How do I prove this statement about density in $[0,1]$?

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I have the following problem:

Let $\Omega =[0,1]$ the closed unit interval endowed with the standart topology and let $A\subset \Omega$. We denote by $\nu$ the outer lebesgue measure on $\Omega$. We say that $A$ is dense in $\Omega$ if for all $x\in \Omega$ and all $\epsilon >0$ there exists a point $y\in A$ such that $x\in B_\epsilon(y)$. Show if this statements are correct and if not give a counterexample:

  1. $A$ is dense, then $\nu(A)>0$
  2. if $\nu(A) =1$ then $A$ is dense in $\Omega$

In my opinion $1.$ should work and the second point not. So I wanted to prove $1$ but I don't see how do do this. First I wanted to show it by contradiction but then I only know that $A$ is a nullset which doesn't help me much I think. Therefore I wanted to show it directly but also this didn't work. So my Idea was to show that at least one open ball is contained in $A$ and then we would immediately get that $\nu(A)>0$ wouldn't we?

Could someone give me a hint only for the first point?

Thank you a lot

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As it has been stated in the comments, 1. is false: $\mathbb{Q}\cap [0,1]$ is dense in $[0,1]$ (because $\mathbb{Q}$ is dense in the real line). And for 2., it is true and, in fact, it is satisfied that $A$ is [0,1] except for a null set. It should be easy to formalize from there.