# 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.

• Suppose otherwise. Then there are $a,b$ in the complement with $a<b$ and $(a,b) \subset A$. Can you conclude? May 15, 2017 at 21:03
• $A^c$ is dense in $[0,1]$ means that $\lim_{j\to\infty} a_j = x \in [0,1]$ for all $a_j \in A^c$. So, if $A^c$ would not be dense in $[0,1]$, then there would exist some interval $(a,b) \subset [0,1]$ such that $(a,b) \not\subset A^c \Leftrightarrow (a,b) \subset A$. Since $A$ contains at least one non-empty open sub-interval, $\mu(A) > 0$, which contradicts the assumption $\mu(A) = 0$. Is this correct? I am still a bit uncartain about why the fact that there is no sequence $a_j \in A^c$ that converges to some $x \in [0,1]$ would imply that there is such an open interval... May 15, 2017 at 21:57

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 $$[0,1]$$. $$\qquad\blacksquare$$

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.$$\qquad\blacksquare$$