$\mathcal{A}_1 \cup \mathcal{A}_2$ is a $\sigma$-algebra $\iff$ $\mathcal{A}_1 \subseteq \mathcal{A}_2$ Given $\mathcal{A}_1$ and $\mathcal{A}_2$ two $\sigma$-algebras over $\Omega$, their union $\mathcal{A}_1 \cup \mathcal{A}_2$ is a $\sigma$-algebra iff $\mathcal{A}_1 \subseteq \mathcal{A}_2$ or $\mathcal{A}_2 \subseteq \mathcal{A}_1$
$\Longleftarrow$
Assume w.l.o.g that $\mathcal{A}_1 \subseteq \mathcal{A}_2$


*

*$\Omega \in \mathcal{A}_1 \cup \mathcal{A}_2$

*Let $A \in \mathcal{A}_1 \cup \mathcal{A}_2$ then $A \in \mathcal{A}_1$ or $A \in \mathcal{A}_2$. Either case, $A^{c} \in \mathcal{A}_1 \cup \mathcal{A}_2$ since $\mathcal{A}_1$ and $\mathcal{A}_2$ are $\sigma$-algebras.

*Let $A_1, A_2, A_3,.... \in \mathcal{A}_1 \cup \mathcal{A}_2$ Then for every $k \in \mathbb{N}$ : $A_k \in \mathcal{A}_2$ or $A_k \in \mathcal{A}_1 \subseteq \mathcal{A}_2$ which implies that $A_k \in \mathcal{A}_2$. Since $\mathcal{A}_2$ is $\sigma$-algebra, $\bigcup_{k=1}^{\infty} A_{k} \in \mathcal{A}_2$ and therefore $\in \mathcal{A}_1 \cup \mathcal{A}_2$
$\Longrightarrow$
I tried to prove this by contradiction.


*

*Assume that $\mathcal{A}_1 \cup \mathcal{A}_2$ is a $\sigma$-algebra over $\Omega$ and that $\mathcal{A}_2 \nsubseteq \mathcal{A}_1$ Let $A_1 \in \mathcal{A}_1 \setminus \mathcal{A}_2$ and $A_2 \in \mathcal{A}_2 \setminus \mathcal{A}_1$. Then $A_1, A_2 \in \mathcal{A}_1 \cup \mathcal{A}_2$ but $A_1 \cup A_2 \notin \mathcal{A}_1 \cup \mathcal{A}_2$ since their union is neither in $\mathcal{A}_1$ or $\mathcal{A}_2$.


but there could be that there is some set in $\mathcal{A}_2$ or in $\mathcal{A}_1$ that when taking $\mathcal{A}_1 \cup \mathcal{A}_2$ ends up being equal to $A_1 \cup A_2$ so this reasoning is most likely wrong. I'm not sure what would be correct here.
 A: Suppose $A_1 \in \mathcal A_1 \backslash \mathcal A_2$ and $A_2 \in \mathcal A_2 \backslash \mathcal A_1$. Then also $A_1^c \in \mathcal A_1 \backslash \mathcal A_2$ and $A_2^c \in \mathcal A_2 \backslash A_1$.
$A_1 \cap A_2$ must be in at least one of $\mathcal A_1$ and $\mathcal A_2$: wlog suppose it is in $\mathcal A_1$.  Then $A_2 \backslash A_1 $ can't be in $\mathcal A_1$ because otherwise $A_2 = (A_1 \cap A_2) \cup (A_2 \backslash A_1)$ would be in $\mathcal A_1$.  So $A_2 \backslash A_1$ must be in $\mathcal A_2$, and then $A_1 \cap A_2 = A_2 \backslash (A_2 \backslash A_1) \in \mathcal A_2$.  Thus $A_1 \cap A_2 \in \mathcal A_1 \cap \mathcal A_2$. 
But now consider $A_1^c \cap A_2^c$.  This can't be in $\mathcal A_1$, else $A_2^c = (A_1^c \cap A_2^c) \cup (A_1 \backslash (A_1 \cap A_2)) \in \mathcal A_1$, and it can't be in $\mathcal A_2$, else $A_1^c = (A_1^c \cap A_2^c) \cup (A_2 \backslash A_1) \in \mathcal A_2$.  So we have a contradiction.
A: Let $A_1\in\mathcal A_1\setminus\mathcal A_2$ and $A_2\in\mathcal A_2\setminus\mathcal A_1$. Suppose that $\mathcal A_1\cup\mathcal A_2$ is a $\sigma$-algebra. Then every combination of $A_1$ and $A_2$ is contained in $\mathcal A_1\cup\mathcal A_2$, and hence in one of those two.
WLOG, suppose that $A_1\cup A_2\in\mathcal A_1$. Then $A_1\cap A_2\in\mathcal A_2$, since $A_2 = (A_1\cup A_2)\setminus(A_1\cap A_2)$. And since $A_1 = (A_1\setminus A_2)\cup(A_1\cap A_2)$, you have $A_1\setminus A_2\in\mathcal A_1$. But then
$$
A_2 = (A_1\cup A_2)\setminus(A_1\setminus A_2)\in \mathcal A_1,
$$
a contradiction!
