Can you give an example of what you mean? $\endgroup$
    – Mike Battaglia
    Commented Sep 13, 2015 at 8:22
  • $\begingroup$ In general, sums such as $\sum_n a_n$ do not necessarily make sense in these fields, even when $a_n\leq \frac1{2^n}$, say. So it isn't clear what you would mean by countable additivity, if you had disjoints sets of measure $a_n$. $\endgroup$
    – Joel David Hamkins
    Commented Sep 13, 2015 at 8:29
  • $\begingroup$ Yes, I see. Well, that's frustrating. Is there some other additivity property that is somehow analogous for larger real-closed fields? This whole thing started because I was wondering about surreal-valued metrics on the surreals, but now I'm a bit frustrated at how "clunky" ordinary measure theory seems to be when you try to generalize anything. $\endgroup$
    – Mike Battaglia
    Commented Sep 13, 2015 at 8:36

2 Answers 2

Reset to default
3
$\begingroup$

You say that $\kappa$ admits a $[0,1]$-valued measure, but actually, it is a $\{0,1\}$-valued measure; every subset of $\kappa$ has measure $0$ or $1$. If I am understanding your suggestion, you intend to measure subsets of the hyperreal unit interval $[0,1]$, which has size $\kappa$, by using that measure. Thus, every subset will continue to have measure either $0$ or $1$, and such a measure is never translation invariant. For example, if we cut the interval in half, then one half will have measure $1$ and the other will have measure $0$.

Perhaps your proposal is more sensible with real-valued measurable cardinals.

Improve this answer
$\endgroup$
9
  • $\begingroup$ My understanding is that every measurable cardinal is also real-valued measurable, which is what I was driving at. Is that wrong? $\endgroup$
    – Mike Battaglia
    Commented Sep 12, 2015 at 23:22
  • 1
    $\begingroup$ Yes, that is correct, using a 2-valued measure you can construct such measures. For example, give yourself $\omega$ many sets of size $\kappa$, and decree that the $n^{th}$ one has value $2^{-n}$, using the original 2-valued measure to measure subsets; now measure subsets of the disjoint union by adding up. $\endgroup$
    – Joel David Hamkins
    Commented Sep 12, 2015 at 23:37
  • 1
    $\begingroup$ Such a measure has the property, however, that there are sets of measure $\frac12$, such that all subsets of it have measure either $0$ or $\frac12$, and this will not be true of any translation invariant measure of the kind you seek. $\endgroup$
    – Joel David Hamkins
    Commented Sep 12, 2015 at 23:39
  • 1
    $\begingroup$ The point of my question is to note that every measurable cardinal is also real-measurable. That is, in addition to the usual 2-valued measure, it also admits a real-valued probability measure. I am curious if there exists any such choice of real-valued measure and bijection onto the hyperreal unit interval which is translation invariant. Obviously the {0,1}-valued measure and the one I linked to above won't do the trick, but I'm asking if there exists some other measure which does. $\endgroup$
    – Mike Battaglia
    Commented Sep 12, 2015 at 23:43
  • 1
    $\begingroup$ I would suggest that you ignore the close votes and edit your question, since it has now become a little clearer what you want. (But I would add that the 2-valued measure IS a probability measure, so saying that "it also admits a real-valued probability measure" is not adding much.) $\endgroup$
    – Joel David Hamkins
    Commented Sep 12, 2015 at 23:50
3
$\begingroup$

This is NOT an answer but a question related to the original question. Take $R^*=R$, the usual reals, then $R^{**}$ is $R^*$, a non-Archimedean extension. By your 3 and 4, the measure of the whole of $R$ is a sum of infinitely many ($\omega$) values $1$. How this can be a value in $R^*$?

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .