Proof that the complement of zero-measure set on $[0,1]$ is dense in $[0,1]$ As a problem during introduction to measure theory, I am asked to prove the following.
Let $A \subset [0,1]$ such that $A$ is measurable with $\mu(A) = 0$ ($\mu$ being the Lebesgue measure). Prove that $A^c =
 [0,1] \setminus A$ is dense in $[0,1]$.
This makes intuitive sense to me $-$ since $\mu(A) = 0$, using that $\mu(X) = \mu(X \cap A) + \mu(X \cap A^c)$ for every $X \subseteq \mathbb{R}$ we get $\mu([0,1]) = \mu(A) + \mu(A^c) = \{\mu(A)=0\} = \mu(A^c)$. Thus, $A^c$ has the same measure as $[0,1]$. I am aware that this is not the definition of $A^c$ being dense in $[0,1]$ though, and I do not know how to formalize this proof.
Any advice would be appreciated.
 A: Your comment suggests you might want to review the definition of dense.
Recall that $A^{c} \subseteq [0,1]$ is dense if and only if for each $x \in [0,1]$ and each $\delta > 0$, there is a $b \in A^{c}$ such that $b \in (x - \delta, x + \delta)$.
Now we prove $\mu(A) = 0$ implies $A^{c}$ is dense (with $\mu$ denoting Lebesgue measure).  Suppose $x \in [0,1]$ is arbitrary and $\delta >0$.  Then $\mu((x - \delta,x + \delta)) = 2 \delta$.  Thus,
$$\begin{align}
   \mu(A^{c} \cap (x - \delta, x + \delta))
   &= \mu((x - \delta,x + \delta)) - \mu(A \cap (x - \delta,x + \delta))\\
   & = \mu((x - \delta,x + \delta))\\
   & = 2 \delta
\end{align}$$
by additivity of the measure: $\mu(B)=\mu(A^c\cap B)+\mu(A\cap B)$, and since $\mu(A) = 0$.  Since $A^{c} \cap (x -\delta,x + \delta)$ has positive measure, it is necessarily non-empty.  In particular, we can pick $b \in A^{c} \cap (x - \delta,x+\delta)$.
Since $x$ and $\delta$ were arbitrary, this proves $A^{c}$ is dense in $[0,1]$.  $\qquad\blacksquare$
A: If $A^c$ is not dense then $\exists a\in\mathbb{R}$ $\colon$ $(a-\delta,a+\delta)$ does not contain any element $b$ of $A^c$ where $b\ne a$.
So $(a-\delta,a+\delta)\setminus\{a\}\subset E$. 
$\begin{align}\text{But, } m^*((a-\delta,a+\delta)\setminus\{a\})&=m^*((a-\delta,a)\cup(a,a+\delta))\\&=m^*((a-\delta,a))+ m^*((a+\delta,a))\\&=2\delta
\end{align}$
$\text{So, } m^*(A)\ge2\delta\implies m^*(A)\ne0
$
So we arrive at a contradiction.$\qquad\blacksquare$
