# Can the characteristic function of a Borel set be approached by a sequence of continuous function through a certain convergence in $L^\infty$?

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.

• I don't think this is weak convergence, the dual of $L^{\infty}$ is finitely additive set functions, not just measures. – Christian Remling Mar 13 at 17:14
• @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. – Nate Eldredge Mar 14 at 0:01
• Similar question: math.stackexchange.com/questions/2482042/… – Dap Mar 14 at 7:07
• @Dap yeah, I agree with the use of Dirac measure. So, we cannot talk about the convergence in all Borel measure. – Bruno Mar 14 at 9:37

(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.
• 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.) – Gro-Tsen Mar 13 at 12:48
• @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? – Yemon Choi Mar 13 at 13:00
• @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? – Gro-Tsen Mar 13 at 14:09
• 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. – Dirk Werner Mar 13 at 19:19
• @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. – Gro-Tsen Mar 13 at 19:42