Can we show $\sigma(\sigma(\mathcal F\cup\mathcal G)\cup\mathcal H)=\sigma(\mathcal F\cup\mathcal G\cup\mathcal H)$?

Question

Let $\mathcal F,\mathcal G,\mathcal H$ be $\sigma$-algebras on a common set. Are we able to show $$\sigma(\sigma(\mathcal F\cup\mathcal G)\cup\mathcal H)=\sigma(\mathcal F\cup\mathcal G\cup\mathcal H)?\tag1$$ Clearly, $\sigma(\mathcal F\cup\mathcal G)\cup\mathcal H\supseteq\mathcal F\cup\mathcal G\cup\mathcal H$ and hence "$\supseteq$" in $(1)$. How can we show the other inclusion?

Answer 1

$\sigma(\mathcal F \cup \mathcal G) \subset \sigma(\mathcal F \cup \mathcal G \cup \mathcal H)$ because $\mathcal F \cup \mathcal G \subset \mathcal F \cup \mathcal G \cup \mathcal H$. Also $\mathcal H \subset \sigma(\mathcal F \cup \mathcal G \cup \mathcal H)$. Hence $\sigma(\mathcal F \cup \mathcal G) \cup \mathcal H \subset \sigma(\mathcal F \cup \mathcal G \cup \mathcal H)$ from which the result follows.