There exists no map $\mu: \mathcal{P}(\mathbb{R})\rightarrow [0,\infty]$ satisfying:
(1) $\mu ((a,b])=b-a$ for all $a,b\in\mathbb{R}$ (this is what I mean by non-trivial in the title),
(2) Translation invariance,
(3) Countable additivity.
We call one such map a measure on $\mathcal{P}(\mathbb{R})$. A reason for its non-existence lies in the properties of the Vitali set $V$, that imply that $V$ cannot belong in any $\sigma$-algebra on which a measure satisfying (1), (2) and (3) exists.
As a relaxation of (3) consider
(3)* Finite additivity.
My question is whether a map $\mu: \mathcal{P}(\mathbb{R})\rightarrow [0,\infty]$ satisfying (1), (2) and (3)* exists.
Let's call this a finitely additive measure on $\mathcal{P}(\mathbb{R})$. If one such map $\mu$ existed then the Vitali set $V$ would have to satisfy $\mu(V)=0$.
A non-exitence proof here could be something along the lines of the Banach Tarski Paradox, but in $\mathbb{R}$.
Bonus question: What about a map $\mu: \mathcal{P}(\mathbb{R})\rightarrow [0,\infty]$ satisfying (1) and (3)?
Finally, I wonder what's so great about countable additivity that someone decided that all measures should satisfy it.
Related results:
If two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system) and satisfy that $\mu_1(X)=\mu_2(X)<\infty$, then $\mu_1=\mu_2$ on the $\sigma$-algebra generated by $\mathcal{A}$ (by the $\pi-\lambda$ Theorem). For example two finite measures that agree on all open sets also agree on all Borel sets.
[Generalization of the above] Suppose two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system). Moreover suppose there exists a countable nested subfamily of $\mathcal{B}\subseteq \mathcal{A}$ of sets that cover $X$ and have finite $\mu_1$-(and $\mu_2$-) measure. Then $\mu_1=\mu_2$ in the $\sigma$-algebra generated by $\mathcal{A}$ (again using the $\pi - \lambda$ Theorem). So for example the Lebesgue measure in all Borel sets in $\mathbb{R}^n$ is uniquely determined by its values on boxes/intervals.
Up to a multiplicative constant, Lebesgue measure is the only translation-invariant measure on the Borel sets that puts finite measure on the unit interval. This can be generalised to higher dimensions considering the unit box $[0,1]^n$ instead of the unit interval. You can find proofs here using, you guessed it, the $\pi - \lambda$ Theorem. The proof using the equivalent Monotone Class Theorem also seems quite straightforward here.
Let $\mu_1$ and $\mu_2$ be $\sigma$-finite measures on a measurable space $(X,\mathcal{S})$, $\mathcal{A}$ an algebra which generates $\mathcal{S}$ and suppose that, for each $A\in\mathcal{A}$, $\mu_1(A)=\mu_2(A)$. Then $\mu_1(B)=\mu_2(B)$ for each $B\in\mathcal{S}$ (still haven't seen a proof for this).
In relation to the bonus question, by the second (or third) related result any measure $\mu$ satisfying (1) is equal to the Borel measure on all Borel sets. And also, if all Lebesgue measurable sets are $\mu$-measurable, equal to the Lebesgue measure on said sets, since the completion of a measure space is unique. So the question really has to do with whether the Lebesgue measure can be extended to a measure on all of $\mathcal{P}(\mathbb{R})$ (see t.b.'s comment here).
