Sigma algebra generated by specific functions

Question

For $\Omega = [-1/2, 1/2] $ and $X(\omega) = \omega^2 $ one shows that $\sigma(X) =\{B \in {\cal B}(\Omega): B=-B\} $ where ${\cal B}(\Omega) $ is the Borel $\sigma$-algebra on $\Omega. $

For $\Omega = [0, 1] $ and $X(\omega) = 1 - |2\omega -1| $ one proves that $\sigma(X) =\{B \in {\cal B}(\Omega): B=1-B\} $ where by $1-B $ i mean the points of $\Omega $ equal to $1-\omega $ for some $\omega \in B. $

The two functions above have something in common; i.e., they are both symmetric about the midpoint of their domain $\Omega. $

I cannot quite see what happens if, say, $\Omega = [-1/2, 1/2] $ and $X(\omega) = -3\omega\cdot 1_{[-1/2,0]}(\omega) + \omega\cdot 1_{(0, 1/2]}(\omega). $ In this case, $X $ is not symmetric over its domain of definition although it shares some similarities with the previous two functions on parts of the domain. And I know that it would be foolish to try to describe in any details $\sigma(X) $ whenever $X $ is a generic function. The last one is an example taken from Borodin's Stochastic Processes textbook.

Thank you.

Answer 1

As we know, $\sigma$-algebra generated by random variable $X$ is defined as: $$\sigma(X) = \{X^{-1}(B): B \in \mathscr{B}\} = X^{-1}(B).$$ Let us consider open interval $(-a, a)$, where $a > 0$.

It's quite easy to show that such an interval belongs to $\sigma(X)$ and it is the form of $B = -B, \ B \in \mathscr{B}(\Omega)$.

Now we check all conditions given in definition of $\sigma$-algebra:

1. Empty set is a form of $B = -B, \ B \in \mathscr{B}(\Omega)$.
2. Complement of such an interval is also a form of $B = -B, \ B \in \mathscr{B}(\Omega)$
3. Countable union of such intervals has also the same property.

what means $\sigma(X)$ is a set of Borel sets having the property $B = -B, \ B \in \mathscr{B}(\Omega)$.

Worth to mention $\sigma$-algebra is closed under complement, countable intersections, unions.

I think the second task can be solved in the same way.