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It seems that Weak topology and weak convergenge in probability spaces claims in the first answer that if $M(X)$ is the space of Borel measures on a Polish space $X$, then it's dual is $C_b(X)$, the space of continuous bounded functions on $X$. Is this correct? It doesn't seem right to me because this would essentially be saying that if $K$ is compact, the double dual of $C(K)$ is itself.

user124910
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    In the link, $M(X)$ is endowed with the weak(star) topology so this is true. In general, the dual of $(X^,w^)$ is (isometric to) $X$. This is not true if $M(X)$ is endowed with the total variation norm – Evangelopoulos Foivos Jan 17 '25 at 20:33
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    Ah, that makes sense. Thank you! Do you bychance have a reference for the result in that question? I see your point, but it seems to apply when $X$ is a compact Hausdorff space, but we can't use the Riesz representation theorem to establish the dual of $C(X)$ when $X$ is not compact. – user124910 Jan 17 '25 at 20:36
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    Do you mean why $C_b(X)^*=M(X)$? This is usually done by first identifying $C_b(X) = C(K)$ (isometrically) where $K$ is compact Hausdorff. In fact, $K=\beta X$ is the Stone-Cech compactification. Now you can apply Riez-Markov-Kakutani's theorem – Evangelopoulos Foivos Jan 18 '25 at 07:26
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    @user124910: The space of Borel complex measures $\mathcal{M}(X)$, acting as linear functionals separates points of, $C_b(X)$. Then the $C_b(X)$ under the weak topology $\sigma(C_n(X),\mathcal{M}(X))$ becomes a locally convex topological vector space and its dual is precisely $\mathcal{M}$. Weak cionvergence of measures (usually restricted to Probability measures) is topologies by the weak* topology $\sigma(\mathcal{M}(X),C_b(X))$ on $\mathcal{M}(X)$. That is what I stated in the answer to which you are alluding. (*to be continued...*) – Mittens Jan 19 '25 at 04:04
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    @user124910: (*...continuation) Using the Stone-Czech compactification $(\beta X,\iota)$ of $X$, then $C(\beta X)$ as a Banach space has $\mathcal{M}(\beta X)$ as its dual (Banach space under the variation norm). Here is natural to use the weak topology $(\sigma(\mathcal{M}(\beta X),C(\beta X))$. Then any Borel complex measure on $\mu$ on $X$ induces a Borel measure on $\beta X$ via the pushforward $\mu\circ\iota^{-1}$. In this setting a net $\mu_\alpha$ in $\mathcal{M}$ converges weakly to $\mu$ iff $\mu_\alpha\circ\iota^{-1}$ converges to $\mu\circ\iota^{-1}$ (in the weak* topology) . – Mittens Jan 19 '25 at 04:34
  • @EvangelopoulosFoivos but wouldn't that give that $C_b^(X)\cong C^(\beta X)\cong M(\beta X)$? Why would one have $C_b^(X)\cong M(X)$? My understanding is, you want to say $C_b^(X)\cong M(X)$, and then, when $M(X)$ is equipped with the weak* topology, $M(X)\cong C_b(X)$ by the theorem you cited. However, we can't use Riesz representation theorem directly on $C_b(X)$, and noting $C_b^(X)\cong C^(\beta X)$ seems to suggest the dual of $C_b(X)$ are measures on $\beta X$, not $X$. – user124910 Jan 20 '25 at 20:42

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