Measurable Functions Determine Measurable Sets

Theorem

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

Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.

Then the following sets are measurable:

$\set {f < g}$

$\set {f \le g}$

$\set {f = g}$

$\set {f \ne g}$

where, for example, $\set {f < g}$ is short for $\set {x \in X: \map f x < \map g x}$.

Now we have the following evident identities:

$\set {f < g} = \set {f - g < 0}$

$\set {f \ge g} = \set {f - g \le 0}$

$\set {f = g} = \set {f - g = 0}$

$\set {f \ne g} = \set {f - g \ne 0}$

Subsequently, using the preimage under $f - g$, the latter may respectively be expressed as:

$\set {f - g < 0} = \paren {f - g}^{-1} \sqbrk {\openint {-\infty} 0}$

$\set {f - g \ge 0} = \paren {f - g}^{-1} \sqbrk {\hointl {-\infty} 0}$

$\set {f - g = 0} = \paren {f - g}^{-1} \sqbrk {\set 0}$

$\set {f - g \ne 0} = \paren {f - g}^{-1} \sqbrk {\R \setminus \set 0}$

Now the sets:

$\openint {-\infty} 0$

$\hointl {-\infty} 0$

$\set 0$

$\R \setminus \set 0$

Since $f - g$ is $\Sigma$-measurable, it follows that:

$\set {f < g}$

$\set {f \le g}$

$\set {f = g}$

$\set {f \ne g}$

$\blacksquare$