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Many of us are familiar with the transfinite numbers as representing different levels of infinity. I was wondering if there were a similar system for categorizing null quantities? My motivation for this question is based on the following:

  1. $x \equiv \emptyset \implies x\subset \{0\}$, so in a set theoretic sense, $\emptyset$ is more "zero" than $0$.
  2. Also, if $m([a,b])=Leb([a,b]) \implies m(\emptyset)=m(0)=m(a)=0, \forall a$ so in a measure theoretic sense, null sets are a superset of both $\emptyset$ and $0$.

I understand that this a a bit vaguely defined, hence why I am asking if there is a rigorous way to categorize nulls just like we do infinities? It seems like there might if one were clever enough.

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As you say, your question is a bit muddled up and you confused different notions of size (the set theoretic notion of cardinality is very different than the measure theoretic notion of length). None the less, there is at least one way in which different sizes of infinities correspond to different sizes of "nullities". Extensions of the reals exist in which infinitesimals and infinitely large numbers are part of the system. Such systems are called nonstandard models of the reals, and they still form a field. Now, in any such field a number $t$ is infinitely large if, and only if, $1/t$ is infinitesimal (and vice versa). Moreover, the usual ordering of the reals extends to the nonstandard reals so you can compare infinities and infinitesimals and it is then the case that $t\mapsto 1/t$ is order reversing. This may be what you are looking for or wondering about.

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  • $\begingroup$ Thanks Ittay. Yes, I know I was mashing together a bunch of different concepts. I felt that would be helpful to provide my motivation to would-be answerers, even if it makes me look like a pretty second-rate mathematician - SE is about answers, not egos. I think you've provided a nice answer to this. I think the infinitesimal concept from non-standard analysis is the rigorous notion behind my befuddled post. Thanks! $\endgroup$
    – user237392
    Commented Sep 21, 2015 at 3:36
  • $\begingroup$ Your question is just fine! I did not mean to imply anything else. Muddled up questions are very often very good questions. And yes, SE is very much about elucidating ideas through questions and answers. Certainly a (+1) from me for you question. $\endgroup$ Commented Sep 21, 2015 at 3:38
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EDIT: I think this is based on a misreading of the question - I interpreted the question as asking for ways of comparing null measure sets. I'm leaving it up since I think it might still be valuable to the OP.


There are, as it turns out, many different kinds of null-ness.

Recall that a set $X\subseteq\mathbb{R}$ is null if, for any $\epsilon>0$, there is a cover of $X$ by open intervals the sum of whose lengths is $<\epsilon$. We can refine this by demanding covers whose intervals decrease in size rapidly:

$X\subseteq\mathbb{R}$ is strong measure zero if, for any sequence of positive reals $(\epsilon_n)_{n\in\mathbb{N}}$, there is a cover $\{I_n: n\in\mathbb{N}\}$ of $X$ by open intervals with $I_n$ having measure $<\epsilon_n$.

This is a really strong condition - it is consistent with ZFC that the only sets with strong measure zero are countable!

Another variation on null-ness:

$X\subseteq\mathbb{R}$ is microscopic if for every $\epsilon>0$, there is a cover $\{I_n: n\in\mathbb{N}\}$ of $X$ where the measure of each $I_n$ is $<\epsilon^n$.

And, more generally, we can consider a notion of measure zero-ness for any family of fast-growing functions. See http://www.sav.sk/journals/uploads/0721132912Horbac.pdf.

Finally, here's a way to measure (hehe) "how null" a null set is:

For $X\subseteq\mathbb{R}$ null, let $scope(X)=\{f: \mathbb{N}\rightarrow\mathbb{R}_{>0}: \exists (I_n: n\in\mathbb{N})[X\subseteq\bigcup I_n, \vert I_n\vert<f(n)]\}.$

I just made up the term "scope," I have no idea how it's actually referred to. This definition leads to a natural preordering on null sets:

For $X, Y\subseteq\mathbb{R}$ null, say $X\le_{null}Y$ if $scope(X)\subseteq scope(Y)$.

I don't know anything about this preorder, but it might be interesting.

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  • $\begingroup$ Wow...there's definitely a non-null amount of information here (sorry...I JUST couldn't help myself...never ignore a bad pun!) ;-) Yes, this is good stuff. It's not directly related to the question, but I use probability theory a bit so these kinds of things interest me. Thanks for sharing (+1)! $\endgroup$
    – user237392
    Commented Sep 21, 2015 at 3:50

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