Let $X$ be a compact metric space and $\mathcal{M}(X)$ be the space of variational-bounded, signed Borel measures equipped with the Kantorovich–Rubinshtein norm, cf. [Section 8.3, 1]:
$$||\mu||\_0:= \sup\Bigg\{\int\_X fd \mu: f \in \mathrm{Lip}\_1(X), \sup\_{x \in X}|f(x)|\leq 1 \Bigg\}$$
where $\mathrm{Lip}\_1(X)$ deontes the set of Lipschitz functions on $X$ with Lipschitz constant smaller or equal to $1$.


Let $K\subseteq X$ be closed (with nonemtpy interior) and $M>0$ be a constant. Are the sets
$$\{\mu \in \mathcal{M}(X) \mid \mu(K)< M\} \text{ and } \{\mu \in \mathcal{M}(X) \mid |\mu|(K)< M\}$$
open in $\mathcal{M}(X)$?


Remark: Bogachev shows that the topology induced by the KR-norm on the set of all nonnegative Borel-measures $\mathcal{M}\_+(X)$ coincides with the topology of weak convergence. Hence, one can verify that the above sets restricted to $\mathcal{M}\_+(X)$ are open in $\mathcal{M}\_+(X)$.


References: [1]
*Bogachev, V. I.*, Measure theory. Vol. I and II, Berlin: Springer (ISBN 978-3-540-34513-8/hbk). xvii, 500 p./v.1; xiii, 575 p./v.2. (2007). [ZBL1120.28001](https://zbmath.org/?q=an:1120.28001).



Open sets in the space of signed measures equipped with the Kantorovich–Rubinshtein norm
No, those sets are not open. Indeed, take any non-isolated point $x$ in $X$ and a sequence $(x\_n)\_{n\in\omega}$ that converges to $x$. For every $n$, consider the sign measure $\mu\_n=\delta\_{x}-\delta\_{x\_n}$, where $\delta\_p$ is the Dirac measures at a point $p\in X$. The definition of the Kantorovich-Rubinshtein norm guarantees that $\|\mu\_n\|\_0$ tends to zero as $n\to\infty$. For $M=\frac 12$ and $K=\{x\}$, the zero measure belongs to the sets $\{\mu\in\mathcal M(X):|\mu|(K)<M\}\subseteq\{\mu\in\mathcal M(X):\mu(K)<M\}$ but for every $n\in\omega$ the measure $\mu\_n$ does not belong to those two sets, which implies that these sets are not open in $\mathcal M(X)$.


