Vtyjkyuk

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 $(Omega, mathcalF, P)$ be a probability triple. Let $mathcalI_1$, $mathcalI_2$, and $mathcalI_3$ be three $pi$-systems on $Omega$ such that for $k=1,2,3$,
$$mathcalI_KsubseteqmathcalFquad textand quadOmegain mathcalI_k.$$
Prove that if $$P(I_1cap I_2cap I_3) = P(I_1) cdot P(I_2) cdot P(I_3)$$ whenever $I_kin mathcalI_k$ (k=1,2,3), then $sigma(mathcalI_1)$, $sigma(mathcalI_2)$, and $sigma(mathcalI_3)$ are independent. Why did we require the $OmegainmathcalI_k$?

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 $OmegainmathcalI_k$ for $k=1,2$. So I am confused by his assumption in E4.1, and question as to "Why did we require the $OmegainmathcalI_k$?" 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 (again, this is just a mockery if a previous Williams Lemma)...

Fix $I_1inmathcalI_1$ and $I_2inmathcalI_2$. Consider the maps
$$ J_3 mapsto P(I_1cap I_2cap J_3) quad textandquad J_3mapsto P(I_1)cdot P(I_2)cdot P(J_3),$$
for $J_3insigma(mathcalI_3)$. One can verify that these are measures on the measure space $(Omega, sigma(mathcalI_3))$, 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
$$P(I_1cap I_2)quad textandquad P(I_1)cdot P(I_2),$$
respectively. If $Omeganotin mathcalI_3$, 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 $OmegainmathcalI_k$ for $k=1,2$.

This begs the following question...is the requirement that $Omegain mathcalI_k$ 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 $OmegainmathcalI_k$ for $k=1,2$ for that proof).

Regarding your new question: The hypothesis $Omegaincal I_k$ 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 $P(Omegacap H)=P(Omega)P(H)$ 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.

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.