Let $(X,d)$ be a metric space equipped with a probability measure $\mu$ (defined on the Borel $\sigma$-algebra on the topology induced by the metric $d$). I am interested in the different values that the following can take

$\sup\lbrace\mu(A):A\subset X\text{ is measurable and }\mu(A)\leq1/2\rbrace~~~~~~~~~~(\ast)$.

If there exists a measurable set $A\subset X$ such that $\mu(A)=1/2$, then it clearly is the case that $(\ast)$ is $1/2$.

If there is no set of measure $1/2$, then it is possible for $(\ast)$ to be different than $1/2$ (Dirac probability measure comes to mind). What I am wondering is if it could still be $1/2$.

In other words, I'm wondering if we could prove the existence (ideally with an example) or the nonexistence of a probability measure on a metric space $(X,d)$ such that for every measurable $A\subset X$, $\mu(A)\neq1/2$ and such that there is a sequence $\lbrace A_n\rbrace$ where $\mu(A_n)\to1/2$.

A few obvious remarks, if we were to find a probability measure as described in the above paragraph, the sequence $A_n$ cannot be nested, otherwise continuity from above or continuity from below would imply that the union or the intersection of the $A_n$ has a measure of $1/2$. Furthermore, the $A_n$ cannot be pairwise disjoint, otherwise $\mu$ cannot be a probability measure: for all $\epsilon>0$, infinitely many of the $A_n$ are contained in $(1/2-\epsilon,1/2+\epsilon)$, which implies by countable additivity that $\mu(X)\geq\infty$.