Weak$^*$-convergence of vector-valued measures implies weak$^*$-convergence in $X^*$?

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Let $K$ be a compact Hausdorff space and $X$ be a Banach space. By the Riesz-Singer representation theorem, we know that there exists a linear isometry from $C(K,X)^*$ onto $rcabv(K,X^*)$, the Banach space of all regular, countably additive, Borel $X^*$-valued measures in $K$ with bounded variation, equipped with the norm $$\|\mu\| = |\mu|(K), \forall \mu \in rcabv(K,X^*).$$ (Here, $|\mu|$ denotes the variation of $\mu$)

Suppose $(\mu_n)_{n \in \mathbb{N}}$ is a weak$^*$-null sequence in $rcabv(K,X^*)$, that is, for each $f \in C(K,X)$, $$\lim_{n \to \infty} \int_K f d\mu_n =0.$$

Let $U$ be an open set in $K$ and $x \in X$. Is it true that $$\lim_{n \to \infty} \mu_n(U)(x) = 0?$$

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If I missed nothing, this is not even true in much simpler situations:

Let $K = (0,1)$ and $X = \mathbb{R}$. Then, the sequence of Dirac differences $$\delta_0 - \delta_{1/n}$$ converges weak-* to $0$, but $$(\delta_0 - \delta_{1/n})(\{0\}) = 1.$$