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The counting measure on $\mathbb{R}^n$ is a map that takes a subset $A$ of $\mathbb{R}^n$ and returns its cardinality if it is finite or the symbol $\infty$ if it is infinite.

So, if $A\subseteq\mathbb{R}^n$ is countably infinite and $B\subseteq\mathbb{R}^n$ is uncountable, then the two sets have different cardinality but the same counting measure, which is a bit odd.

We could solve this "problem" (if one considers this a problem) by redefining the counting measure to be just the cardinality, i.e. for finite sets nothing changes, but infinite sets are mapped to cardinal numbers. But the problem is that this "measure" is no longer a $[0,\infty]$-valued map, but the codomain would include infinite cardinal numbers.

Okay, now turn to a more interesting measure, namely the $1$-dimensional Hausdorff-measure on $\mathbb{R}^2$. This measure assigns to a measurable set the length of this set. So, if the set is a rectifyable curve, it gets its usual length, and a straight line gets the symbol $\infty$. Now, as in the counting measure case, there are different versions of set of infinite measure. A line in the plane is of infinite length, but it is a countable union of segments of finite length. The same is true for a countable union of lines. BUT the whole plane is in a way "more" infinite, as it is not a countable union of segments of finite length. So, one could argue that there are different sizes of infinite measure in this case as well.

So, one could say that the $1$-dimensional Hausdorff-measure of a straight line is $\aleph_0$, while the Hausdorff-measure of the whole plane is the continuum (even though both sets contain the same number of points).

So, one could extend the idea of a $[0,\infty]$-valued $1$-dimensional Hausdorff-measure to a "measure" that takes non-negative real-numbers and cardinal numbers as values.

The same should work for all $d$-dimensional Hausdorff-measures on arbitrary metric spaces and I can finally come to my question:

Question Part 1) Is a measure taking non-negative real-numbers as well as infinite cardinal numbers a useful thing to consider or is there a good reason why one should not do that?

Question Part 2) Has this been considered before?

Thanks to all who read this text up this point! :-)

Edit: Maybe, to make it more clear what I mean. My idea was along the lines that the measure of set should be $\leq \kappa$ if the set can be written as the union of $\kappa$-many sets of finite measure. But I already can see a few problems with this approach: The whole idea of a measure is build on countability, uncountable union of null-sets need not be null-sets and so on, that probably this could somehow lead to problems?

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For the Hausdorff measure this is basically solved by its intrinsic dimension, i.e. you need two numbers to decide the "size of a set": its Hausdorff dimension $s$ and the $s$-dimensional Hausdorff measure of the set. So if I understand correctly, your question is basically "Can we do the same with just one number?", correct?

Not sure what counts as "one number", but you could do this with a construction similar to the "long line": For only integer dimensions and arbitrarily dimensions set $$\mathbb{L} = \mathbb{N}\times [0,\infty[$$ and equip it with the order topology from the lexicographic order.

For subsets in $\mathbb{R}^d$ and integer dimensions you can use $\{0,1,2,\dots,d\}\times [0,\infty[$ and for fractional dimensions $[0,d]\times[0,\infty[$ (with the order topology for the lexicographic order) would work.

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  • $\begingroup$ If $\infty$ is excluded, what measure would you assign to $\mathbb R^d$ itself? And how would this solution distinguish the counting measure of a countably from an uncountably infinite set when $d = 1$, especially if there are cardinalities between $\aleph_0$ and the continuum? $\endgroup$ Commented 9 hours ago
  • $\begingroup$ @Dirk: Thanks for the answer. It is however not directly the generalization I am looking for. If $\kappa$ is a cardinal between $\aleph_0$ and the continuum (assume that those cardinals exist), then the disjoint union of $\kappa$ many lines should have (generalized) 1dim Hausorff-measure $\kappa$ ; but I guess the Hausdorff-dimension should be $1$, wouldn't it? $\endgroup$ Commented 9 hours ago

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