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.

  • 3
    $\begingroup$ Suppose otherwise. Then there are $a,b$ in the complement with $a<b$ and $(a,b) \subset A$. Can you conclude? $\endgroup$ May 15, 2017 at 21:03
  • $\begingroup$ $A^c$ 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... $\endgroup$
    – 0ffset
    May 15, 2017 at 21:57

2 Answers 2


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.$\qquad\blacksquare$


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