Example of a set and two $\sigma$ algebras such that union is not a $\sigma$-algebra

What is an example of a set $X$ and two $\sigma$-algebras $\mathcal{A}_1$ and $\mathcal{A}_2$, each consisting of subsets of $X$, such that $\mathcal{A}_1 \cup \mathcal{A}_2$ is not a $\sigma$-algebra?

Assume $X$ has more than $2$ elements, and let two of those elements by $a,b\in X$. Then $\mathcal{A}_1 = \{\emptyset,\{a\},X\setminus\{a\},X\}$ and $\mathcal{A}_2= \{\emptyset,\{b\},X\setminus\{b\},X\}$ are both $\sigma$-algebras over $X$.

However, $\mathcal{A}_1\cup \mathcal{A}_2 = \{\emptyset,\{a\},\{b\},X\setminus\{a\},X\setminus\{b\},X\}$ is not a $\sigma$-algebra, as $\{a\}\cup \{b\}=\{a,b\}\notin \mathcal{A}_1\cup \mathcal{A}_2$, for example.

• Darn, beaten by seconds ... – Henning Makholm Jun 30 '16 at 16:41

For example, $X=\{1,2,3\}$ and $$\mathcal A_1 = \{\varnothing,\{1\},\{2,3\},X\} \\ \mathcal A_2 = \{\varnothing,\{2\},\{1,3\},X\}$$

For a more natural example, you could consider $\mathscr{B}([0,1])$, The $\sigma-$algebra generated by the open sets on $[0,1]$ (called the Borel sets) and $\mathscr{B}([-1,0])$.

Note that $$(-1,0) \in \mathscr{B}([-1,0]) \qquad \text{and} \qquad (0,1) \in \mathscr{B}([0,1]).$$ And $\{0\}$ is contained in both.

Though each is a $\sigma-$algebra, Their union, $$\mathscr{B}([-1,0]) \cup \mathscr{B}([0,1])$$ Fails to contain $(-1,1)$, even though $$(-1,0) \cup \{0\} \cup (0,1)=(-1,1)$$

So it fails to even be a Ring, much less a $\sigma-$algebra.