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In this paper the authors recall the definition of weak$^{\star}$-convergence in the footnotes on page 2. The problem setting is as follows:

Let $(S, \Sigma)$ be a measurable space, where $S \neq \emptyset$, and $\Sigma$ is a $\sigma$-algebra on $S$. Denote by $\Delta(S, \Sigma)$ the set of all probabilities (here in this case defined as additive capacity). Then, the statement is as follows:

A net $\{ P_{\alpha}\}_{\alpha \in I}$ converges to $P$ in the weak$^{\star}$-topology if and only if $P_{\alpha}(A) \longrightarrow P(A)$ for all $A \in \Sigma$.

How does this characterization fit the usual definition of convergence wrt to the weak$^{\star}$-topology from functional analysis? Any help is very much appreciated.

Galois1763
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Let $X$ be the space of all bounded measurable functions on $(S,\Sigma)$ with the sup norm. This is Banach space and any probability measure $P$ on $(S,\Sigma)$ can be considered as an element of the dual space (via $f \mapsto \int f dP$). Convergence as defined here is precisely weak* convergence in $X^{*}$.

For a justification you have to note that any bounded measurable function is a uniform limit of simple functions. [Some details are in my comment below].

Kavi Rama Murthy
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  • Thank you for the answer. I would expect that (let us assume for simplicity that we have sequences instead of nets) we have $P_n \rightarrow P$ (weak$^{\star}$) iff $\int f dP_n \rightarrow \int f dP$ for all bounded and measurable functions $f$. I do not understand why it is sufficient for the "iff" statement in the question to assume set wise convergence of the probability measures. – Galois1763 Jun 07 '23 at 12:35
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    Suppose $P_{\alpha}(A) \to P(A)$ for every $A$. $|f-g|<\epsilon$ implies $|\int fdP_{\alpha}-\int gdP_{\alpha}| <\epsilon$ for every $\alpha$ and $|\int fdP-\int gdP| <\epsilon$. Also $\int gdP_{\alpha} \to \int gdP$. Now use triangle inequality to finish the proof of $\int f dP_{\alpha} \to \int fdP$. – Kavi Rama Murthy Jun 07 '23 at 12:38
  • This somehow would justify why $P_{\alpha}(A) \rightarrow P(A)$ for every $A$ implies that ${P_{\alpha}}$ converges to $P$ with respect to the weak$^{\star}$-topology, but why is the other direction "obvious"? – Galois1763 Jun 07 '23 at 16:04
  • Ah, for the other direction we just say $f = 1_{A}$ for $A \in \Sigma$, right? – Galois1763 Jun 07 '23 at 17:43
  • Yes, that is right. @Galois1763 – Kavi Rama Murthy Jun 07 '23 at 23:10