The Lebesgue measure is conventionally defined as
$$\mu(X) = \inf\{\sum_{n \in \mathbb{N}}(b_n-a_n) | X \subseteq \bigcup_{n \in \mathbb{N}}(a_n,b_n) \}$$
Which can be thought of intuitively as covering the set with translated and stretched copies of the set $(0,1)$. I am wondering whether we could equally well define it with any other measurable set, and get the same measure, just by dividing by the Lebesgue measure of that basic set. I.e. for some set $B$
$$\mu'(X) = \mu(B) \inf\{\sum_{n \in \mathbb{N}}s_n | X \subseteq \bigcup_{n \in \mathbb{N}}(r_n+s_nB) \}$$
Where $r_n$ are aritrary elements of $\mathbb{R}$ and $s_n$ are arbitrary elements of $\mathbb{R}_{>0}$. The thing that got me thinking about this was trying to figure out a proof of $$\mu(AX) = |\det A|\mu(X)$$ Where it seems like it would be very helpful to be able to define the measure in terms of sets of the form $AP$ where $P$ is a product of open intervals. I do not know if this is the usual proof. But thinking about this, it seemed like a) this would work and b) this isn't anything particular to sets of the form $AP$. Other than $0$ and $\infty$ measure sets, it seems like any set could work. Edit: meagre sets also won't work. Moreover, it seems clear that a sets like a disconnected union of a positive measure meagre set and another set wouldn't be able to cover $(0,1)$ without significant overlap, so it seems some kind of 'nowhere meagre' condition will be required. My best guess is that this condition should be that the intersection of any (non-empty) open set with $B$ is non-meagre.
E.g. after imposing this restriction, even unbounded sets have to have almost all of their substance in a bounded interval, so the unboundedness shouldn't cause a problem. Highly disconnected sets like fat Cantor sets seem trickier to deal with, but there isn't any obvious proof they couldn't work. @fedja pointed out that this construction won't work for any meagre sets, so fat cantor sets won't work.
It is clear that this at least produces a valid outer measure, and Caratheory's criterion turns it into a measure on some $\sigma$ alebra. This measure would also have to be translation invariant, so as long as the Caratheodory criterion produces the same $\sigma$ algebra, it I think the measure has to be the same.
I haven't proven this, but I think if stretched and shifted copies of $B$ can cover $(0,1)^n$ so that the set of 'overlap' has Lebesgue measure $0$, that would be enough to show that $(0,1)$ is in the $\sigma$ algebra generated. And it seems like equality would follow from that. But not being meagre anywhere doesn't seem like a strong enough condition to guarantee the former, and proving either of these things rigorously seems above my ability level.
Although the $n$-dimensional case motivated this, I would be happy with proofs for the $1$ dimensional case. I would ideally like a criterion for which sets this construction works for, though this may be rather optimistic. Failing that, I would like any results about what classes of sets the construction does or doesn't work for, and particularly classes of sets that this does work for.
