I am currently taking a course on probability theory for mathematicians where we're doing some measure theory. I've been thinking about how it is that if $\mathcal{L}$ is a $\lambda$-system and a $\pi$-system, then $\mathcal{L}$ is a $\sigma$-algebra (the converse is very straightforward).

Before explaining what I mean, I should point out that we have taken the following definition of $\lambda$-system. $\mathcal{L} \in \mathcal{P}(\Omega) $ is a $\lambda$-system iff

  1. $\Omega \in \mathcal{L}$
  2. $A, B \in \mathcal{L}$ and $A \subseteq B \Rightarrow B \setminus A \in \mathcal{L}$
  3. $A_1, A_2, \ldots \in\mathcal{L}$ such that $A_n \uparrow A \Rightarrow A \in \mathcal{L}$

Suppose $\mathcal{L}$ is a $\lambda$-system. For $\mathcal{L}$ to be a $\sigma$-algebra, in addition to the easy-to-check fact that for any set in $\mathcal{L}$, its complement is also in $\mathcal{L}$, the enumerable union of an arbitrary collection of sets in $\mathcal{L}$ must also be in $\mathcal{L}$. So I started by taking two arbitrary sets.

Let $A, B \in \mathcal{L}$. Suppose $D_1 = A, D_j = A \cup B, \forall j \in \mathbb{N}-\{1\}$. Then $D_1 \subseteq D_2 \subseteq D_3 \subseteq \ldots $ and clearly $\cup_{j \in \mathbb{N}} D_j = A \cup B$. This would mean that $D_j \uparrow (A \cup B)$, so property $(3)$ above would imply that $A \cup B \in \mathcal{L}$.

But then if I already had this for two sets, I could generalize for finite unions. Moreover, if $A_1, A_2, \ldots \in L$, then $$\bigcup\limits_{j=1}^n A_j \uparrow \bigcup\limits_{j=1}^\infty A_j. $$ Again, property $(3)$ would imply that $\bigcup\limits_{j=1}^\infty A_j \in \mathcal{L}$.

Clearly there is something wrong, since not all $\lambda$-systems are $\sigma$-algebras. I would really appreciate that you point out any mistakes in the reasoning above.

  • 1
    $\begingroup$ Your $D_j$ are equal to $A\cup B$ (second paragraph after definition); how do you know $D_j$ is in $\mathcal{L}$? You seem to be trying to use the $D_j$ to prove that $A\cup B$ is in $\mathcal{L}$, but you seem to be assuming that it is. $\endgroup$ May 1, 2020 at 2:23

1 Answer 1


Your error is in your argument that if $A,B\in\mathcal{L}$ then $A\cup B\in\mathcal{L}$.

You define $D_j=A\cup B$ for $j\geq 2$. Then you try to apply (3) for $\cup D_j$. But in order to apply (3) you need to know that $D_j\in\mathcal{L}$ for all $j$, in particular, you need to know that $A\cup B\in\mathcal{L}$... which is what you are trying to prove.

For an example of a $\lambda$-system that is not closed under finite unions, take $\Omega=\{1,2,3,4,5\}$, and let $\mathcal{L}$ consist of:

  1. $\Omega$;
  2. $\{1,2,3\}$;
  3. $\{2,3,4\}$;
  4. $\{4,5\}$;
  5. $\{1,5\}$;
  6. $\varnothing$.

It contains $\Omega$. The case where we have $A\subseteq B$ is if $B=\Omega$, in which case we just get the complement; of $A=\varnothing$, in which case we just get $B$; or if $A=B$, in which case we get $\varnothing$. The complement of (2) is (4), the complement of (3) is (5), the complement of (1) is (6). So the system satisfies the second property. Similarly, the only increasing sequences have exactly one term, or else they are one set and $\Omega$, one set and the empty set, or the empty set, a set, and $\Omega$; in any case, the unions are in the system. Thus, this is a $\lambda$-system.

If you try to use your argument to show that $\{1,2,3\}\cup\{2,3,4\}\in\mathcal{L}$, you’ll notice that while $D_1\in\mathcal{L}$, none of $D_k=\{1,2,3,4\}$ are in $\mathcal{L}$. You were assuming what you wanted to prove in that step.

  • $\begingroup$ Thanks for your answer! I had proven that for $A, B \in \mathcal{L}$ that are mutually disjoint, $A \cup B \in \mathcal{L}$ via other method. But then I thought that my rationale above could generalize it for arbitrary $A, B$. Anyway, missed that condition due to my enthusiasm I guess. $\endgroup$ May 3, 2020 at 1:16
  • $\begingroup$ @NicolasGutierrez: Oops; I was missing the empty set in the list of elements (obtained by looking at the complement of $\Omega$ in itself). $\endgroup$ May 3, 2020 at 1:18

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