Characterization of Measurable Functions

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $f: X \to \overline \R$ be an extended real-valued function.

Then the following are all equivalent:

$\text {(1)}: \quad$

$\ds $

$\ds f$ is measurable 

$\text {(2)}: \quad$

$\ds $

$\ds \forall \alpha \in \R: \set {x \in X: \map f x \le \alpha} \in \Sigma$

$\text {($2'$)}: \quad$

$\ds $

$\ds \forall \alpha \in \Q: \set {x \in X: \map f x \le \alpha} \in \Sigma$

$\text {(3)}: \quad$

$\ds $

$\ds \forall \alpha \in \R: \set {x \in X: \map f x < \alpha} \in \Sigma$

$\text {($3'$)}: \quad$

$\ds $

$\ds \forall \alpha \in \Q: \set {x \in X: \map f x < \alpha} \in \Sigma$

$\text {(4)}: \quad$

$\ds $

$\ds \forall \alpha \in \R: \set {x \in X: \map f x \ge \alpha} \in \Sigma$

$\text {($4'$)}: \quad$

$\ds $

$\ds \forall \alpha \in \Q: \set {x \in X: \map f x \ge \alpha} \in \Sigma$

$\text {(5)}: \quad$

$\ds $

$\ds \forall \alpha \in \R: \set {x \in X: \map f x > \alpha} \in \Sigma$

$\text {($5'$)}: \quad$

$\ds $

$\ds \forall \alpha \in \Q: \set {x \in X: \map f x > \alpha} \in \Sigma$

$\blacksquare$