In measure theory $\emptyset$ and $\{a\}$ are considered as intervals . Why? I recently started self learning measure theory.
While reading a note , I came across this ,
In measure theory $\emptyset$ (empty set) and $\{a\}$ (singleton set) are considered as intervals . Why are they considered as intervals ? 
 A: Let $a \in \mathbb R$.
The empty set can be written as an open interval:
$$\emptyset = (a,a) = \{b \in \mathbb R: a<b<a\}.$$
The singleton $\{a\}$ can be written as a closed interval:
$$\{a\}= [a,a] = \{b \in \mathbb R: a\leq b \leq a\}.$$
This is the 'how'. The 'why' is the following:
Consider the set $\mathcal A'$ of all subintervals (excluding singletons and the empty set) of $\mathbb R$, so every $I$ in $\mathcal A'$ is either of the form $(a,b), [a,b], (a,b], [a,b)$, where $a<b$ and $\pm \infty$ is allowed as a value for $a$ and/or $b$ if its end is open.
A natural operation on sets like intervals is to consider unions and intersections. If we perform operations on elements of a set, it would be nice if the result would be an element of the set as well (we don't perform divisions on the set of integers, for example).
But the the intersection of $[0,1)$ and $(1,2]$ is the empty set (not an interval in the strict sense), and the intersection of $(0,1]$ and $[1,2)$ is $\{1\}$ (not an interval as well). So we extend our notion of intervals to these two special cases (which is the same as relaxing the $a<b$ to a $a \leq b$ in the definition above) and end up with $\mathcal A$ of all subintervals (includung singletons and the empty set) of $\mathbb R$.
This set $\mathcal A$, equipped with the operations union and intersection behaves similarly to the integers equipped with addition and multiplication, so we call it an algebra.
A: An interval is a set $I$ with the property that if $a,b\in I$ and $a<c<b$, then $c\in I$. This is vacuously satisfied by both of the sets you mention.
A: This is merely for simplifying terminology and avoiding tedious repetitions.
For example, in constructing the Lebesgue measure on the Borel $\sigma$-algebra over the bounded half-closed interval $(0,1]$, it is an important fact that the collection of finite disjoint unions consisting of the empty set and half-closed intervals of the form $(a,b]$ is an algebra over $(0,1]$ that generates the Borel $\sigma$-algebra. It is much easier to say “finite disjoint union of intervals” instead, with the tacit understanding that the empty set is considered as an interval.
