A $\lambda$-system $\mathcal{L}$ that is a $\pi$-system is automatically a $\sigma$-field.

Question

Let $\Omega$ be a set. Let $\mathcal{L}$ be a $\lambda$-system, that is:

1. $\Omega \in \mathcal{L}$.
2. $A \in \mathcal{L} \implies A^c \in \mathcal{L}$.
3. $A_n \in \mathcal{L}, n\geq 1$ and $A_m \cap A_n = \varnothing$ when $n \neq m \ \implies \cup_{n} A_n \in \mathcal{L}$.

A $\pi$-system just means $\mathcal{L}$ is also closed under finite intersection.

A $\sigma$-field is a set of subsets of $\Omega$ that contains $\varnothing$ and is closed under complement and countable union.

Clearly, I only have to show the last property (countable union) as the first two are immediate from the definitions.

Let $A_n \in \mathcal{L}, n \geq 1$ be a countable collection of sets in $\mathcal{L}$. I've tried this out:

$B_n = A_n \setminus (\cup_{n \neq m} A_m)$ satisfies property (3) but doesn't lead to a proof, nor does $\cup A_n = \cup B_n$.

Yeah... sort of stuck here.

Answer 1

If $\{E_n\}$ is a countable collection in $\mathcal{L}$, then define the sets \begin{align*} F_1 &\doteq E_1 \\ F_n &\doteq E_n \backslash (E_1 \cup E_2 \cup \dots \cup E_{n-1}) \end{align*} Then $\{F_n\}$ are disjoint and $\cup_1^{\infty} E_n = \cup_1^{\infty} F_n$. We also have each $F_n \in \mathcal{L}$ since $\mathcal{L}$ is closed under complementation and finite intersection. Hence, $\cup_1^{\infty} E_n \in \mathcal{L}$.