Stability by complement. Why is it needed in measured space? Let $E$ be a set. Let $\Sigma \subseteq P(E)$ be a subset that satisfy the "non-emptiness" and the "stable by countable union" requirements of a $\sigma$-algebra. And finally let $\mu \in {\bar {\mathbb R}}^{\Sigma}$ such as $\mu$ satisfies the $3$ properties of a measure.
The question is why does $\Sigma$ need to meet the "stable by complementary" requirement in order for ($E$,$\Sigma$, $\mu$) to be a measured space?
What would happen otherwise?
 A: Really, you can view measure as let's say a tool to measure sizes. Naively, say that we are using this tool on an object which size is finite. And we measure the sizes of one subpart of said object which we call $O$. So far measure means "size", the parts of the object our tool can work on are the sigma algebra.
$O$ has a finite size so $size(O)=S$ for example, it means our tool can work on it. $P$ is measurable when our tool can measure its size. The stability by complementary means you should be able to tell the size of "the rest" of O. It is quite a natural property that if you know the size of something, and the size of a part of it, you can know the size of the remaining part.
What happens when it fails? You have $O = P + P^c$. But if $P^c$ is not measurable that means your $size(O)$ and $size(P)$ are defined but not $P^c$. If you transpose it to Lebesgue measure for example, it is like being able to tell the distance between $a$ and $b$ $\in \mathbb{R}$, and being able to tell the distance between $a$ and $c=0.5(a+b)$  but not being able to tell the distance between $c$ and $b$...
A: Ok I think I got it.
Suppose that the property "stability by complementation" is not met for $\Sigma$.
Then one would say, "Ok but in that case $E$ is not measurable (and this is bad, blablabla)". So let's add the weaker property: $E$ belongs to $\Sigma$.
So in that case we have three properties:

*

*$\emptyset \in \Sigma$

*$E \in \Sigma$

*Stability by infinite countable union

But noting that for any $A\in \Sigma$
$$E = A \cup (E\backslash A)$$
and using the $\sigma$-additivity of $\mu$ we have:
$$\mu(E) - \mu(A) = \mu(E\backslash A) $$
The left-hand side is defined while the RHS is not necessarily defined. Especially for the $A$ such as $A^c \notin \Sigma$. Contradiction.
