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


2 Answers 2


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.

  • $\begingroup$ correct me if I'm wrong, but L doesn't appear to be closed under relative complement... $\endgroup$ Commented Oct 28, 2010 at 15:49
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    $\begingroup$ I think you only need relative complement of included sets? I.e., $A \subseteq B \Rightarrow B \setminus A \in L$. $\endgroup$
    – Neil G
    Commented Oct 28, 2010 at 16:07
  • $\begingroup$ @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. $\endgroup$ Commented Oct 28, 2010 at 16:35
  • $\begingroup$ 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. $\endgroup$
    – Neil G
    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.)


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