Trace Sigma-Algebra of Measurable Set

Theorem

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

Let $A \in \Sigma$.

Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.

Then:

$\Sigma_A = \set {B \in \Sigma : B \subseteq A}$

That is, the elements of $\Sigma_A$ are precisely the $\Sigma$-measurable sets that are subsets of $A$.

Proof

Let:

$\Sigma' = \set {B \in \Sigma : B \subseteq A}$

We first show that:

$\Sigma_A \subseteq \Sigma'$.

Let $B \in \Sigma_A$, then there exists $S \in \Sigma$ such that:

$B = A \cap S$

From Sigma-Algebra Closed under Finite Intersection, we have:

$B \in \Sigma$

while, from Intersection is Subset, we have:

$A \cap S \subseteq A$

so:

$B \subseteq A$

giving:

$B \in \Sigma'$

So we have:

$\Sigma_A \subseteq \Sigma'$

by the definition of set inclusion.

Now let $B \in \Sigma'$.

From Intersection with Subset is Subset, we have:

$B = B \cap A$

while $B \in \Sigma$, so:

$B \in \Sigma_A$

from the definition of trace $\sigma$-algebra.

So we have:

$\Sigma' \subseteq \Sigma_A$

by the definition of set inclusion.

So, we have:

$\Sigma_A = \Sigma' = \set {B \in \Sigma : B \subseteq A}$

$\blacksquare$