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 $\set {x \in X: \map f x < \map g x}$.

Proof
From Pointwise Difference of Measurable Functions is Measurable, $f - g: X \to \overline \R$ is $\Sigma$-measurable.

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:


 * $\openint {-\infty} 0$
 * $\hointl {-\infty} 0$
 * $\set 0$
 * $\R \setminus \set 0$

are all open or closed in the Euclidean topology.

Hence by definition of Borel $\sigma$-algebra, and by Closed Set Measurable in Borel Sigma-Algebra, they are $\map \BB \R$-measurable.

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


 * $\set {f < g}$
 * $\set {f \le g}$
 * $\set {f = g}$
 * $\set {f \ne g}$

are all $\Sigma$-measurable sets.