Sigma-Algebra Closed under Symmetric Difference

Theorem

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

Let $A, B \in \Sigma$.

Then the symmetric difference $A \Delta B$ is contained in $\Sigma$.

Proof

From Sigma-Algebra Closed under Set Difference, we have:

$A \setminus B \in \Sigma$

and:

$B \setminus A \in \Sigma$

Since $\sigma$-algebras are closed under countable union, we have:

$\paren {A \setminus B} \cup \paren {B \setminus A} \in \Sigma$

From the definition of symmetric difference, we have:

$A \Delta B = \paren {A \setminus B} \cup \paren {B \setminus A}$

so:

$A \Delta B \in \Sigma$

$\blacksquare$