Definition: Length of an Interval (Convex Set) in $\mathbb R$ I think I know the answer, but I'm looking for a reference that confirms it.
A "proper" interval of an ordered set X as defined by Munkres Topology p.84 is a set limited by $a, b \in X$ with $a < b$ and contains all points $x \in X$ with $ a \lt/\le x \lt/\le b$ with $\lt/\le$ giving open, closed, or half-open intervals.
Other authors extend the usage of an interval of $\mathbb R$ to include all types of convex sets. So as well as the four "proper" intervals they add the four rays (upward / downward pointing, open/closed), $\mathbb R$ itself, singleton sets and the empty set.
By my understanding, the length of a "proper" interval is the difference of its sup and inf (i.e. $b - a$ whether present in the set or not); the length of singletons and the empty set is zero; and the length of $\mathbb R$ and the rays is $\infty$.
Maths is a subject of precision, and as noted I'm looking for an authoritative confirmation (or correction).
 A: The field of mathematics that deals with concepts like "length" is called Measure theory.
Measure theory takes a family of sets that satisfies some properties (called a $\sigma$-algebra and assigns a "measure" to each of the sets in the family.
A very specific measure called the Lebesgue measure is very often used on sets of $\mathbb R$, and this particular measure defines the length of an interval $[a,b]$ as $b-a$, so yes, you are correct.
A: Maths may be a subject of precision, but it is not a subject of authority. Better than authoritative confirmation, you can verify for yourself what the length of the empty set must be, by applying some of the basic abstract properties that are expected of length.
For example, two of the basic properties of length are: 


*

*Given an interval $I$, we have $\text{Length}(I) \ge 0$.

*Given two intervals, if $I \subset J$ then $\text{Length}(I) \le \text{Length}(J)$.


So if you admit the emptyset $\emptyset$ into the pantheon of intervals, then for any interval $I$, since $\emptyset \subset I$ we may apply property 2 to conclude 
$$\text{Length}(\emptyset) \le \text{Length}(I)
$$
From the existence of intervals of any positive length $r$, such as $I=(0,r)$, we conclude that
$$\text{Length}(\emptyset) \le 0
$$
Then from property 1 we have 
$$\text{Length}(\emptyset) \ge 0
$$
Putting these together, we deduce $\text{Length}(\emptyset)=0$.
By the way, the negative number $-2$ cannot be the length of anything.  When you wrote that the length of $(1,-1)$ can be $-2$, you were misapplying the length formula, which only applies to intervals of the form $(a,b)$ where $a<b$.
