# Finding a set which has non-full outer measure on every interval, and so does its complement. [duplicate]

I've been struggling with this exercise, and was hoping someone here might be able to point me in the right direction.

Question. Let $m^*$ denote the Lebesgue outer measure on $[0,1]$. Show that there exists $A \subseteq [0,1]$ s.t. for every interval $I \subseteq [0,1]$ we have

1. $m^*(A \cap I)<m^*(I)$,
2. $m^*(A^\complement \cap I)<m^*(I).$

EDIT: I guess I misunderstood the conditions, and this is indeed a duplicate. Apologies.

• chech this math.stackexchange.com/questions/57317/… – Dimitris Jan 24 '14 at 0:41
• Thanks! I suppose I didn't read that question and its answers carefully enough. Mine is indeed a duplicate. Not sure how to proceed from here. Should I delete my question? – doodle Jan 24 '14 at 23:04