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?



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.

  1. $\Omega\in \mathcal{L}$
  2. $A\in \mathcal{L} \Rightarrow A^c\in \mathcal{L}$
  3. 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:

  1. 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.

  • $\begingroup$ Thanks! How do you know "You know that L is a λ-system"? $\endgroup$ – Tim Jan 14 '13 at 18:03
  • $\begingroup$ @Tim Have you read my answer? See what I wrote after: "If you look at the proof for $\mathcal{L}$ being a $\lambda$-system". You see exactly why $\mathcal{L}$ is a $\lambda$ system in case of two probability measures. $\endgroup$ – math Jan 14 '13 at 18:17
  • $\begingroup$ My question is how $L$ being $\lambda$ system can be proved from $\lambda-\pi$ theorem? What follows "If you look at the proof for L being a λ-system" isn't a proof for $L$ being $\lambda$ system, is it? $\endgroup$ – Tim Jan 14 '13 at 18:25
  • $\begingroup$ @Tim You did not understand what Nate Eldrege said / you wrote down! The statement, which Nate Eldrege refers to is: You know that $\mathcal{L}$ is $\lambda$ system. If $P,Q$ agree on a $\pi$ system, which generates $\mathcal{F}$, then the two measure coincide. Furthermore, I showed generally when $\mathcal{L}$ is a $\lambda$-system (just use, if $P,Q$ are prob. measure, $P(\Omega)=1=Q(\Omega), P(A^c)=1-P(A)=1-Q(A)=Q(A^c)$ and as I said, point 3. is jsut $\sigma$ additivity. $\endgroup$ – math Jan 14 '13 at 18:29
  • $\begingroup$ Nate said "It is a good exercise to verify that L is a λ-system. This is a common application of the π-λ theorem", which means it is a good exercise to verify L being a λ-system, by application of the π-λ theorem. In other words, given the π-λ theorem, L being a λ-system can be proved, but I don't know how, which is my question. $\endgroup$ – Tim Jan 14 '13 at 18:32

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.


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