Union of Sigma-Algebras may not be Sigma-Algebra

Theorem

Let $X$ be a set.

Let $\struct {X, \Sigma_1}$ and $\struct {X, \Sigma_2}$ be $\sigma$-algebras.

Then $\Sigma_1 \cup \Sigma_2$ may not be a $\sigma$-algebra on $X$.

Proof

Let $X = \set {1, 2, 3}$.

Let:

$\Sigma_1 = \set {\O, \set 1, \set {2, 3}, \set {1, 2, 3} }$

and:

$\Sigma_2 = \set {\O, \set 2, \set {1, 3}, \set {1, 2, 3} }$

Then $\Sigma_1$ and $\Sigma_2$ are $\sigma$-algebras on $X$.

We have:

$\Sigma_1 \cup \Sigma_2 = \set {\O, \set 1, \set 2, \set {1, 3}, \set {2, 3}, \set {1, 2, 3} }$

However, this does not contain:

$\set 1 \cup \set 2 = \set {1, 2}$

Since $\sigma$-algebras are closed under countable union, $\Sigma_1 \cup \Sigma_2$ therefore cannot be a $\sigma$-algebra.

$\blacksquare$