1. ## Measurability question

Let $\displaystyle X \, : \, (\Omega, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be a random variable. Let
$\displaystyle \mathcal{F} = \{A \, : \, A = X^{-1}(B), \mbox{ some } B \in \mathcal{B}(\mathbb{R})\}$.

Show that $\displaystyle X$ is measurable as a function from $\displaystyle (\Omega, \mathcal{F}) \mbox{ to } (\mathbb{R},\mathcal{B}(\mathbb{R}))$.

Here $\displaystyle \mathcal{B}(\mathbb{R})$ is the Borel $\displaystyle \sigma$-algebra on $\displaystyle \mathbb{R}$.

Is it sufficient just to say that for every $\displaystyle B \in \mathcal{B}(\mathbb{R}), \, A = X^{-1}(B) \in \mathcal{F}$, so $\displaystyle X$ is measurable from $\displaystyle (\Omega, \mathcal{F}) \mbox{ to } (\mathbb{R},\mathcal{B}(\mathbb{R}))$ by definition?

2. ## Re: Measurability question

It's sufficient. Maybe the most interesting part of this exercise is to show that $\displaystyle \mathcal F$ is a $\displaystyle \sigma$-algebra.

3. ## Re: Measurability question

Thank you.

I guess $\displaystyle \mathcal{F}$ is a $\displaystyle \sigma$ follows from that $\displaystyle X^{-1}$ commutes with complements, countable intersections and unions.

4. ## Re: Measurability question

Originally Posted by TheProphet

I guess $\displaystyle \mathcal{F}$ is a $\displaystyle \sigma$ follows from that $\displaystyle X^{-1}$ commutes with complements, countable intersections and unions.
Yes.