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A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC.

(2)+Countable Choice for Finite Sets (CCFS) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CCFS (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). Since DC implies CCFS, (2) is false in that model.

A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC.

(2)+Countable Choice for Finite Sets (CC(fin)) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CC(fin) (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). Since DC implies CC(fin), (2) is false in that model.

A different way to see that (2) doesn't follow from ZF is this. Take a model where there is an infinite Dedekind-finite set but CC(fin) still holds (the Consequences of AC website lists a bunch of models where Form 10 is true and Form 9 is false).

Now if $\Omega$ is an infinite Dedekind-finite set and $\mu$ satisfies (2), then every point of $\Omega$ will have to be an atom of $\mu$ (or else $\mu(\Omega - \{ x \}) = \mu(\Omega)$). But by CC(fin), the set of atoms is countable, which is impossible for a Dedekind-finite set.

I'm curious if (2) implies CC(fin).

A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC.

(2)+Countable Choice for Finite Sets (CCFS) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CCFS (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). So (2) is false in that model.

A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC.

(2)+Countable Choice for Finite Sets (CCFS) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CCFS (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). Since DC implies CCFS, (2) is false in that model.

A quick note that (2) doesn't follow from ZF (assuming ZF is consistent).  (2)+Countable Choice for Finite Sets (CCFS) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CCFS (there are at most $n$ $x$s such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). So (2) is false in that model.

A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC. 

(2)+Countable Choice for Finite Sets (CCFS) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CCFS (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). So (2) is false in that model.