Let $m$ be Lebesgue measure on $\mathbb R$, and let $m\_i$ and $m\_o$ be the inner and outer measures respectively. 



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> Is it the case that for all $A \subset \mathbb R$ and all $x \in [m\_i(A), m\_o(A)]$ there exists a countably additive extension $m^+$ of $m$ to the powerset of $\mathbb R$ such that $m^+(A)=x$?
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A result of Solovay's says that the existence of a measurable cardinal is equiconsistent with the weaker assertion that 



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> There exists a countably additive extension of $m$ to the powerset of $\mathbb R$.
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So an affirmative answer to my question will need to assume the existence of a measurable cardinal. Is anything more needed?


I'll note that if we allow $m^+$ to be merely finitely additive, then the answer to the question is affirmative without any large cardinal assumptions.



On the existence of a family of countably additive extensions of Lebesgue measure
Suppose $\kappa$ is the least real-valued measurable cardinal and $\nu:\mathcal{P}(\kappa) \to [0, 1]$ is a witnessing $\kappa$-additive probability measure. Gitik and Shelah showed that the Maharam type of the measure algebra $\mathbb{M}$ of $\nu$ is $\geq \kappa^+$ (See Theorem 2.6 in Gitik-Shelah, Forcing with ideals and simple forcing notions, Israel J. Math. 68 (1989), 129-160 or Theorem 3G in Fremlin's survey on real-valued measurable cardinals [here](https://www1.essex.ac.uk/maths/people/fremlin/rvmc.pdf)). In particular, forcing with $\mathbb{M}$ adds $\geq \kappa^+$ random reals. Let $G$ be an $\mathbb{M}$-generic filter over $V$. Then $V[G]$ has a set $X \subseteq [0, 1]$ of size $\kappa$ all of whose Lebesgue null subsets are countable. In $V[G]$, let $N$ be the transitive collapse of the well founded $G$-ultrapower of $V$ and suppose $j: V \to N \subseteq V[G]$ is the generic ultrapower embedding with critical point $\kappa$ (See Foreman's chapter in Handbook of Set Theory for background on generic utrapowers). Since the null ideal of $\nu$ is countably saturated, $N$ contains every $\kappa$-sequence of ordinals from $V[G]$ (Proposition 2.14 in Foreman's chapter in Handbook of Set Theory). In particular, $X \in N$. Since $N$, $V[G]$ have the same reals, $N \models$ "$X$ is Lebesgue non null and every Lebesgue null subset of $X$ is countable". Therefore $N \models$ "There is a Lebesgue non null set of cardinality $\aleph\_1$". By elementarity of $j$, $V \models$ "There is a Lebesgue non null set of cardinality $\aleph\_1$". For a forcing-free proof of this, see Proposition 6F in Fremlin's survey linked above.


So we can fix $A \subseteq [0, 1]$ of positive Lebesgue outer measure such that $|A| = \aleph\_1$. Let $\mu:\mathcal{P}([0, 1]) \to [0, 1]$ be a total extension of Lebesgue measure. Since $\kappa$ is the least real-valued measurable cardinal, the additivity of the null ideal of $\mu$ is at least $\kappa$. Now $\kappa$ is weakly inaccessible (and much more) and so $\kappa > \aleph\_1$. It follows that $\mu(A) = 0$. Hence no total extension of Lebesgue measure can assign positive measure to $A$.


