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Let $K$ be a compact Hausdorff space (not necessarily metric or even separable). Let $M(K)$ be the space of all Radon measures on $K$ (that is, finite signed regular Borel measures) endowed with the variation norm. By the Riesz representation theorem $M(K)$ is isometrically isomorphic to the dual space $C(K)^*$ of the space $C(K)$ of all continuous real-valued functions on $K$ endowed with the supremum norm. By the weak topology on $M(K)$ I mean the topology given by the subbase consisting of the sets $V(\mu;x^*;\varepsilon)=\{\nu\in M(K)\colon\ |x^*(\mu)-x^*(\nu)|<\varepsilon\}$, where $\mu\in M(K)$, $x^*\in M(K)^*$ ($\cong C(K)^{**}$), and $\varepsilon>0$.

Let $(\mu_n)$ be a sequence in $M(K)$. It is well-known that $(\mu_n)$ converges to some $\mu\in M(K)$ with respect to the weak topology if and only if $\lim_{n\to\infty}\mu_n(B)=\mu(B)$ for every Borel set $B\subseteq K$. My question is thus the following:

QUESTION: Can we define the weak topology on $M(K)$ by means of the subbase given by the sets of the form $W(\mu;B;\varepsilon)=\{\nu\in M(K)\colon\ |\mu(B)-\nu(B)|<\varepsilon\}$, where $\mu\in M(K)$, $B\subseteq K$ Borel, and $\varepsilon>0$?

(If it helps, one may assume that $K$ is totally-disconnected.)

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    $\begingroup$ Quick thoughts: the topology you describe is the weak one defined by the space of step functions with Borel subsets as sets of constancy. Hence its dual is precisely this linear space. But the dual of $M(K)$ with the weak topology is, as you say, $C(K)^{\ast \ast}$. $\endgroup$
    – user131781
    Commented Sep 17, 2020 at 13:13
  • $\begingroup$ Oh, I see, so it actually means that the new topology (given by the sets $W(\mu;B;\varepsilon)$) is strictly weaker than the weak topology (given by the sets $V(\mu;x^*;\varepsilon)$... $\endgroup$ Commented Sep 17, 2020 at 13:31

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