Restriction of Measurable Function is Measurable on Trace Sigma-Algebra
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable functions.
Let $E \in \Sigma$.
Let $\Sigma_E$ be the trace $\sigma$-algebra of $E$ in $\Sigma$.
Then the restriction $f \restriction_E$ is $\Sigma_E$-measurable.
Proof
From the definition of a $\Sigma_E$-measurable function, we aim to show that:
- $\set {x \in E : \map f x \le \alpha} \in \Sigma_E$
for each $\alpha \in \R$.
Let $\alpha \in \R$.
We have:
- $\set {x \in E : \map f x \le \alpha} = \set {x \in X : \map f x \le \alpha} \cap E$
Since $f$ is $\Sigma$-measurable, we have:
- $\set {x \in X : \map f x \le \alpha} \in E$
So, from the definition of trace $\sigma$-algebra, we have:
- $\set {x \in X : \map f x \le \alpha} \cap E \in \Sigma_E$
So we have:
- $\set {x \in E : \map f x \le \alpha} \in \Sigma_E$
for each $\alpha \in \R$.
So:
- $f \restriction_E$ is $\Sigma_E$-measurable.
$\blacksquare$