# Thread: Measurability question

1. ## Measurability question

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

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

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

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

2. ## Re: Measurability question

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

3. ## Re: Measurability question

Thank you.

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

4. ## Re: Measurability question

Originally Posted by TheProphet

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