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:


 * $A \setminus B \in \Sigma$