Here, "measure" always means Lebesgue measure on $\mathbb{R}$. This question is partly motivated by my answer https://math.stackexchange.com/questions/1444498/is-there-a-categorizaiton-system-for-null-quantities/1444524#1444524.

Some measure zero sets are more measure zero than others:

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

  • A set $X\subseteq\mathbb{R}$ is microscopic if for any sequence $(\epsilon_i)_{i\in\mathbb{N}}$ of positive reals where for some positive real $\delta$ we have $\epsilon_i=\delta^{i+1}$, there is a cover $\{I_n: n\in\mathbb{N}\}$ of $X$ by intervals with $\mu(I_n)<\epsilon_n$.

  • Etc. (See e.g. http://www.sav.sk/journals/uploads/0721132912Horbac.pdf.)

More generally, if $F$ is a family of functions from $\mathbb{N}$ to $\mathbb{R}_{>0}$, say $X\subseteq\mathbb{R}$ is $F$-microscopic if for every $f\in F$, there is a cover $\{I_n: n\in\mathbb{N}\}$ of $X$ by open intervals such that $\mu(I_n)<f(n)$. To measure how null a given null set is, we can look at its scope (I'm not sure what this is actually called, I couldn't find a reference to it): $scope(X)=\{f: \mathbb{N}\rightarrow\mathbb{R}_{>0}: X\text{ is $\{f\}$-microscopic}\}.$ This leads to a natural preorder on the set of null subsets of $\mathbb{R}$: $$X\le_{null}Y\iff scope(X)\subseteq scope(Y).$$

My question is:

What does the resulting degree structure $\mathfrak{N}=(Null/\equiv_{null}, \le_{null})$ look like?

I'm particularly interested in the extent to which set-theoretic hypotheses such as large cardinals or forcing axioms are relevant; based on the fact that even when just studying strong measure zero sets, set-theoretic hypotheses become important, I suspect there is in fact some relevance.

Comment 1: Arguably my definition of "scope" is wrong, and we should instead look at something slightly more well-behaved, like $\{f: X\text{ is $\{\alpha f: \alpha\in\mathbb{R}_{>0}\}$-microscopic}\}$ or similarly. If tweaking the definition of scope would lead to a better result, feel free to do so.

Comment 2: Of course we can work in much more generality than $\mathbb{R}$ with Lebesgue measure, but already this case seems really interesting.

EDIT: It seems like a good first step would be to try to understand forcings which add sets of a prescribed scope. Ideally, this could be used to prove e.g. that it's consistent that there are null sets with incomparable scopes.

The following forcing notion might be a good first try. For $F$ a set of functions from $\mathbb{N}$ to $\mathbb{R}_{>0}$, let $\mathbb{P}_F$ be the set of ordered pairs $(D, C)$ where

  • $D$ is a countable set of reals, and

  • $C$ is a finite set of infinite families of intervals $\mathcal{C}_{f_0}, . . . ,\mathcal{C}_{f_n}$ where $f_i\in F$ and each $\mathcal{C}_{f_i}=\{I_n^{f_i}: n\in\mathbb{N}\}$ satisfies $\mu(I_n^{f_i})<f_i(n)$, such that

  • $D\subseteq \bigcap_{\mathcal{C}_{f_i}\in C} (\bigcup_{n\in\mathbb{N}} I_n^{f_i})$, and

  • the set $\bigcap_{\mathcal{C}_{f_i}\in C} (\bigcup_{n\in\mathbb{N}} I_n^{f_i})$ has positive measure (to prevent the forcing from being trivial).

ordered by $(D, C)\le (D', C')$ if $D\supseteq D'$ and $C\supseteq C'$. Forcing with $\mathbb{P}_F$ yields a set of reals which is uncountable, and is $F$-microscopic. It also, unfortunately, does a fair bit of damage to the ground reals. A countably closed forcing would be nicer, but it's not clear to me how to make that work without accidentally building a set of strong measure zero.

