Sigma-Algebra Closed under Set Difference

Theorem

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

Let $A, B \in \Sigma$.

Then the set difference $A \setminus B$ is contained in $\Sigma$.

Proof

Since $\sigma$-algebras are closed under relative complement, we have:

$\relcomp X B \in \Sigma$

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

$A \cap \relcomp X B \in \Sigma$

From Set Difference as Intersection with Relative Complement, we have:

$A \setminus B = A \cap \relcomp X B$

so:

$A \setminus B \in \Sigma$

$\blacksquare$