Measurable function definition

Question

In my lecture notes for stochastic processes we define measurable functions as

Here $\mathcal{A}$ is a $\sigma-$algebra on $\Omega$. However, when I read on other sites (e.g. Wolfram), the set that should be measurable is defined in another way, specifically the inequality sign should be flipped. Is the definition of measurable function incorrect in my picture? Or is there some detail that I've missed?

Answer 1

$\{\omega: X(\omega) \leq x \} \in \mathcal A$ for every real number $x$ if and only if $\{\omega: X(\omega) \geq x \} \in \mathcal A$ for every real number $x$. So the definitions are equivalent.

Proof is based on the following: $\{\omega: X(\omega) \geq x \}=(\bigcup_n \{\omega: X(\omega) \leq x -\frac 1 n\})^{c}$ and $\{\omega: X(\omega) \leq x \}=(\bigcup_n \{\omega: X(\omega) \geq x +\frac 1 n)^{c}$.

Answer 2

For any $x \in \mathbb{R}$, we can show that following are equivalent:

1. $\{\omega : X(\omega) \ge x\}$ is measurable.
2. $\{\omega : X(\omega) \le x\}$ is measurable.
3. $\{\omega : X(\omega) > x\}$ is measurable.
4. $\{\omega : X(\omega) < x\}$ is measurable.

For the proof of this, we note that since a set A is measurable if and only if its complement is measurable so we see that 1 $\iff$ 4 holds and likewise 2 $\iff$ 3 holds. Now we prove that 1 $ \iff$ 3 holds. This can be seen from the facts that $\{ \omega : X(\omega) \ge x\} = \cap_{n =1}^{\infty}\{\omega : X(\omega) > x - \frac{1}{n} \} $ and $\{ \omega : X(\omega) > x\} = \cup_{n =1}^{\infty}\{\omega : X(\omega) \ge x + \frac{1}{n} \} $. Hence now you know that you can take any definition of a measurable function.