Suppose that $U\subseteq S^{1}$ is open where $S^{1}=\{z\in\mathbb{Z}:|z|=1\}$. Then define $\mu_{n}(U)=\max_{t\in S^{1}}\frac{1}{n}\cdot|\{k\in\{1,...,n\}|t\cdot e^{\frac{2\pi ik}{n}}\in U\}|$. Define $\mu(U)=\lim_{n}\mu_{n}(U)$ (here this limit is with respect to the divisibility ordering on the natural numbers. This limit converges since if $n|m$, then $\mu_{m}(U)\leq\mu_{n}(U)$). Of course, this measure can be defined on all subsets of $S^{1}$, but I am only interested in how this measure behaves on open sets.
$\textbf{Properties of the measure $\mu$}$
Let me now outline some interesting properties that this measure satisfies and some counterexamples as well along with proofs of why the counterexamples work. This measure has the property that $\mu(U)=\mu((\overline{U})^{\circ})$ for each open set $U$, so one can consider this algebra to be a measure on the Boolean algebra of regular open sets. Furthermore, we have $\mu(U\cup V)\leq\mu(U)+\mu(V)$. In particular, if $U,V$ are regular open, then $\mu(U\vee V)\leq\mu(U)+\mu(V)$ where $\vee$ denotes the least upper bound in the Boolean algebra of regular open sets. If $m$ denotes the Lebesgue probability measure, then it is easy to see that $m(U)\leq\mu(U)\leq m(\overline{U})$. In particular, if $m((\partial\overline{U})^{\circ})=0$, then $\mu(U)=m(\overline{U})$.
$\bullet$ There are open sets $U$ where $m(U)<\mu(U)$. For instance, if $C\subseteq S^{1}$ is a closed nowhere dense set with $m(C)>0$ and $U,V$ are disjoint open sets with $C=\partial U=\partial V$, then $m(U)+m(V)<1=\mu(S_{1})=\mu(U\cup V)\leq\mu(U)+\mu(V)$, so $m(U)<\mu(U)$ or $m(V)<\mu(V)$.
$\bullet$ There are open sets $U$ where $\mu(U)<m(\overline{U})$.
Let $S^{+}=\{e^{i\pi x}|x\in(0,\pi)\},S^{-}=\{e^{i\pi x}|x\in(-\pi,0)\}.$ Let $C\subseteq\overline{S}^{+}$ be a closed nowhere dense set with $1,-1\in C$ and $m(C)>1/4$. Let $R,S\subseteq\overline{S^{+}}$ be open subsets of $S^{1}$ where $\partial R=\partial S=C$.
Let $U=R\cup\{-x|x\in S\}$ and let $V=S\cup\{-x|x\in R\}$. Then $\mu_{n}(U)=\mu_{n}(V)\leq\frac{n}{2}-1$. Therefore $\mu(U)=\mu(V)\leq\frac{1}{2}$. However, we have $m(\overline{U})\geq\mu(C\cup\{-x|x\in C\})=2\mu(C)>2\cdot\frac{1}{4}=\frac{1}{2}$. Thus, $m(\overline{U})>\mu(U)$.
$\bullet$ $\mu$ is not finitely additive. Suppose that $C$ is a fat cantor set with $m(C)>\frac{1}{2}$ (i.e. $C$ is a closed nowhere dense subset). Let $U=S^{1}\setminus C$ and suppose that $U=\bigcup\mathcal{U}$ where $\mathcal{U}$ is a collection of disjoint open intervals. Let $C^{\sharp}=C\setminus\bigcup_{O\in\mathcal{U}}\partial O$. Then $m(C^{\sharp})=m(C)>\frac{1}{2}$. Now we shall select open sets $(A_{n,i})_{n>0,1\leq i\leq \lceil n/2\rceil}$ by induction on $n$. Suppose that we have selected open sets $A_{m,i}$ for $m<n$. Now there is some $x\in S^{1}$ where $|\{k\in\{1,...,n\}|x \exp(\frac{2\pi i k}{n})\in C^{\sharp}\}|>\frac{n}{2}.$ Now let $A=\{x\cdot\exp(\frac{2\pi i k}{n}|x\cdot\exp(\frac{2\pi i k}{n}\in C^{\sharp}\}$. Then there is some open neighborhood $V$ of $1$ such that $A\cdot U\cap\bigcup_{m<n,i\leq \lceil m\rceil}A_{m,i}=\emptyset$. Furthermore $\{y\in V|A\cdot y\subseteq U\}$ is dense in $V$. Therefore select $y\in V$ so that $A\cdot y\subseteq U$ and where each element in $A\cdot y$ is contained in a distinct element in $\mathcal{U}$. This can be achieved simply by taking $y$ to be sufficiently close to zero. Now let $(A_{n,i})_{i\leq\lceil n/2\rceil}$ be disjoint open sets in $\mathcal{U}$ containing members of $A\cdot y$. Then let $A_{n}=\bigcup_{i=1}^{\lceil n/2\rceil}A_{n,i}$. Then the sets $(A_{n})_{n\geq 1}$ are pairwise disjoint open sets. Let $B_{n}=A_{n!}$ for all $n\geq 1$. Then if $L\subseteq\mathbb{N}$ is an infinite subset, then $\mu(\bigcup_{n\in L}B_{n})>\frac{1}{2}$.
