monotone class theorem when $\Omega\not\in\mathcal A$, but $\Omega=\bigcup_{n\in\mathbb N}Ω_n$ for some $(Ω_n)_{n\in\mathbb N}\subseteq\mathcal A$

Question

Remember the following monotone class theorem: Let

• $\Omega$ be a set;
• $E$ be a normed $\mathbb R$-vector space;
• $\mathcal A\subseteq2^\Omega$ be a $\pi$-system with $\Omega\in\mathcal A$;
• $\mathcal H\subseteq E^\Omega$ be a $\mathbb R$-vector space with
  1. If $A\in\mathcal A$ and $x\in E$, then $1_Ax\in\mathcal H$.
  2. If $(f_n)_{n\in\mathbb N}$ and $f:\Omega\to E$ with $$\|f_n\|_E\le\|f\|_E\;\;\;\text{for all }n\in\mathbb N\tag1$$ and $f_n\xrightarrow{n\to\infty}f$, then $f\in\mathcal H$.

Then, $$\{f:\Omega\to E\mid f\text{ is strongly }\sigma(\mathcal A)\text{-measurable}\}\subseteq\mathcal H.\tag2$$ If (2.) holds only for all bounded $f$, then $$\{f:\Omega\to E\mid f\text{ is bounded and strongly }\sigma(\mathcal A)\text{-measurable}\}\subseteq\mathcal H.\tag3$$

Now I've got a situation where everything holds true, but $\Omega\not\in\mathcal A$. Instead, it only holds $$\Omega=\bigcup_{n\in\mathbb N}\Omega_n\tag4$$ for some $(\Omega_n)_{n\in\mathbb N}\subseteq\mathcal A$. Is this still enough to obtain $(2)$ and $(3)$?