Proof that indicator function is random variable (Borel-measurable)

Question

I realize this highly trivial but it's precisely why I decided to ask it. Also my explanation doesn't quite match the correct one.

On a standard probability triple $(\Omega, \mathcal{F}, P)$ we define a function $$ X({\omega}) = \Bigg\{ \begin{array}{lr} 1 & \text{ if } A \in \mathcal{F}\\ 0 & otherwise \end{array} $$ We need to find $\{\omega: X(\omega) = 1\} \Leftrightarrow X^{-1}(\{1 \}) = A$ and obviously $A \in \mathcal{F}$ by construction of $X$, so $X$ is a Borel-measurable random variable.

I'm sure my solution is solid, but it's a bit different from the correct one. Also it seems somewhat trivial.

Answer 1

Well, the correct definition is

\begin{align*} \chi_A(\omega):=\left\{\begin{array}{clc} 1 & \textrm{if}& \omega \in A\\ 0 & \textrm{if} & \omega\in\Omega\setminus A \end{array}\right., \end{align*} where it is supposed $A\in\mathcal{F}$. Take any interval $(a, b]$ in $\mathbb R$. Then: $$\chi_A^{-1}((a, b]))=\left\{\begin{array}{lcl} \Omega & \textrm{if}& 0\in (a, b]\ \textrm{and}\ 1\in (a, b]\\ \Omega\setminus A & \textrm{if} &0\in (a, b]\ \textrm{and}\ 1\not\in (a, b] \\ A & \textrm{if} &0\not\in (a, b]\ \textrm{and}\ 1\in (a, b]\\ \phi & \textrm{if} &\textrm{if}\ 0\not\in (a, b]\ \textrm{and}\ 1\not\in (a, b] \end{array}\right..$$

In any case, $\chi_A^{-1}((a, b])\in \mathcal{F}$. Since the Borel sigma algebra on $\mathbb R$ is generated by intervals of the kind $(a, b]$, $\chi_A$ is measurable.

Answer 2

More generally you can prove that the indicator function is a borel function (and thus a r.v.). Here my prove but remember that I am just a student so I hope it is correct.

Let $(\Omega ; F )$ be a probabilisable space and $A \in F$ with $F$ a $\sigma$-algebra . The indicator function $\mathbb{1}_A:(\Omega; F)\to (\{0,1\} ; \sigma(\{0,1\})$ is defined as follows: $$ \mathbb{1}_A(\omega)= \begin{cases} 1 & \text{if } \omega \in A\\ 0 & \text{if } \omega \notin A \end{cases} $$ We want to show that $\mathbb{1}_A$ is a Borel function, that means to prove that, the preimage of every element from the Borel set on $\{0,1\}$ (wrotte $\sigma(\{0,1\})$)under $\mathbb{1}_A$ is an element of $F$.

Let $B$ and element of the $\sigma$-algebra on$\{0,1\}$, in other words $B \in \sigma(\{0,1\}) $.
If $B=\emptyset$ so $\mathbb{1}_A^{-1}(B)=\emptyset \in \Omega$
If $B=\{0\}$, then $\mathbb{1}_A^{-1}(B)=\Omega \; without \; A \in F$ by definition of a $\sigma$-algebra the complement of $A$ belongs too to $F$.
If $B=\{1\}$, then $\mathbb{1}_A^{-1}(B)=A $, which is also an element of $\Omega$ by definition.
Finally, if $B=\{0,1\}$, then $\mathbb{1}_A^{-1}(B)=\Omega \in \sigma(\{0,1\})$.

Therefore, the preimage of every element of $\sigma(\{0,1\})$ under $\mathbb{1}_A$ is an element of the $\sigma$-algebra $F$ on $\Omega$, and hence $\mathbb{1}_A$ is a Borel function.