There is a finitely additive, translation invariant measure on the entire power set of $\mathbf C$ (see, for example, Amenability, the ping-pong lemma, and the Banach–Tarski paradox). Clearly, it satisfies your desired condition (1). Since half-open ‘intervals’ in $\mathbf C$ are finite, disjoint unions of $1/n$-equal parts that are intervals for various $n$, and generate the Borel $\sigma$-algebra, and since Lebesgue (I would rather say Haar) measure assigns measure $1/n$ to a $1/n$-equal part that is an interval, we have that any countably additive extension as in (2) extends Haar measure. I suspect that a Vitali-type argument shows that no extension as in (2) exists, but I do not currently see it.

There is a finitely additive, translation invariant measure on the entire power set of $\mathbf C$ (see, for example, Amenability, the ping-pong lemma, and the Banach–Tarski paradox). Clearly, it satisfies your desired condition (1). Since rational-length, half-open ‘intervals’ in $\mathbf C$ are finite, disjoint unions of $1/n$-equal parts that are intervals for various $n$, and generate the Borel $\sigma$-algebra, and since Lebesgue (I would rather say Haar) measure assigns measure $1/n$ to a $1/n$-equal part that is an interval, we have that any countably additive extension as in (2) extends Haar measure. I suspect that a Vitali-type argument shows that no extension as in (2) exists, but I do not currently see it.

The Hahn–Banach theorem shows that there is a finitely additive, translation invariant measure on the entire power set of $\mathbf C$. Clearly, it satisfies your desired condition (1). Since half-open ‘intervals’ in $\mathbf C$ are finite unions of $1/n$-equal parts that are intervals for various $n$, and generate the Borel $\sigma$-algebra, since Lebesgue (I would rather say Haar) measure assigns measure $1/n$ to a $1/n$-equal part that is an interval, we have that any countably additive extension as in (2) extends Haar measure. I suspect that a Vitali-type argument shows that no extension as in (2) exists, but I do not currently see it.

There is a finitely additive, translation invariant measure on the entire power set of $\mathbf C$ (see, for example, Amenability, the ping-pong lemma, and the Banach–Tarski paradox). Clearly, it satisfies your desired condition (1). Since half-open ‘intervals’ in $\mathbf C$ are finite, disjoint unions of $1/n$-equal parts that are intervals for various $n$, and generate the Borel $\sigma$-algebra, and since Lebesgue (I would rather say Haar) measure assigns measure $1/n$ to a $1/n$-equal part that is an interval, we have that any countably additive extension as in (2) extends Haar measure. I suspect that a Vitali-type argument shows that no extension as in (2) exists, but I do not currently see it.