Suppose that $(X,\tau)$ is a topological space and let $\mathscr B$ denote the Borel $\sigma$-algebra on it. Moreover, let $\mu:\mathscr B\to[0,\infty]$ be a regular Borel measure, that is,
- $\mu(K)<\infty$ for any compact set $K\subseteq X$;
- $\mu(B)=\inf\{\mu(U)\,|\,B\subseteq U,\text{ $U\subseteq X$ is open}\}$ for any $B\in\mathscr B$; and
- $\mu(B)=\sup\{\mu(K)\,|\,K\subseteq B,\text{ $K\subseteq X$ is compact}\}$ for any $B\in\mathscr B$.
Now let $E\in\mathscr B$ be a Borel set such that
- $E$ is dense: $\operatorname{cl}E=X$; and
- $E$ has no interior: $\operatorname{int}E=\varnothing$.
Conjecture: Either $\mu(E)=0$ or $\mu(E^{\mathsf c})=0$.
Is this conjecture true?
If counterexamples exist, can one rule them out by imposing additional restrictions either on the topological structure (Hausdorff, local compactness, separability, metrizability, etc.) or on the measure-theoretic structure (e.g., $\sigma$-finitude)?
In particular, does it hold for the Euclidean topology on $\mathbb R$ where $\mu$ is the Lebesgue measure?