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Suppose that $(X,\tau)$ is a topological space and let $\mathscr B$ denote the Borel $\sigma$-algebra on it. Moreover, let $\mu:\mathscr B\to[0,\infty]$ be a regular Borel measure, that is,

  • $\mu(K)<\infty$ for any compact set $K\subseteq X$;
  • $\mu(B)=\inf\{\mu(U)\,|\,B\subseteq U,\text{ $U\subseteq X$ is open}\}$ for any $B\in\mathscr B$; and
  • $\mu(B)=\sup\{\mu(K)\,|\,K\subseteq B,\text{ $K\subseteq X$ is compact}\}$ for any $B\in\mathscr B$.

Now let $E\in\mathscr B$ be a Borel set such that

  • $E$ is dense: $\operatorname{cl}E=X$; and
  • $E$ has no interior: $\operatorname{int}E=\varnothing$.

Conjecture: Either $\mu(E)=0$ or $\mu(E^{\mathsf c})=0$.


Is this conjecture true?

If counterexamples exist, can one rule them out by imposing additional restrictions either on the topological structure (Hausdorff, local compactness, separability, metrizability, etc.) or on the measure-theoretic structure (e.g., $\sigma$-finitude)?

In particular, does it hold for the Euclidean topology on $\mathbb R$ where $\mu$ is the Lebesgue measure?

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    $\begingroup$ No. $\;\;\;\;\;\;\; ([-1,\hspace{-0.03 in}0]\cap \mathbb{Q}) \: \cup \: ([1,\hspace{-0.04 in}2]-\mathbb{Q}) \;\;\;$ is a counterexample. $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ $\endgroup$ – user57159 Aug 20 '15 at 23:43
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It's definitely not true.

You can even strengthen "empty interior" to "meager" (every meager set has empty interior by the Baire category theorem). A fat Cantor set is meager but has positive measure. (You can even take a countable union of such sets to get a meager set with full measure, but that isn't what you asked for here.)

Now to make it dense, just take the union of your fat Cantor set with the rationals. It's still meager so it still has empty interior, and the measure didn't change.

In general, you shouldn't expect relationships to hold between measure-theoretic and topological notions of "largeness" and "smallness"; they're sort of orthogonal. I guess there is the almost trivial fact that under Lebesgue measure, a set of full measure must be dense, but the converse is certainly not true.

I don't think it's likely that any interesting topological assumptions can get you closer here. $\mathbb{R}$ with Lebesgue measure is about as nice a topological measure space as you could desire (you can take $[0,1]$ instead if you like compact spaces).

Okay, here's one case where it works. Suppose $X$ has the discrete topology. Then there aren't any sets which are dense and have empty interior, so your claim is vacuously true for any measure. (Exception: $X = \emptyset$, but in this case the claim holds for a different reason.)

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  • $\begingroup$ “Now to make it dense, just take the union of your fat Cantor set with the rationals. It's still meager so it still has empty interior, and the measure didn't change.” I actually tried this, but I didn’t realize why meager subsets of $\mathbb R$ had empty interior (due to Baire’s category theorem: a countable intersection of dense open sets is dense.) $\endgroup$ – triple_sec Aug 21 '15 at 0:12
  • $\begingroup$ FYI: I formulated this conjecture in an attempt to give a topological solution to this interesting measure-theoretic problem. As you explained, however, such an attempt is doomed to failure. $\endgroup$ – triple_sec Aug 21 '15 at 0:33

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