Timeline for Speed of convergence in Lebesgue's density theorem

10 events

when toggle format	what		by	license	comment
Jun 25, 2016 at 17:04	history	edited	Christian Remling	CC BY-SA 3.0	added 17 characters in body
Jun 24, 2016 at 23:52	history	undeleted	Christian Remling
Jun 24, 2016 at 23:52	history	edited	Christian Remling	CC BY-SA 3.0	deleted 52 characters in body
Jun 24, 2016 at 16:40	history	deleted	Christian Remling		via Vote
Jun 24, 2016 at 14:35	comment	added	user240643		@FedorPetrov I think you're right. Let $K_N$ be the set you defined, then you get $a_n(K_N)=1/4$ for all $n<N$ and $a_n(K_N)=0$ for all $n\geq N$, so $\sum_n a_n(K_N)=(N-1)/4$. But there really is no "limit-analogue" for the sets $K_N$ as $N\rightarrow\infty$. 'Infinite' sets can't be that balanced. I'm going to think about your and ChristianRemling's answer now, thanks so far!
Jun 24, 2016 at 7:57	comment	added	Fedor Petrov		@ChrtistianRemling Hm, what if $K$ is a union of segments $[2i/2^N,(2i+1)/2^N]$ for $i=0,\dots,2^{N-1}-1$? On the first glance, all $a_n$ for $n< N$ are bounded from below by $1/8$ or something like that?
Jun 24, 2016 at 6:34	history	edited	Christian Remling	CC BY-SA 3.0	added 158 characters in body
Jun 24, 2016 at 6:02	comment	added	Fedor Petrov		It is bit unclear what exactly does your argument prove. That $\sum_{n\le N} a_n$ always does not exceed, say, 100 for any set?
Jun 24, 2016 at 1:28	history	edited	Christian Remling	CC BY-SA 3.0	added 270 characters in body
Jun 23, 2016 at 22:13	history	answered	Christian Remling	CC BY-SA 3.0