# Proof set of content zero has void interior

Given a set of content zero $$A \subset R^{n}$$, how can be proved that the interior of $$A$$ is void?

• Can you show that the content of every open ball of positive radius is positive? Open sets are precisely the unions of open balls. Do you see how this does the trick? – MPW Apr 17 at 20:58
• Yeah, that would do, but I'm still unsure how to prove formally that the content of every open ball is positive. Maybe by reduction to absurdity...? – math_noob Apr 17 at 21:08
• Prove that the unit ball has positive content. Every other ball can be obtained by dilating and translating it. Translation doesn't affect content, and dilation by a factor of $c$ will scale the content by a factor of $c^n$. – MPW Apr 17 at 21:10
• If you don't want to work with balls you can work with cubes instead. It is easier. If $A$ has a point in the interior then it contains an open cube. – Mark Apr 17 at 21:15
• Obviously, the way to go would be to proof that the volume of the open ball is a lower bound of the sum of the volumes of any finite covering of the ball, but still, I can't figure out a way to proof it in a synthetic way. – math_noob Apr 17 at 21:19