Why countable union and not countable disjoint union in definition of $\sigma$-algebra. I was trying to find the motivation behind the definition of $\sigma$-algebra.The idea actually came from the fact that we can measure the whole set and if we could measure a set $A$,then we could measure its complement as well by the virtue of the formula $\mu(X\setminus A)=\mu(X)-\mu(A)$ and if we are given a set which is composed of disjoint pieces and if we can measure the pieces we can measure the whole by $\mu(\bigcup\limits_n A_n)=\sum\limits_n \mu(A_n)$.So,we need to assure that the whole set $X$ belongs to $\mathcal S$,the $\sigma$-algebra and if $A\in \mathcal S$ then $X\setminus A\in \mathcal S$ and third one we should have if $A_1,A_2,...$are disjoint sets in $\mathcal S$,then $\bigcup\limits_n A_n\in \mathcal S$.But in the original definition we do not assume disjoint,we only assume that countable union of measurable sets is measurable.WHat is the reason behind not assuming disjoint.Although it is ok because assuming countable union is sufficient for disjoint union.
 A: An algebra of sets $\mathcal{A}$ on a set $X$ is closed under finite unions and complements, so also under finite intersections and relative complements (via De Morgan's laws). The algebra $\mathcal{A}$ is said to be a $\sigma$-algebra if it is closed under countable unions.
In particular, a $\sigma$-algebra is closed under “countable disjoint unions” (that is unions of countable families consisting of pairwise disjoint members of $\mathcal{A}$).
Is the converse true? Let's assume the algebra $\mathcal{A}$ is closed under countable disjoint unions and consider any countable family $(A_n)_{n\ge0}$.
We can define a new family by
$$
B_{n}=A_n\setminus\bigcup_{k<n}A_k=
A_{n}\setminus(A_0\cup A_1\cup\dots\cup A_{n-1})
$$
In particular $B_0=A_0$ (because we're subtracting the empty set) and $B_n\subseteq A_n$. Suppose $m<n$; then we can show that $B_m\cap B_n=\emptyset$. Indeed if $x\in B_m$ we have $x\in A_m$, but as $m<n$ we see that $x\in A_0\cup\dots\cup A_{n-1}$ and so $x\notin B_n$.
Moreover, we clearly have
$$
\bigcup_{n\ge0}B_n\subseteq \bigcup_{n\ge0}A_n
$$
Suppose $x\in\bigcup_{n\ge0}A_n$. Thus, for some $n$, we have $x\in A_n$ and we can take $n$ to be minimal with respect to this property. Thus $x\notin A_m$ for $m<n$ and therefore $x\in B_n$. So we have proved that
$$
\bigcup_{n\ge0}A_n=\bigcup_{n\ge0}B_n\in\mathcal{A}
$$
because we assumed closure under countable disjoint unions.
A: 
The idea actually came from the fact that we can measure the whole set
and if we could measure a set $A$,then we could measure its complement
as well by the virtue of the formula $\mu(X\setminus A)=\mu(X)-\mu(A)$

This is not true unless otherwise $\mu(A) <\infty$.
For an example Consider the Lebesgue measure space $(\Bbb{R}, \mathcal{L}(\Bbb{R}), m) $
Then $\mu(\Bbb{R}\setminus (0, \infty)) =mu(\Bbb{R})-\mu(0,\infty)=\infty -\infty$
But $\infty -\infty$ is not defined! (It can be my CGPA Or your CGPA)

A $\sigma$ algebra is closed under countable union so obvious for countable union of disjoint sets.
If a collection of subsets of $X$ satisfy the first two conditions of sigma algebra ( closed under trivial set inclusion and set complement) and colsed under disjoint union of countable sets, then this collection is called a Dynkin system.

A Dynkin system (or $\lambda $ -system) $D$ is a collection of subsets
of X that contains X and is closed under complement and under
countable unions of disjoint subsets.

So your question is about the motivation of Dykin system. See here.
