Normally, when we're dealing with intervals $[a,b]$, it is at least implied that $a\leq b$. In this case, there are multiple equivalent ways to define what the interval actually is:
- $[a,b]=\{x\in\mathbb{R}\colon a\leq x\leq b\}$
- $[a,b]:=\operatorname{conv}(\{a,b\})$.
While the former definition is the one usually taught, the latter seems rather more elegant to me, though the convex-hull operator is itself defined by a more complicated set-builder notation. As both are equivalent, the distinction is unnecessary.
It gets interesting when we have $[a,b]$ with $a>b$. Usually, I would consider this simply wrong, but sometimes you have a general interval $[a,b]$ and then it can be tedious and contraproductive to always have to require $b\geq a$. What such an interval is comes out differently from both definitions:
- Always the empty set. This is the usual interpretation according to Wikipedia, but it's not necessarily very useful.
- The set with "correctly-ordered" borders $[b,a]$.
The latter result seems more useful to me, for instance we have always $$\mu([a,b]) = \mu([b,a])$$ $$ \int\limits_a^b\mathrm{d}x\ f(x) = \int_{[a,b]}\!\!\!\mathrm{d}x\ f(x)\cdot\operatorname{sgn}(b-a) $$ and so on.
What reasons are there to not just always define intervals this way?