The set of measurable subsets where two measures agree From Nate Eldredge's reply

Let $(\Omega, \mathcal{F})$ be a measurable space, and let $P,Q$ be two probability measures on $\mathcal{F}$.  It is a good exercise to verify that $$\mathcal{L} := \{ A \in \mathcal{F} : P(A) = Q(A) \}$$ is a $\lambda$-system.  (This is a common application of the $\pi$-$\lambda$ theorem : if one can show that $P$ and $Q$ agree on a $\pi$-system that generates $\mathcal{F}$, then $P$ and $Q$ must be the same.)

I can't see and therefore was wondering how $L$ being a $\lambda$ system is an application of $\pi$-$\lambda$ theorem, i.e. $L$ being a $\lambda$ system can be proved from $\pi$-$\lambda$ theorem?
In order for $L$ to be a $\lambda$ system, can $P$ and $Q$ be not necessarily probability measures, but $\sigma$-finite, or arbitrary?
Thanks!
 A: The usual application is that if two probability measures agree on a $\pi$-system, then they agree on the $\sigma$-algebra generated by that $\pi$-system and the reason is that the family of sets where the measures agree form a $\lambda$-system as Nate mentioned. 
It is not possible to weaken the assumption to $\sigma$-finiteness: Lebesgue measure $\lambda$ and twice Lebesgue measure $2\lambda$ agree on the sets of the form $(r,\infty)$ and these form a $\pi$-system.
A: Suppose $P,Q$ agree on a $\pi$ System $\mathcal{C}\subset\mathcal{F}$, with $\sigma(\mathcal{C})=\mathcal{F}$. That means $P(C)=Q(C)\forall C\in\mathcal{C}$. You know that $\mathcal{L}$ is a $\lambda$-system and contains the $\pi$-system $C$. Hence by the $\lambda-\pi$ Theorem, 
$$P(B)=Q(B)\forall B\in \sigma(\mathcal{C})=\mathcal{F}$$
They agree on $\mathcal{F}$, i.e. they are the same. 
To your second question: You must still guarantee that $\mathcal{L}$ is a $\lambda$ System, i.e. 


*

*$\Omega\in \mathcal{L}$

*$A\in \mathcal{L} \Rightarrow A^c\in \mathcal{L}$

*For every sequence $(A_i)_{i\ge 1}$ of pairewise disjoint elements of $\mathcal{L}$, you have $(\bigcup_{i\ge 1}A_i)\in \mathcal{L}$


If you look at the proof for $\mathcal{L}$ being a $\lambda$-system:


*

*is true because $P(\Omega)=1=Q(\Omega)$. Generally, you must have two finite measures $P,Q$ with $P(\Omega)=Q(\Omega)$. Same for 2., since you are using $P(A^c)=P(\Omega)-P(A)$, which again forces $P,Q$ to be finite measures with $P(\Omega)=Q(\Omega)$. 3. is just the $\sigma$ additivity and is true for every measure. 


This means that in general a $\sigma$-finite measure does not work.
