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

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 $$\lim_{n \to \infty} \mu_n(U)(x) = 0?$$

• I mean the computation $x^*_n(x)$, where $x^*_n = \mu_n(U) \in X^*$. That is, $\mu_n(U)(x) = \langle \mu_n(U), x \rangle$. Sorry for the confusion. – Vinícius Morelli Mar 4 '16 at 21:02
• Each $\mu_n$ belongs to $rcabv(K,X^*)$. Therefore, given a Borel set $A \subset K$, $\mu_n(A) \in X^*$. – Vinícius Morelli Mar 4 '16 at 21:06
• Right, I missed the point that you want them to be vector-valued. – Tomek Kania Mar 4 '16 at 21:08

Let $K = (0,1)$ and $X = \mathbb{R}$. Then, the sequence of Dirac differences $$\delta_0 - \delta_{1/n}$$ converges weak-* to $0$, but $$(\delta_0 - \delta_{1/n})(\{0\}) = 1.$$
• That is correct; however, $\{0\}$ is not open. – Vinícius Morelli Mar 4 '16 at 21:53
• I apologize, your example does settle the question. It suffices to notice that $$(\delta_0 - \delta_{\frac{1}{n}}) (]0,1[) = -1, \forall n \geq 2.$$ Thank you very much for the example. Again, I apologize for not noticing and accepting your answer earlier. – Vinícius Morelli Mar 5 '16 at 18:22