Let me preface this by saying that basically the same question has been asked before on the StackExchange. However, there is one small detail in an exercise that I cannot reconcile.
The following question is Exercise 4.1 in "Probability with Martingales" by David Williams:
Let $(\Omega, \mathcal{F}, P)$ be a probability triple. Let $\mathcal{I}_1$, $\mathcal{I}_2$, and $\mathcal{I}_3$ be three $\pi$-systems on $\Omega$ such that for $k=1,2,3$, $$\mathcal{I}_K\subseteq\mathcal{F}\quad \text{and} \quad\Omega\in \mathcal{I}_k.$$ Prove that if $$P(I_1\cap I_2\cap I_3) = P(I_1) \cdot P(I_2) \cdot P(I_3)$$ whenever $I_k\in \mathcal{I}_k$ (k=1,2,3), then $\sigma(\mathcal{I}_1)$, $\sigma(\mathcal{I}_2)$, and $\sigma(\mathcal{I}_3)$ are independent. Why did we require the $\Omega\in\mathcal{I}_k$?
It seems to me that one can simply mock the proof directly from the Lemma on page 39 in his book where $k=2$. I believe I successfully did the proof. However, in that Lemma, there was not the assumption that $\Omega\in\mathcal{I}_k$ for $k=1,2$. So I am confused by his assumption in E4.1, and question as to "Why did we require the $\Omega\in\mathcal{I}_k$?" Is this simply a callously worded exercise where the answer is "We did need this assumption." Or am I in fact missing something?
The proof I have is as follows (mimicking a previous lemma in Williams):
Fix $I_1\in\mathcal{I}_1$ and $I_2\in\mathcal{I}_2$. Consider the maps $$ J_3 \mapsto P(I_1\cap I_2\cap J_3) \quad \text{and}\quad J_3\mapsto P(I_1)\cdot P(I_2)\cdot P(J_3),$$ for $J_3\in\sigma(\mathcal{I}_3)$. One can verify that these are measures on the measure space $(\Omega, \sigma(\mathcal{I}_3))$, that they both have a total mass of...
OH WAIT! Just as I was typing the "..." above I realized something. I think the technical difficulty here is because William's earlier Lemma used a result that if two finite measures agree on a $\pi$-system and have the SAME total mass, then they agree on the $\sigma$-algebra generated by the $\pi$-system.
Therefore, the technical issue is that the total mass of the two measures above are $$P(I_1\cap I_2)\quad \text{and}\quad P(I_1)\cdot P(I_2),$$ respectively. If $\Omega\not\in \mathcal{I}_3$, it is not necessarily the case that the above two numbers equal (since the condition in the problem is specifically for all three $\pi$-systems at once). Using this approach by creating measures two more times, the same issue will come up, requiring that $\Omega\in\mathcal{I}_k$ for $k=1,2$.
This begs the following question...is the requirement that $\Omega\in \mathcal{I}_k$ only an issue for this particular proof? The previous question here does not require this for the same question. However, the proof there made use of the $\pi-\lambda$ theorem, which was not directly used in the earlier Lemma in Williams (for the $k=2$ case; although, recall that there they actually did not need the requirement that $\Omega\in\mathcal{I}_k$ for $k=1,2$ for that proof).
Regarding your new question: The hypothesis $\Omega\in{\cal I}_k$ for each $k$ is necessary when we're talking about independence of sigma-algebras generated from more than two $\pi$-systems, for the reason that you discovered. (The other question that you linked should have required the hypothesis. Note that the accepted answer does assume that $\Omega$ is among the sets for which the independence holds.)
When there are two $\pi$-systems, then the assertion $P(\Omega\cap H)=P(\Omega)P(H)$ holds vacuously, so including the hypothesis would be unnecessary.
If you look closely at the proof of Lemma 1.6 in Williams' Appendix A, you'll see that the requirement $\mu_1(\Omega)=\mu_2(\Omega)$ is necessary before we can apply Dynkin's lemma.