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Recently I have been thinking about certain aspects of measure theory and its applications, and have the following question:

Does there exist a measure space $(\mu, \mathcal{A}, \mathbb{R})$, such that the corresponding Sigma-algebra $\mathcal{A}$ contains all countable, dense subsets of the real numbers and such that the measure satisfies the following two properties:

$1.)$ all infinite subsets have positive measure.

$2.)$ If $A,B\in \mathcal{A}$, $A\subset B$, and $B\setminus A$ is infinite, then $\mu(A)<\mu(B)$.

For instance, the measure of the rationals should be bigger than the measure of the dyadic rationals.

If the standard notion of measure does not suffice, can this be made to work for measures with values in the surreal numbers?

Added: In the partial answer by Severin Schraven it was indeed ruled out that the measures of some of such sets should be infinite, which leaves us to deal with measures taking values in the surreal numbers or Hardy fields.

Anixx
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    Why not take a bijection $f:\Bbb N\to X$ and consider $\mu(S)=\sum_{j\in S} 3^{-f^{-1}(j)}$ ? – Sassatelli Giulio Apr 20 '25 at 18:42
  • @SassatelliGiulio there can be absolutely huge number of different bijections from naturals to a dense set. – Anixx Apr 20 '25 at 18:46
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    Yes. So why not? – Sassatelli Giulio Apr 20 '25 at 18:49
  • @SassatelliGiulio your measure would depend on bijection. – Anixx Apr 20 '25 at 18:51
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    I don't understand: is there a requirement? – Sassatelli Giulio Apr 20 '25 at 19:02
  • @SassatelliGiulio measure is a function of set, not of set and bijection. – Anixx Apr 20 '25 at 19:43
  • Countable dense sets? "Dense" where? – jjagmath Apr 20 '25 at 19:50
  • @jjagmath https://en.wikipedia.org/wiki/Dense_set – Anixx Apr 20 '25 at 20:53
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    @Anixx jiagmath is asking where the set is dense. For example the rationals are dense in the reals, but not in the complex number. – Severin Schraven Apr 20 '25 at 21:00
  • @SeverinSchraven subsets of reals are meant. – Anixx Apr 20 '25 at 21:01
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    @Anixx While we are at it. What is your objection to Giulio's suggestion? You fix some bijection and get a measure. That gives you an example. – Severin Schraven Apr 20 '25 at 21:03
  • @SeverinSchraven measure is a function of a set. How do you fix bijection for all possible countable sets? I do not see a suggestion. What is his measure for rationals compared dyadic rationals or triadic rationals? If you define a measure anew for any new argument you encounter, it has no meaning, it is not a function or mapping. – Anixx Apr 20 '25 at 21:06
  • @Anixx Your question, as I (and seemingly Giulio) understand it, seems to be: Fix a countable set $X$ and find a measure $\mu$ on $X$ such that infinite subsets of $X$ have nonnegative measure and if $A\subseteq B\subseteq X$ and $B\setminus A$ has infinitely many elements, then $\mu(A)<\mu(B)$. If you work on a fixed countable set $X$, then you fix some bijection $f:\mathbb{N}\rightarrow X$ and use Giulio's construction to get a fixed measure. If you actually mean something else, then you should edit your question. – Severin Schraven Apr 20 '25 at 21:11
  • @SeverinSchraven no, I am asking for a mapping from countable dense subsets of reals to real numbers (or maybe surreals if necessary). – Anixx Apr 20 '25 at 21:18
  • @Anixx Then you should not phrase it like that. A measure takes elements in a Sigma-algebra and maps it to nonnegative real numbers. – Severin Schraven Apr 20 '25 at 21:20
  • @SeverinSchraven I am asking for a measure defined on all countable dense subsets of reals. A measure of their density. – Anixx Apr 20 '25 at 21:21
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    @Anixx Then write precisely what you mean in your question. It is not our job to guess what you mean. – Severin Schraven Apr 20 '25 at 21:22
  • @SeverinSchraven I added clarification. – Anixx Apr 20 '25 at 21:24
  • @Anixx Can you please check whether I have understood/stated your question correctly? If yes, your question might benefit from a more precise rephrasing. If you improve your question, it might get reopened and more (better) answers could get added. – Severin Schraven Apr 20 '25 at 21:53
  • @SeverinSchraven I am in the process, trying to understand. Thanks for your answer. – Anixx Apr 20 '25 at 21:54

1 Answers1

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Disclaimer: This is only a partial answer as I don't manage to rule out the case of measures with values in the surreal numbers (or the completion thereof).

Precise question:

The precise question (as per discussion in the comments): Does there exist a measure space $(\mu, \mathcal{A}, \mathbb{R})$, such that the corresponding Sigma-algebra $\mathcal{A}$ contains all countable, dense subsets of the real numbers and such that the measure satisfies the following two properties:

$1.)$ all infinite subsets have positive measure.

$2.)$ If $A,B\in \mathcal{A}$, $A\subset B$, and $B\setminus A$ is infinite, then $\mu(A)<\mu(B)$.

If needed, the OP would allow for measures with values in surreal numbers.

Proof of non-existence (real-valued measure): This proof works for measures in nonnegative numbers.

Such a measure cannot exist (if it takes values in the extended real line). Consider the sets $$S_x=x+\mathbb{Q}.$$ There exists an uncountable set $\Lambda\subseteq \mathbb{R}\setminus \mathbb{Q}$ such that $S_x\cap S_y=\emptyset$ for $x,y\in \Lambda$ and $x\neq y$.

By assumption, we have for $x\in \Lambda$ that $$\mu(S_x)>0.$$ Thus, there exists $\varepsilon>0$ and a countably infinite subset $\Lambda_1$ such that for all $x\in \Lambda_1$ holds $$\mu(S_x)>\varepsilon.$$ Now pick one element $x_0\in \Lambda_1$ and define the sets $$B=\bigcup_{x\in \Lambda_1} S_x, A=\bigcup_{x\in \Lambda\setminus \{x_0\}} S_x.$$ Both $A,B$ are countable, dense subsets of the reals and $\mu(A)=\infty=\mu(B)$ (as they are the disjoint union of infinitely many sets with measure $>\varepsilon$). However, $B\setminus A=S_{x_0}$ is infinite, thus, $\mu(A)$ would need to be strictly smaller than $\mu(B)$.

Added: It is not clear to me how to deal with measures with values in the surreal numbers (as I don't know anything about surreal numbers). It seems that they are not complete (Completion of surreal numbers), thus, one would probably need to complete them first. I would then try to show that for every $A\subset \mathbb{R}$ countable, dense, and contains an some singelton with infinite measure, we have $$\mu(A)=\sum_{x\in A}\mu(\{x\})=\sup\{ \mu(\{ x\}) \ : \ x\in A\}.$$ Ond could hope to run a similar argument as for the real case. However, that would require somebody who is better versed in surreal numbers than me.

Severin Schraven
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Shaun Apr 22 '25 at 09:41