It's a known fact (see for example Bartoszynski and Judah: Set Theory, On the Structure of the real Line) that for a filter $F$ on $\omega$ the following conditions are equivalent:
$F$ is Lebesgue measurable.
$\exists$ a family $\{A_n: n\in \omega \}$ such that
- $A_n$ consists of finitely many finite subsets of $\omega$ $\forall n \in \omega$
- $\cup A_n {}\cap{} \cup A_m = \emptyset$ if $n\neq m$
- $\sum_{n=1}^{\infty}\ \mu(\{X \subseteq \omega: \exists a \in A_n$ $a \subseteq X\}) < \infty$ where $\mu$ denotes the Lebesgue measure
- $\forall X \in F$ $\exists^{\infty} n$ $\exists a \in A_n$ such that $a\subseteq X$
Let's define $B_n = \{a \in [\omega]^n: \exists k$ $a\in A_k\}$
How can we deduce, that all sets $B_n$ are finite?
Thank you in advance for your help!
I suppose by Lebesgue measure you mean the standard product measure on $\mathcal{P}(\omega)$ (identified with $2^\omega$). If $A_n$ contains a singleton $\{i\}$, then
$$\mu\{X\subseteq\omega:\exists a\in A_n\ a\subseteq X\}\geq\mu\{X\subseteq\omega:i\in X\}=\frac{1}{2}$$
So there cannot be infinitely many $A_n$ that contain singletons, since otherwise the sum would be infinite. Similarly there cannot be infinitely many $A_n$ that contain $2$-sets, $3$-sets, etc.