A non-trivial example of $\cal F$- measurable function? Given a measurable space $(\Omega, \cal{F})$, $f:(\Omega, \cal F) \to (\Bbb{B},\cal B)$, where $\cal B$ is the Borel $\sigma$-algebra of $\Bbb R$, is said to be $\cal {F}$-measurable if $f^{-1}(B)\in \cal F $ for any $B\in \cal B$.
Any function $f:\Omega\to\Bbb R$ is trivially $2^\Omega$ measurable. A constant function on $\Omega$ is $\{\emptyset, \Omega\}$ measurable. If $\Omega$ is Lebesgue measurable and $\cal F$ is the set of all Lebesgue measurable subsets of $\Omega$, then all Lebesgue measurable functions are $\cal F$-measurable.
My question is can anyone help provide an example of $\cal F$-measurable function besides the above three examples? Thank you!
 A: *

*For any $\sigma$-algebra $\mathcal{F}$ on $\Omega$, and any $A \in \mathcal{F}$, the indicator function $\Bbb 1_{A}(x) : = \begin{cases} 1 & x \in A \\ 0 & x \not \in A \end{cases}$ is $\mathcal{F}$-measurable.  It's a good (and easy) exercise to prove this.

*Also, if $\Omega$ is a topological space, and $\mathcal{F}$ contains $\mathcal{B}(\Omega)$, the Borel $\sigma$-algebra on $\Omega$, then every continuous function (defined to be a function where the pre-image of each open set is open) is $\mathcal{F}$-measurable.  This requires a tiny bit of work to prove, but it's not hard.  You will have to use the fact that $\{ A \subseteq \Bbb R \mid f^{-1}(A) \in \mathcal{B}(\Omega) \}$ is a $\sigma$-algebra (which contains the open sets if $f$ is continuous...thus it contains $\mathcal{B}(\Bbb R)$...which is good because?).
A: Consider the symmetric Borel sets 
$$\{B\in \mathcal{B}(\mathbb{R})| - B = B\}.$$
The measurable functions in this $\sigma$-algebra are the even measurable functions.
