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 $\kappa > 2^\omega$?



Finitely additive, $\kappa$-additive atomless measures in ZFC
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.


