Absolute Value of Measurable Function is Measurable

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

Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.

Then:


 * $\size f$ is a $\Sigma$-measurable function.

Proof
From Characterization of Measurable Functions, it suffices to show that for each real number $t \in \R$, we have:


 * $\set {x \in X : \size {\map f x} \le t} \in \Sigma$

If $t < 0$, we have:


 * $\set {x \in X : \size {\map f x} \le t} = \O$

So, from Properties of Algebras of Sets, we have:


 * $\set {x \in X : \size {\map f x} \le t} \in \Sigma$

if $t < 0$.

If $t \ge 0$, we can write:

Since $f$ is $\Sigma$-measurable, we have that both:


 * $\set {x \in X : -t \le \map f x} \in \Sigma$

and:


 * $\set {x \in X : \map f x \le t} \in \Sigma$

from Characterization of Measurable Functions.

From Properties of Algebras of Sets, the intersection of any two sets in $\Sigma$ is contained in $\Sigma$.

So:


 * $\set {x \in X : -t \le \map f x} \cap \set {x \in X : \map f x \le t} \in \Sigma$

if $t \ge 0$.

That is:


 * $\set {x \in X : \size {\map f x} \le t} \in \Sigma$

if $t \ge 0$.

So:


 * $\set {x \in X : \size {\map f x} \le t} \in \Sigma$

for all $t \in \R$.