Does the following condition implies full outer measure? Let $X \subseteq 2^{\omega}$ be a set of positive Lebesgue measure. Suppose that for every $\eta, \nu \in 2^{<\omega}$ of the same length, the measure of $X$ above $\eta$ is the same as the measure of $X$ above $\nu$.
Is it necessarily true that the outer measure of $X$ is $1$?
If the answer is positive, do we need countable choice for the proof?
 A: Yes. Suppose not, look at $X^c$ which is of positive measure. By Lebesgue Density Theorem, there exists $\sigma, \tau\in 2^{\omega}$ (WLOG might assume they have the same length) such that $X$ and $X^c$ has measure $>\frac{1}{2}$ above $\tau, \sigma$ respectively. By hypothesis, $X$ has measure $>\frac{1}{2}$ above $\sigma$ too. But then above $\sigma$, $X$ intersects with $X^c$ which is absurd. I don't think AC is required in the proof of Lebesgue density theorem (at least in the setting of Cantor space).
A: The measure theoretic part was answered, so let me complement it by answering the choice related question. The axiom of countable choice is needed on a far more fundamental level when you talk about measure theory. 
It is consistent that the real numbers are a countable union of countable sets. In that case there is no $\sigma$-additive Borel measure (except the $0$ measure), not even on the Borel sets. 
But countable choice ensures that countable unions of countable sets are countable, and that we're fine. Of course, you could still do measure theory in the absence of any choice by working with sets that have Borel codes (really countable choice ensures that every Borel set has a code), but this is just terrible for your health. You can look in Fremlin's "Measure Theory" (vol. 5) for more information on measure theory without choice. 