  • 3
    $\begingroup$ You might want to look at Borel's "rarefaction" classification of measure zero sets. See Winfried Just and Claude Laflamme's 1990 Trans. AMS paper Classifying sets of measure zero with respect to their open covers. Laflamme wrote at least two more papers related to this, and you can find earlier work by Frechet, Zenon Moszner, Léonard Urbanek, Claude Tricot, Frédéric Roger, and some others. (I have a bibliography on this topic if you're interested.) $\endgroup$ Sep 21, 2015 at 21:59
  • $\begingroup$ @DaveLRenfro I'd be very interested! Can you email me that bibliography? My email address is in my profile. $\endgroup$ Sep 22, 2015 at 3:22
  • $\begingroup$ The bibliography would also be a great answer! $\endgroup$ Sep 24, 2015 at 22:22
  • 2
    $\begingroup$ Looking through the bibliography, I'm not sure it actually answers the question - the poset given by rarefaction (ams.org/journals/tran/1990-321-02/S0002-9947-1990-0967315-8/…) seems different than what I'm describing. But it is very close, and extremely interesting! $\endgroup$ Sep 24, 2015 at 22:49

1 Answer 1


$\newcommand{\scope}{\mathrm{Scope}}$ $\newcommand{\res}{\upharpoonright}$ The edit makes it seem like already the question of whether incompatible null sets can exists is of interest to you, I hope that is correct. I will construct $\subseteq_{\mathrm{Null}}$-incompatible null sets (without any additional assumptions). (I use the definition of $\subseteq_{\mathrm{Null}}$ in the question, not the one in Comment 1).

All the null sets here will be "Cantor-like" sets. As a first step, we will analyse (to some extend) for which $f$ the Cantor set is $f$-microscopic. Recall the usual "take away the middle third" construction of the Cantor set: Let $I_\emptyset=[0, 1]$ and if $I_s$ is defined let $I_{s^\frown 0}$ be the (closed) "first third" of $I_s$ and $I_{s^\frown 1}$ the (closed) "last third" of $I_s$. The Cantor set is $$\mathcal C=\bigcup_{x:\mathbb N\rightarrow\{0, 1\}}\bigcap_{n\in\mathbb N}I_{x\res n}$$ (with $n=\{0,\dots, n-1\}$). Set $\mathcal C^\ast=\mathcal C\setminus\{1\}$. Via proof by picture, there is a (red) interval cover $\langle J_n\mid n<\omega\rangle$ of $\mathcal C^\ast$ with $\mu(J_n)=3^{-(n+1)}$:

Proof by picture I:

Curiously, there is no such cover for $\mathcal C$ itself, but there is one after any one point is removed from $\mathcal C$.

Lemma: If $f(n)\leq\frac{2}{3^{n+1}}$ for all $n$, then $\mathcal C^\ast$ is $f$-microscopic iff $f(n)>\frac{1}{3^{n+1}}$ for all $n\in\mathbb N$.

Proof: We already know that the "if" part holds. For the other direction, let us first assume that for some $\epsilon>0$ and all $n$, $f(n)\leq\frac{2-\epsilon}{3^{n+1}}$. Find $N\geq 1$ large enough so that $3^{-N}\leq\epsilon$ for all $n$. Exemplarily, we deal with the case $N=2$ in the pictures. Let $\langle J_n\mid n\in\mathbb N\rangle$ witness that $\mathcal C^\ast$ is $f$-microscopic. By induction we will find $A_n\subseteq {}^{N+n}\{0, 1\}$ so that

  • $\vert A_n\vert\geq 2^{N-1}$
  • If $s\in A_n$ and $m<n$ then $s\res N+m\in A_m$
  • If $s\in A_n$ then $I_s\cap J_k=\emptyset$ for all $k\leq n$.

We start with $n=0$. Since $\mu(I_0)<\frac{2}{3}-\frac{1}{3^N}$, $J_0$ must be disjoint from at least $2^{N-1}$ of the $2^N$-many intervals $I_s$ with $s$ of length $N$. We proof this by picture and imagination: Imagine one slides the red interval $J_0$ to the left so that its lower endpoint is $0$. If it is now slid to the right, then it disjoint from the first interval on the third level before it meets the next interval:

Proof by Sliding

And this pattern continues. Thus $A_0=\{s:N\rightarrow 2\mid I_s\cap J_0=\emptyset\}$ has size at least $2^{N-1}$. Suppose we have managed to get to stage $n$. Since every interval $I_s$, $s\in A_n$ splits into two smaller intervals on the next level, there are at least $2^N$-many of those. By a similar argument as above, $J_{n+1}$ can cover at most half of those. Hence $$A_{n+1}=\{s:n+1\rightarrow \{0, 1\}\mid s\res n\in A_n\wedge I_{s}\cap J_{n+1}=\emptyset\}$$ works.

