David Williams "Probability with Martingales" Exercise 4.1

Question

Let me preface this by saying that basically the same question has been asked before on MathStackExchange. 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:

---

Exercise 4.1: 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 appears that one can simply mock the proof directly from the lemma on page 39 in his book where $k=2$. However, in that lemma, there was no assumption that $\Omega\in\mathcal{I}_k$ for $k=1,2$. Therefore, I am confused by the assumption in E4.1 and the last question in the exercise. Is this simply a callously worded exercise or am I missing something? The proof I have is as follows (mimicking a previous lemma in Williams):

Proof of Exercise 4.1: Fix $I_1\in\mathcal{I}_1$ and $I_2\in\mathcal{I}_2$. Consider the mappings $$ J_3 \mapsto P(I_1\cap I_2\cap J_3) \quad \text{and}\quad J_3\mapsto P(I_1)\cdot P(I_2)\cdot P(J_3),$$ 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! As I was typing the "..." above, I realized something. Recall that 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 here 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. 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 same question in this post does not require this assumption. 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 they did not require that $\Omega\in\mathcal{I}_k$ for $k=1,2$ for that proof).

Answer 1

What if $\mathcal I_3$ consists of just the empty set. Then hypothesis holds for any $\mathcal I_1$ and $\mathcal I_2$ ; $\mathcal I_1$ and $\mathcal I_2$ need not be independent.

Answer 2

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.