# What is an example of a lambda-system that is not a sigma algebra?

What is an example of a lambda-system that is not a sigma algebra?

Here is a somewhat more natural example.

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.)

However, $\mathcal{L}$ need not be a $\sigma$-algebra. For instance, consider a sample space consisting of two coin flips: $$\Omega = \{ HH, HT, TH, TT \}, \quad \mathcal{F} = 2^\Omega.$$ Let $P$ be the probability measure under which the coins are independent and unbiased, and let $Q$ be the measure under which the first coin is unbiased but the second coin is stuck to the first so that they always come up the same. Explicitly, $$P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}$$ $$Q(HH)=Q(TT)=\frac{1}{2}; \quad Q(HT)=Q(TH)=0.$$ Then one can check that the events on which $P$ and $Q$ agree are those which only look at one of the coins (or none), so that $$\mathcal{L} = \{ \{HH,HT\}, \{HH,TH\}, \{HT,TT\}, \{TH,TT\}, \emptyset, \Omega\}.$$ This is not a $\sigma$-algebra since it is not closed under intersections.

• correct me if I'm wrong, but L doesn't appear to be closed under relative complement... Commented Oct 28, 2010 at 15:49
• I think you only need relative complement of included sets? I.e., $A \subseteq B \Rightarrow B \setminus A \in L$. Commented Oct 28, 2010 at 16:07
• @Neil G: Right, and note that there are no nontrivial inclusions among the sets of $\mathcal{L}$. That's also why the "increasing unions" axiom is satisfied. Commented Oct 28, 2010 at 16:35
• Even though both answers are correct, I'm going to mark this answer because it gives me a bit of intuition about what's going. The other answer would have been a better solution to an exam question. Commented Oct 28, 2010 at 17:56

For another example, let $$(\Omega, \mathcal{F}, P)$$ be a probability space and fix some event $$A \in \mathcal{F}$$. Let $$\mathcal{L}$$ be the collection of all events which are independent of $$A$$, i.e. $$\mathcal{L} = \{ B \in \mathcal{F} : P(A \cap B) = P(A) P(B)\}.$$ It is not hard to check that $$\mathcal{L}$$ is a $$\lambda$$-system. To see it need not be a $$\sigma$$-algebra, take as in my other answer a probability space $$\Omega = \{HH, HT, TH, TT\}$$, $$\mathcal{F} = 2^\Omega$$, $$P(A) = \frac{1}{4} |A|$$ consisting of two independent fair coin flips. Set $$A = \{HH, HT\}$$, the event that the first coin is heads. Then $$\{HH, TH\}, \{HH, TT\}$$ are in $$\mathcal{L}$$ but their union $$\{HH, TH, TT\}$$ is not.

Incidentally, this is really of the same form as my other answer if we take $$Q$$ to be the conditional probability measure $$Q(E) = P (E \mid A) = P(E \cap A)/P(A)$$. (Except when $$P(A)=0$$, but that case is trivial.)