Proving two outer measures defined on the unit circle $C$ are the same measure. Denote by $H^1(A)$ the 1-dimensional Hausdorff measure defined on $\mathbb{R}^n$. Now take $C$ to be the unit circle in $\mathbb{R}^2$. Now define an outer measure $\mathcal{H}$ on $C$ by
$$
\mathcal{H}(A):= \frac{1}{2\pi}H^1(A)
$$
Where $A \subset C$.
I am asked to prove that $\mathcal{H}$ is the only outer measure on $C$ which satisfies the following 3 conditions:


*

*$\mathcal{H}(C) = 1$.

*$\mathcal{H}$ is Borel regular.

*$\mathcal{H}$ is invariant under rotations.


I've already managed to prove that $\mathcal{H}$ satisfies the conditions but I'm having trouble with proving the uniqueness. That is, that there exists no other outer measure $\mathcal{M}$ on $C$ which also satisfies the conditions.
 A: Suppose $\mu$ is a measure that satisfies the given conditions. Then $\mu$ is an honest measure when restricted to each open, half-open, or closed arc by regularity, and in particular, it is additive on these arcs. By translation-invariance and additivity, one may show that $\mu$ must assign to each arc of length $\frac{2\pi}{n}$ the measure $\frac{1}{n}$. By additivity, this implies that for any arc of length $\frac{2a\pi}{b}$ with $a,b\in \Bbb N$ and $a\leq b$, $\mu$ assigns this arc measure $\frac{a}{b}$.
From here, one can show that any open interval on $S^1$ must be mapped to it's length divided by $2\pi$ by $\mu$ by a least-upper-bound argument. Now you know the value of $\mu$ on a collection of subsets which generate the $\sigma$-algebra of Borel sets: this determines the measure, and in fact it must coincide with the given measure.
The reason it must coincide is that by regularity, every measurable set $A$ contains a Borel subset $B$ with $\mu(A)=\mu(B)$. Since the value of $\mu$ on each arc is known, we may write $B$ in terms of the $\sigma$-algebra operations on the arcs, apply the measure to both sides of this relation, and then we get the value for $B$.
