Want to find $f_n$ a sequence of continuous functions, so that for all Borel regular measure $\mu$, we have
$\int f_n d\mu\rightarrow\int \chi_\Delta d\mu$, as $n$ goes to infinity, where $\Delta$ is a Borel set.
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2$\begingroup$ I don't think this is weak convergence, the dual of $L^{\infty}$ is finitely additive set functions, not just measures. $\endgroup$– Christian RemlingCommented Mar 13, 2019 at 17:14
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1$\begingroup$ @ChristianRemling: And moreover, the dual of $L^\infty$ is only those finitely additive set functions which are absolutely continuous to your reference measure (which has not been specified). I think this question needs to be clarified. $\endgroup$– Nate EldredgeCommented Mar 14, 2019 at 0:01
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1$\begingroup$ Similar question: math.stackexchange.com/questions/2482042/… $\endgroup$– DapCommented Mar 14, 2019 at 7:07
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$\begingroup$ @Dap yeah, I agree with the use of Dirac measure. So, we cannot talk about the convergence in all Borel measure. $\endgroup$– BrunoCommented Mar 14, 2019 at 9:37
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(EDITED) Let $\Delta$ be the rationals in $[0,1]$. If your condition is satisfied, in particular $f_n$ converges pointwise to $\chi_\Delta$. Let $C_n = \{x \in [0,1]: \forall m > n, \;|f_n(x) - f_m(x)|\le 1/3\}$. Then $C_n$ are closed and their union is $[0,1]$. By the Baire category theorem some $C_n$ has nonempty interior. But this is impossible since $f_n$ is continuous and both $\Delta$ and its complement are dense.
answered Mar 13, 2019 at 12:16
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$\begingroup$ Of course, this raises just another question now: is it nevertheless true that the weak closure in $L^\infty$ of the set of continuous functions contains the characteristic functions of Borel sets? (The topology is not metrizable, so limits of sequences do not define the closure.) $\endgroup$– Gro-TsenCommented Mar 13, 2019 at 12:48
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1$\begingroup$ @Gro-Tsen If by weak closure you mean $\sigma(L^\infty, (L^\infty)^*)$-closure then isn't this just equal to the norm closure? (Mazur's theorem.) Or is this the probabilist's notion of weak convergence? $\endgroup$ Commented Mar 13, 2019 at 13:00
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$\begingroup$ @YemonChoi I'm a poor ignorant algebraist who's a bit lost in a twisty maze of "weak" topologies all alike, but I meant the one which seems to be implicit in the question, namely, the coarsest topology on the set of bounded Borel functions which makes $f \mapsto \int f\,d\mu$ continuous for every finite regular Borel measure. (I guess I shouldn't have written $L^\infty$.) Or, what I hope amounts to the same: what if we change the question slightly to allow converging (Moore-Smith) nets $f_\alpha$ rather than merely sequences $f_n$ of continuous functions? $\endgroup$– Gro-TsenCommented Mar 13, 2019 at 14:09
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$\begingroup$ If $\Delta$ is as suggested by Robert, then, again by one of the corollaries of Baire's theorem, $\chi_\Delta$ has a point of continuity (being Baire-$1$), which it doesn't. I think, Robert's $C_n$ are all empty (hence closed!), since $f_n$ is continuous; still the convergence assumption would force the union of the $C_n$ to be $[0,1]$. -- As for notation, it seems to me that the OP has defined his own ``weakish'' convergence different from weak convergence in Banach spaces. $\endgroup$ Commented Mar 13, 2019 at 19:19
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$\begingroup$ @ChristianRemling I agree that this is not the "weak" topology defined by the dual of $L^\infty$ (it was probably a mistake of mine to even mention $L^\infty$, which is not in the question), but it is legitimate to ask about the topology defined (on the bounded Borel functions I guess) by the seminorms $f\mapsto\left|\int f\,d\mu\right|$ for finite regular Borel measures $\mu$, which seems to be a kind of weak topology, no? I'm seriously confused at this point, but I think this makes sense. $\endgroup$– Gro-TsenCommented Mar 13, 2019 at 19:42