Absolute Value of Measurable Function is Measurable
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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:
\(\ds \set {x \in X : \size {\map f x} \le t}\) | \(=\) | \(\ds \set {x \in X : -t \le \map f x \le t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in X : -t \le \map f x} \cap \set {x \in X : \map f x \le t}\) |
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 $t \in \R$.
$\blacksquare$