what is a measurable set I have a question about the meaning of measurable set.
I was reading that if given a space say $\Omega$ and a sigma algebra say $X$ for that space then we call  $(\Omega, X)$ a measurable space. So does it mean we call $\emptyset$, $\Omega$ measurable set (how about the other sets inside that sigma algebra, do we all them measurable set as well?) and if say $E$ is one of the set in $X$ being measurable, does it  mean we also say $E^c$ is measurable...?  my question is "is it correct to automatically say all the sets in $X$ are measurable"?  
Do we have to worry about the mapping say $F$: $X \longrightarrow Y$, do we have to care about what $F$ is in order to say whether the set in the sigma algebra $X$ are "measurable" or not? 
The reason I am asking this question is because when I was reading the "accepted answer" to this question here:
Why the characteristic function is measurable?
the person who answer the question says this:
$X_E^{-1}(\{0\}) = E^c$ is perfectly measurable. 
My question is where is the "perfectly measurable" assumption arrived from?
is it automatic that if say a set called $E$ is measurable (in a sigma algebra), then it automatically mean $E^c$ is measurable? how about the set of $\emptyset$, and the set of $\Omega$?  are the also considered as "automatically measurable"? 
My question I guess is how does that person who answered that question say "perfectly measurable", is it just by definition? 
Could someone explains? 
 A: Let $(X,\mathcal{A})$ a measurable space, that is, $\mathcal{A}$ is a $\sigma $-algebra defined on $X$. Then a subset $E\subset X$ is measurable if and only if $E\in \mathcal{A}$.
Because $\mathcal{A}$ is a $\sigma $-algebra then if $E$ is measurable it means that $E^\complement $ is also measurable. Moreover: if $(E_k)_{n\in \Bbb N }$ is a sequence of measurable sets then there are also $\bigcup_{n\geqslant 1}E_n$ or $\bigcap_{n\geqslant 1}E_n$. And any other constructions of sets based in complementation and countable union of measurable sets defines also an element of $\mathcal{A}$, that is, a measurable set.
A function $g:(X,\mathcal{A})\to (Y,\mathcal{B})$ between two measurable spaces $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ is said to be measurable if and only if $f^{-1}(D)\in\mathcal{A}$ for any chosen $D\in \mathcal{B}$. When $g:(X,\mathcal{A})\to \Bbb R $, if no other thing is said, we assume that $\Bbb R $ is a measurable space equipped with the Borel $\sigma $-algebra or the Lebesgue $\sigma $-algebra (they are almost the same, and in almost any situation we dont need to distinguish between the Borel or the Lebesgue $\sigma $-algebra, in any case the Borel $\sigma $-algebra is contained in the Lebesgue $\sigma $-algebra).
Then its easy to check that a characteristic function $\chi_K:(X,\mathcal{A})\to \Bbb R $ is measurable if and only if $K\in\mathcal{A}$, because $\chi_K^{-1}(\{1\})=K$ and $\{1\}$ is a Borel subset of $\Bbb R $. Also, by the definition of characteristic function, we have that $\chi_K^{-1}(\{0\})=K^\complement$, and so it is measurable when $K$ is (again note that $\{0\}$ is a Borel set).
I hope you see it more clear now.
A: In order to a triple $(\Omega, \mathcal{A}, \mu)$ be a measurable space, the measure $\mu$ must be defined for every set in $\mathcal{A}$. The reason we need such a $\sigma$-algebra is exactly to distinguish the measurable sets (i.e., the ones in $\mathcal{A}$) of the non-measurable sets.