Now $A=\bigcup_{n\in\mathbb N} A_n$ forms a tree via end-extension with all levels non-empty and finite. By König's Lemma, there is a branch $x:\mathbb N\rightarrow \{0, 1\}$ through $A$ (i.e. $x\res N+n\in A_n$ for all $n$) and the unique real $y\in \bigcap_{n\in\mathbb N} I_{x\res n}$ is in $\mathcal C$ and avoids every interval $J_n$. Thus $y$ must equal $1$ and $x\equiv 1$. This also means that $\vert A_n\vert = 2^{N-1}$ for any $n$: If $A_n$ is larger, after taking $x\res N+n$ out of $A_n$, we can still continue the argument indefinitely and get a branch through $A$ different from $x$, contradiction.

Finally assume, for a contradiction, that for some $m$, $f(m)\leq 3^{-(m+1)}$ so that $\mu(J_m)<3^{-(m+1)}$. Again by a sliding argument, one can be convinced that $J_m$ cannot completely cover $2^{N-1}$-many of the intervals $I_{s^\frown i}$, $s\in A_{m-1}$, $i<2$. Thus there must be some such $t=s^\frown i$ so that $I_t\cap J_m\neq\emptyset (\Rightarrow t\notin{A_m})$ and $I_t\not\subseteq J_m$. In particular, one of the endpoints $z$ of the interval $I_t$ is not covered by $J_m$ (and any earlier $J_n$, $n<m$). (If we are in the nasty case of $z=1$, replace $z$ by a slightly smaller real in $\mathcal C\cap I_t\setminus J_m$) This real $z$ must be accounted for at a larger stage: There is $M>m$ with $z\in J_M$. But this is a big problem: This $z$ is so far away from from the other intervals still in the game that $J_M$ cannot both cover $z$ and enough of them:

Proof by picture II

$J_m$ misses $z$ and the later $J_M$ covers $z$ but cannot reach the $4$ blue intervals of which it should cover $2$.

Thus $\vert A_{M}\vert>2^{N-1}$, contradiction.

The final trick to remove the $\epsilon$ is to diagonalise the above argument, i.e. one increases $N$ along the construction of the $A_n$'s if necessary.$\Box$

To construct incompatible null sets, we relativise the construction of the Cantor set: Let $g:\mathbb N\rightarrow(0, \infty)$ so that

  • $g(0)\leq \frac{1}{3}$
  • $\forall n\ g(n+1)\leq\frac{g(n)}{3}$

Then let $I^g_\emptyset=[0, 1]$ and if $I^g_s$ is constructed $\mathrm{len}(s)=n$, $I^g_{s^\frown 0}$ is the (closed) initial segment of $I^g_s$ of length $g(n+1)$ and $I^g_{s^\frown 1}$ is the (closed) end-segment of $I^g_s$ of length $g(n+1)$. Then $$\mathcal C_g=\bigcup_{x:\mathbb N\rightarrow 2}\bigcap_{n\in\mathbb N} I^g_{x\res n}$$ is an uncountable closed null set. If we set $\mathcal C^\ast_g=\mathcal C_g\setminus\{1\}$, then by the same argument as above we get:

If $f\leq 2g$ then $\mathcal C_g^\ast$ is $f$-microscopic iff $f>g$.

From this we can construct $\subseteq_{\mathrm{Null}}$-incompatible null sets: Take two functions $g_i$, $i<2$ for which $\mathcal C_{g_i}^\ast$ is defined so that

  • For all $n$ there is some $i$ with $g_i(n)\leq g_{1-i}(n)<2g_i(n)$
  • For both $i$ there is $n_i$ with $g_i(n_i)<g_{1-i}(n_i)$

If we let $f_i$ so that $f_i(n_i)\in (g_i(n_i), g_{1-i}(n_i)]$ and for $n\neq n_i$ $$f_i(n)=\mathrm{min}\{2g_0(n), 2g_1(n)\}$$ then $$f_i\in\scope(\mathcal C_{g_i}^\ast)\setminus\scope(\mathcal C_{g_{1-i}}^\ast)$$ as the single mishap at $n_i$ excludes $f_i$ from the scope of $\mathcal C_{g_{1-i}}^\ast$ but not from the scope of $\mathcal C_{g_i}^\ast$. Observe that we only used two integers for "mishaps", but we have countably infinite many possibilities for them to occur. So in a similar manner one can construct a countably infinite antichain in $\mathcal N=(\mathrm{Null}/\equiv_{\mathrm{Null}},\subseteq_{\mathrm{Null}})$.

In my opinion, this construction can either be seen optimistically, in the sense that it gives some insight into the structure of $\mathcal N$, or pesimistically in the sense that it is evidence for the definition of $\scope$ being incorrect (in the same vein as your Comment 1).

I hope there is still some interest in this question after all these years!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.