Suppose that $U\subseteq S^{1}$ is open. Then define $\mu_{n}(U)=\max_{t\in S^{1}}\frac{1}{n}\cdot|\{k\in\{1,...,n\}|t\cdot e^{\frac{2\pi ik}{n}}\}|$. Define $\mu(U)=\lim_{n}\mu_{n}(U)$ (here this limit is with respect to the divisibility ordering on the natural numbers. It is easy to check that this limit always converges). 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 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 $\mu$}$

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.