By replacing $\frac{1}{2}$ with $1-\epsilon$, one can show that for all $\epsilon>0$ one can construct open sets $(B_{n})_{n\in\mathbb{N}}$ such that whenever $L\subseteq\mathbb{N}$ is an infinite subset, then $\mu(\bigcup_{n\in L}B_{n})>1-\epsilon$. In particular, for each $\epsilon>0$, there are infinitely many pairwise disjoint open subsets of $S^{1}$ each with measure greater than $1-\epsilon$. Therefore the finite additivity of $\mu$ fails fairly badly.
$\bullet$ Finite additivity also fails in a much stronger sense. I claim that there are open sets $U,V$ with $\overline{U}\cap\overline{V}=\emptyset$ but where $\mu(U\cup V)<\mu(U)+\mu(V)$. Suppose that $T=\{e^{ix}|x\in[0,\pi/2]\}$. Then there are countably many pairwise disjoint open sets $A_{n}$ so that $A_{n}\subseteq T$ for all $n$ and where if $L\subseteq\mathbb{N}$ is an infinite subset, then $\mu(\bigcup_{n\in L}A_{n})>\frac{1}{5}$. Then let $L,M\subseteq\mathbb{N}$ be pairwise disjoint infinite subsets. Then let $U=\bigcup_{n\in L}A_{n}$ and let $V=-(\bigcup_{n\in M}A_{n})$. Then $U,V$ are open sets where $\mu(U)>\frac{1}{5}$ and $\mu(V)>\frac{1}{5}$, so $\mu(U)+\mu(V)>\frac{2}{5}$. However, for all $x\in S^{1}$, at most one of the elements $x,ix,-x,-ix$ is contained in $U\cup V$. Therefore, we have $\mu_{4}(U\cup V)=\frac{1}{4}$, so $\mu(U\cup V)\leq\frac{1}{4}<\frac{2}{5}<\mu(U)+\mu(V)$
$\textbf{Questions about the measure $\mu$}$
Is there a reference for this measure $\mu$? I would also be satisfied if someone can find a published version a similar measure on the real number line or the Cantor cube or a measure defined slightly differently. This notion appears to be related to the notion of the upper density of a subset of $\mathbb{N}$.
Is there is an interesting equivalent way of defining the measure $\mu$? For instance, can we associate to each open set $U$ some set $F$ such that $\mu(U)=m(F)$.
Does there exist a function $\mathbf{E}$ which maps open sets to subsets of $S^{1}$ such that $\mathbf{E}(U)\subseteq\mathbf{E}(V)$ whenever $U\subseteq V$ and where $\mu(U)=m^{*}(\mathbf{E}(U))$ ($m_{*}(\mathbf{E}(U))$ respectively) for all open sets $U$ where $m^{*}$ denotes the Lebesgue outer measure and $m_{*}$ denotes the Lebesgue inner measure respectively)? Does there still exist a function $\mathbf{E}$ if we require $(\overline{U})^{\circ})\subseteq\mathbf{E}(U)\subseteq\overline{U}$? What if we also require $\mathbf{E}$ to be rotational invariant?
Let $\mu^{\sharp}(U)=\lim_{P}\sum_{O\in P}\mu(O\cap U)$ where $P$ denotes the collection of all partitions of the regular open sets of $S^{1}$ into finitely many open intervals and where the limit is taken of the net where the partitions of $S^{1}$ are ordered by refinement. I am also interested in these same questions for the measure $\mu^{\sharp}$ as well. For example, does there exist a function $\mathbf{E}$ which satisfies the same properties as mentioned in the above paragraph but where $\mu^{\sharp}(U)=\mathbf{E}(U)$?
Furthermore, I am interested if there is a collection $Z$ of regular open subsets of $S^{1}$ such that if $O$ is a neighborhood of $1$ and $U,V$ are regular open sets, then $U\in Z$ iff $V\in Z$ and where if $\mathbf{E}_{Z}(U)=\{x\in X|Ux^{-1}\in Z\}$, then $\mu^{\sharp}(U)=m(\mathbf{E}_{Z}(U))$ (or where $\mu^{\sharp}(U)=m^{*}(\mathbf{E}_{Z}(U))$ or $\mu^{\sharp}(U)=m_{*}(\mathbf{E}_{Z}(U))$) for all open sets $U$. Can we replace the function $\mathbf{E}$ with a function that maps open sets to measurable functions from $S^{1}$ to $[0,1]$ so that $\mu^{\sharp}(U)=\int\mathbf{E}(U)(x)dm(x)$ for each open set $U$ and so that $\mathbf{E}$ satisfies the properties analogous to the ones mentioned in this and the last paragraph?
My original motivation for studying this measure was to construct a counterexample in mathematical logic, but this counterexample is not quite the counterexample I want, but this seems like a very interesting kind of measure to study in its own right.
