If $\sigma$-algebras $\mathscr{F}$ and $\mathscr{G}$ are independent and if sigma-algebras $\mathscr{F}$ and $\mathscr{H}$ are independent then $\mathscr{F}$ and $\mathscr I := \sigma(\mathscr{G},\mathscr{H})$ are independent.
What is a counterexample to this?
If $\sigma$-algebras $\mathscr{F}$ and $\mathscr{G}$ are independent, then
$\forall F \in \mathscr{F}, G \in \mathscr{G}$,
$$P(F \cap G) = P(F)P(G)$$
If $\sigma$-algebras $\mathscr{F}$ and $\mathscr{H}$ are independent, then
$\forall F \in \mathscr{F}, H \in \mathscr{H}$,
$$P(F \cap H) = P(F)P(H)$$
So I need an $I \in \mathscr I$ s.t.
$$P(F \cap I) = P(F)P(I)$$
Consider a probability space $(\Omega, \mathscr{F}, \mathbb{P})$, events A, B and C, pairwise independent but not mutually independent.
Then
$$P(A \cap B) = P(A)P(B)$$
$$P(A \cap C) = P(A)P(C)$$
$$P(C \cap B) = P(C)P(B)$$
$$P(A \cap C \cap B) \ne P(A)P(C)P(B)$$
Let:
$$\mathscr F = \sigma(A)$$
$$\mathscr G = \sigma(B)$$
and $\mathscr H = \sigma(C)$
Then $\mathscr I = \sigma(B, C)$.
$\mathscr F = \sigma(A)$ and $\mathscr G = \sigma(B)$ are independent.
$\mathscr F = \sigma(A)$ and $\mathscr H = \sigma(C)$ are independent.
We must show that $\mathscr F = \sigma(A)$ and $\mathscr I = \sigma(B, C)$ are not independent i.e. $\exists I \in \mathscr I$ s.t.
$$P(A)P(I) \ne P(A \cap I)$$
Now choose $I = B \cap C$ which works because $\mathscr G = \sigma(B)$ and $\mathscr H = \sigma(C)$ are independent.