Given a countable collection of open intervals in $\mathbb{R}$, can you always make it so that (i) the intervals are disjoint (ii) all intervals have a positive minimum distance between them and (iii) have the same total measure (let's say it's finite to begin with)?
I was thinking yes because the first is a property of open sets in $\mathbb{R}$ and if they're already disjoint, then the second can be done because if $A$ and $B$ are intervals with 0 distance between them, then
$$\overline{A} \cap \overline{B} = \{x\}$$ a single point such that $$A \cup B \cup \{x\}$$ is now an open interval with the same measure.
i) Yes. Since R is locally countable any open set is the union of open components which are intervals. Since R is separable, there is countable many such intervals.
ii) No. (0,1) and (1,3) for example.
iii) Yes, the measure of the union of the original sets will
be the same as the total of the measures of the disjoint sets.
Is that what you were asking?