Timeline for Can the characteristic function of a Borel set be approached by a sequence of continuous function through a certain convergence in $L^\infty$?
Current License: CC BY-SA 4.0
16 events
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| Mar 14, 2019 at 13:26 | vote | accept | Bruno | ||
| Mar 14, 2019 at 13:22 | comment | added | Bruno | @Robert Israel, I see, and it's a nice proof. You are actually proving the characteristic function of rationals is not the pointwise limit of continuous functions. Using in particular the Dirac measure, I now have a full answer for my question. | |
| Mar 14, 2019 at 12:06 | comment | added | Robert Israel | @Bruno Suppose $(a,b) \subset C_n$. Thus for every $x \in (a,b)$, $|f_n(x) - \chi_\Delta(x)| = \lim_{m \to \infty} |f_n(x) - f_m(x) | \le 1/3$. Thus $f_n(x) \ge 2/3$ if $x \in \Delta$, $f_n(x) \le 1/3$ if $x \notin \Delta$. If $(a,b)$ contains a member of $\Delta$ and a member of its complement, this contradicts the Intermediate Value Theorem. | |
| Mar 14, 2019 at 8:40 | comment | added | Gro-Tsen | @ChristianRemling Yes, it was a mistake of mine to even mention $L^\infty$ (it was not in the question). But if we consider, instead, all bounded Borel functions (with pointwise equality!), I think it makes sense to take the topology defined by the seminorms $f\mapsto\int f\,d\mu$ for all finite regular Borel measures $\mu$. The original question would then be: "can we find a sequence of continuous functions converging to a given $\chi_\Delta$ in this sense?", but one can also ask about their closure (or convergence of nets). (Maybe this is a stupid question.) | |
| Mar 14, 2019 at 8:26 | comment | added | Bruno | @Robert Israel: Excuse me, I don't see the contradiction between the continuity of the functions and the density of rationals. | |
| Mar 14, 2019 at 8:19 | comment | added | Bruno | @Christian Remling It's my fault. I didn't ask it separately. I just want to know whether it is possible to find a sequence of continuous functions approaching the characteristic function in any regular Borel measure. At the mean time, is such a convergence possible in $L^\infty$ in the weak sense? | |
| Mar 14, 2019 at 4:27 | history | edited | Robert Israel | CC BY-SA 4.0 |
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| Mar 14, 2019 at 0:51 | comment | added | Christian Remling | @Gro-Tsen: Actually, the most relevant point seems to be the one Nate makes above, in a very slightly different context. For a general (not necessarily ac) measure $\mu$, one can't even define $\int f\, d\mu$ as a functional on $L^{\infty}$ since the integral is now sensitive to changes of $f$ on a (Lebesgue) null set. | |
| Mar 13, 2019 at 21:50 | comment | added | Dirk Werner | My apologies; the part of my previous comment that the $C_n$ must be empty was bare nonsense. | |
| Mar 13, 2019 at 21:04 | history | edited | Robert Israel | CC BY-SA 4.0 |
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| Mar 13, 2019 at 19:42 | comment | added | Gro-Tsen | @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. | |
| Mar 13, 2019 at 19:19 | comment | added | Dirk Werner | 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. | |
| Mar 13, 2019 at 14:09 | comment | added | Gro-Tsen | @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? | |
| Mar 13, 2019 at 13:00 | comment | added | Yemon Choi | @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? | |
| Mar 13, 2019 at 12:48 | comment | added | Gro-Tsen | 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.) | |
| Mar 13, 2019 at 12:16 | history | answered | Robert Israel | CC BY-SA 4.0 |