Restriction of Measurable Function is Measurable on Trace Sigma-Algebra

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.