I was going through lecture notes on introductory measure theory and I have a question
Let $(\Omega, \mathcal{F})$ be a measurable space. Let $E$ be an element of $\mathcal{F}$. Define a map to $(\mathbb{R},\mathcal{B}):$
$$(\mathbb{1}_E)(\omega)=\begin{cases} 1 & \omega \in E\\ 0 & \omega \notin E\end{cases}$$
What I understood is that for this map to be measurable we must have that for any element $A$ in $\mathcal{B}$, $(\mathbb{1}_E)^{-1}(A)$ must be in $\mathcal{F}$. I have read explanations why characteristic function is measurable. They define inverse of characteristic function the following way:
$$(\mathbb{1}_E)^{-1}(A)=\begin{cases} E & 1\in A,0\notin A\\ E^c & 1\notin A, 0\in A\\ \Omega & 1\in A, 0\in A\\ \emptyset & 1\notin A, 0\notin A \end{cases}$$
This way I can see indeed that characteristic function is measurable, but how do you know that you should define inverse of characteristic function in this particular way?
What is bothering me for example is that inverse of $[\frac{1}{2},1]$ is $E$, but there is only one point in this interval attained by $(\mathbb{1}_E)(\omega)$
I can't accept this as a proof that characteristic functions are measurable, because this is just one construction of its inverse. Is there some rigorous way to justify that this is indeed inverse of characteristic function?
Best Answer
This is not an inverse of $1_E$, in the normal sense. You're dealing with the more general concept of pre-image here. It's defined - in general, when $f: A \to B$ is a function and $C \subseteq B$, we define $$ f^{-1}(C) = \{ x \in A : f(x) \in C \}. $$ Does that answer your question?