So I found this post about subsets of measure $1$ in $[0,1]$ being necessarily dense. I was wondering if this holds true for more general measure spaces, i.e. are sets of full measure necessarily dense?

I asked a friend this question and we think we have a counterexample, although we don't know if this counts as a properly defined measure space.

Just take $[0,1]$ again but with every singleton defined to be an open set. Then the irrational points have measure 1, but any rational point is open and doesn't intersect the irrationals, so the subset of irrationals in the interval is not dense. Is this allowed?

  • 1
    $\begingroup$ Take the Dirac measure on $[0,1]$. Then $\delta(\{0\}) = 1$ but $\{0\}$ is not dense in $[0,1]$. $\endgroup$
    – copper.hat
    Jul 26, 2020 at 20:57
  • $\begingroup$ Right, that works. Maybe I could add this to my post: what properties can guarantee that a set of full measure is dense? $\endgroup$ Jul 26, 2020 at 21:04

2 Answers 2


Yes, your counter-example is allowed. The topology you are describing is the Discrete Topology on $[0,1]$. In this topology, any subset is both open and closed, so in particular the closure of any subset is itself, hence the only dense subset is the entire space. As you note, there are, however, proper subsets of full measure.

For the type of conclusion you want, the measure on your space should have something to do with the topology on your space. To this end, let $X$ be any topological space. The Borel algebra $\mathcal{B}(X)$ is the $\sigma$-algebra on $X$ generated by all the open sets. A measure $\mu$ on $\mathcal{B}(X)$ is called a Borel measure.

Claim: For a topological space $X$ and a Borel measure $\mu$ on $X$, the properties "any Borel set of full measure is dense" and "any non-empty open set has positive measure" are equivalent.

Proof:"$\Rightarrow:$" Assume $U\neq\emptyset$ is an open subset of $X$ and $\mu(U)=0$. Then $\mu(X\setminus(X\setminus U))=\mu(U)=0$, so $X\setminus U$ has full measure. Since $U$ is open, $X\setminus U$ is closed, hence $\overline{X\setminus U}=X\setminus U$, but $X\setminus U\neq X$, since $U\neq\emptyset$, so $X\setminus U$ is not dense in $X$. The claim follows by contrapositive.

"$\Leftarrow:$" Assume $A\subseteq X$ is Borel and not dense in $X$, i.e. $\overline{A}\neq X$. Then $X\setminus\overline{A}\neq\emptyset$. Since $\overline{A}$ is closed, $X\setminus\overline{A}$ is open, so, by hypothesis, $\mu(X\setminus\overline{A})>0$. By definition, $\overline{A}$ does not have full measure. The claim follows by contrapositive.

Note that the second property is also known as strict positivity. In particular, this property holds for every Haar measure, such as the Lebesgue measure on any $\mathbb{R}^n$.


It seems that the notion you are trying to extract from your examples is the (topological) support of a measure https://en.wikipedia.org/wiki/Support_(measure_theory) i.e. the smallest closed set of points $x$ for which every neighborhood of $x$ has a strictly positive measure.

You need to be able to measure open and closed sets for this to make sense so looking at Borel measures in this context is sufficient (though not necessary).

In your example with every singleton being open means that every subset is open which runs into trouble if your starting point is Lebesgue measure and if you trust in the Axiom of choice which implies the existence of non-measurable subsets of $[0,1]$. You can not measure all open sets in that case.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .