Sigma-Algebra Closed under Set Difference
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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:
- $A \setminus B \in \Sigma$
$\blacksquare$