Finitely additive, $\kappa$-additive atomless measures in ZFC Under Martin's Axiom (and non-CH) the Lebesgue measure is $2^\omega$-additive in the sense that unions of fewer than continuum ($2^\omega$) many null sets are measureable and null. In ZFC we may however extend the Lebesgue measure to a finitely-additive measure on the power set of $[0,1]$ and still call it atomless.
Are there ZFC examples of finitely-additive measures that extend  an atomless measure $\mu$ and are $\kappa$-additive in the above sense w.r.t $\mu$-null sets, where $\kappa > 2^\omega$?
 A: I claim that a non-atomic measure $\mu$ can never be $<{2^\omega}^+$-additive. Then the same applies to any finitely-additve extension.
Let $(\Omega, \frak{A}, \mu)$ be a measure space and let us assume that $\mu$ is non-atomic. It follows that there exists $A \in \frak{A}$ such that $0 < \mu(A) < \infty$. I now want to partition $A$ into $2^\omega$ many null sets. Start by splitting $A$ into $A_0$ and $A_1$ both are sets in $\frak{A}$ of positive measure. This can be done, since $\mu$ is non-atomic. Assume that $A_s$ for $s\in 2^{<\omega}$ has been defined and partition it into $A_{s^\frown 0}$ and $A_{s^\frown 1}$. For every $x \in \,^{\omega}2$ define $A_x:= \bigcap_{n < \omega} A_{x \restriction n}$, which is the first limit step.
First note that $\mu(A_x)=\inf_{n < \omega} \mu(A_{x \restriction n})$ and that some $A_x$ may already have measure 0, while others may still have positive measure. If $A_x$ has measure 0, then we do not have to take care of it anymore. If $A_x$ has positive measure, we continue as before and split it into $A_{x^\frown 0}$ and $A_{x^\frown 1}$. This way we get a (transfinite) binary tree, such that some branches die out at limit stages.
I claim that every branch dies out at countable height, i.e. there is no $\omega_1$-branch. If not, there exists a $\subseteq$-decreasing sequence $(A_\alpha)_{\alpha < \omega_1}$ of length $\omega_1$ such that $\mu(A_\beta) < \mu(A_\alpha)$ if $\beta > \alpha$. But this is impossible, since there cannot exist an uncountable decreasing sequence in $\mathbb{R}$ (separability). Therefore, there are only $2^\omega$ branches and so $A=\bigcup_{\alpha < 2^\omega} A_\alpha$.
This and (much) more can be found in Jech's book in chapter 10.
