Countable Chain Condition How to prove the following statement:

For any class of sets $\{A_\alpha\}\in \mathcal{P}^2(\mathbb{R})$ such that $A_\alpha \cap A_\beta$ has a zero Lebesgue measure for all $\alpha \neq \beta$, there are at most countable number of sets in $\{A_\alpha\}$ having nonzero Lebesgue measure.

Perhaps the requirement $\{A_\alpha\}\in \mathcal P^2 (\mathbb{R})$ can be dropped.
 A: The standard argument goes by assuming otherwise, and then noting that for some $n\in\mathbb Z$, $\mathcal A=\{\alpha:A_\alpha\cap[n,n+1]$ has nonzero Lebesgue measure$\}$ is still uncountable. The point of this move is that the sets $B_\alpha=A_\alpha\cap[n,n+1]$ for $\alpha\in\mathcal A$ are bounded (in fact, they are all contained in an interval of length 1). 
Now argue that for some $m\in\mathbb N$, $\mathcal B=\{\alpha\in\mathcal A:B_\alpha$ has Lebesgue measure $>1/m\}$ is again uncountable. In fact, it is enough that for some $m$ this set is infinite. This is a contradiction, since the union of any $m+1$ of the $B_\alpha$ with $\alpha\in\mathcal B$ has measure larger than $1$ but is contained in $[n,n+1].$
A: Let the index set for the class $\{A_\alpha\}$ be $I$. Define, for nonempty finite $J\subset I$, $B_J:=\cup_{\alpha\in J}A_\alpha$. Let $\mu$ be the standard normal distribution on $\Bbb R$. (The salient features of $\mu$ are that it is a probability measure on $\Bbb R$ equivalent to Lebesgue measure.) Let $c:=\sup_J\mu(B_J)$, the supremum extending over all finite subsets of $I$. Choose a sequence $J_1\subset J_2\subset\cdots\subset I$ of finite subsets of $I$ such that $\lim_n\mu(B_{J_n})=c$. Suppose $\beta$ is an element of the complement in $I$ of the countable set $K:=\cup_n J_n$. Then 
$$
c\ge\mu(A_\beta\cup \left(\cup_{\alpha\in K}A_\alpha\right))=\mu(A_\beta)+\mu(\cup_{\alpha\in K}A_\alpha)=\mu(A_\beta)+c,
$$
forcing $\mu(A_\beta)=0$. It follows that $A_\beta$ is Lebesgue null. (The additivity in the above display results because the intersection of $A_\beta$ and $\cup_{\alpha\in K}A_\alpha$ is Lebesgue null.)
