Timeline for Are there dense sets of positive but not full measure?

7 events

when toggle format	what		by	license	comment
Mar 20, 2015 at 2:32	comment	added	Pietro Majer		Constructing a measurable subset $A$ of $[0,1]$ such that $0 < \mu(A\cap [a,b]) < b-a$ for any $0\le a < b\le 1$ is an old and well-known elementary exercise in Measure Theory; see Rudin's Real and Complex Analysis.
Mar 20, 2015 at 0:53	comment	added	Adam Epstein		One may exhibit cantor subsets of $[0,1]$ with any measure between $0$ and $1$ inclusive, by an explicit construction where at stage $n$ one removes the central part of proportion $a_n$, for a suitable sequence. Take the union of such a set with $\mathbb{Q}\cap[0,1]$.
Mar 20, 2015 at 0:50	history	edited	Bjørn Kjos-Hanssen	CC BY-SA 3.0	added 34 characters in body
Mar 20, 2015 at 0:47	comment	added	Richard Stanley		The second example of Kjos-Hanssen does contain an interval. For an example not containing an interval, simply take $A=([0,1/2]-\mathbb{Q})\cup([1/2,0]\cap \mathbb{Q})$. However, I am guessing that the condition you actually want is that for all $0\leq a<b\leq 1$, we have $0<\mu([a,b]\cap A)<b-a$.
Mar 20, 2015 at 0:42	history	edited	Bjørn Kjos-Hanssen	CC BY-SA 3.0	added 193 characters in body
Mar 20, 2015 at 0:40	comment	added	Stanley Yao Xiao		My apologies, I intended to not consider trivial sets which contain intervals
Mar 20, 2015 at 0:32	history	answered	Bjørn Kjos-Hanssen	CC BY-SA 3.0