Let $f:(X, d)\to (X, d)$ be a homeomorphism on compact metric space $(X, d)$. We define a new metric $d_\infty$ on $X$ by $d_\infty(x, y)= \sup_{n\in\mathbb{Z}}d(f^n(x), f^n(y))$, and $s(x)=\inf \{d_\infty-diam(U): x\in U \text{ open }\}$.
In a paper author claim that $s$ is upper-semicontinuous and invariant function, hence it is constant $\mu$-almost everywhere.
My attempt: By definition of $s(x)$, for every $\epsilon>0$, there is open set $U$ of $x$ such that $d_\infty -diam(U)<s(x)+\epsilon$, hence for every $y\in U$, $s(y)\leq d_\infty-diam(U)<s(x)+\epsilon$, hence $s$ is upper semicontinuous. Also since $d_\infty-diam (f(U))= d_\infty-diam(U)$, the function $s$ is invariant.
But it is not clear for me the following statement: "it is constant $\mu$-almost everywhere"?
Please help me to know it.