Relative null-ness 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.
 A: $\newcommand{\scope}{\mathrm{Scope}}$
$\newcommand{\res}{\upharpoonright}$
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 $\mu(J_n)=3^{-(n+1)}$:

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:

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:

$J_m$ 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!
