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Let $A$ be a compact topological space equipped with the Borel $\sigma$-algebra, and $X=B_b(A)$ be the vector space of bounded measurable functions. Let $Y=\mathcal M(A)$ be the vector space of finite signed measure on $A$. Define the dual pair $<\cdot, \cdot>$ between $(X,Y)$ such that $ <f, \mu >=\int_A f(a)\mu(da) $. Let $\sigma(X,Y)$ be the weakest topology such that for all $\mu\in Y$, the linear map $X\ni f\mapsto <f,\mu>\in \mathbb R$ is continuous.

Define the set $U=\{f\in X\mid \sup_{a\in A}|f(a)|\ge 1\}$. Is the set $U$ closed in the $\sigma(X,Y)$ topology? I am not sure how to proceed to prove or disprove the claim.

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No, assuming $A$ is $T_1$ and infinite. Indeed, just choose a sequence of distinct points $\{a_n\}_{n \in \mathbb{N}} \subset A$. Then $f_n = 1_{\{a_n\}} \in U$, but $f_n \to 0$ pointwise, so as they are uniformly bounded by $1$, by bounded convergence theorem, $\langle f_n, \mu \rangle \to 0$ for all $\mu \in Y$, i.e., $f_n \to 0$ in $\sigma(X, Y)$ topology. But $0 \notin U$.

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  • $\begingroup$ Thank you for pointing out the typo and the answer. $\endgroup$
    – John
    Mar 30 at 23:13
  • $\begingroup$ Is $f_n \rightarrow 0$ pointwise is enough ? Shouldnt the convergence be in something like sup norm. Although for the given set of functions it seems to work. $\endgroup$
    – Balaji sb
    Mar 31 at 1:47
  • $\begingroup$ @Balajisb en.wikipedia.org/wiki/… $\endgroup$
    – David Gao
    Mar 31 at 2:02
  • $\begingroup$ I agree then its better to mention your functions are uniformly bounded also. Because in the space of bounded functions, pointwise convergence in general does not imply dominated convergence theorem i think. Just for book keeping. I am not saying much. $\endgroup$
    – Balaji sb
    Mar 31 at 2:10
  • $\begingroup$ @Balajisb I mean, when I was writing the answer I was under the impression that that was obvious, but sure, if you think that will make it clearer. $\endgroup$
    – David Gao
    Mar 31 at 2:15

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