Let $m$ be a Borel measure on a topological space $(X,\tau)$. Then $m$ is of property:

  1. P1 if there is a sequence $\mathcal F$ of closed subsets of $X$ such that for any $\epsilon >0$ and each $A\in\tau$, there is $F\in\mathcal F$ such that $F\subseteq A$ and $m(A-F)<\epsilon$.

  2. P2 if there is a sequence $\mathcal K$ of compact subsets of $X$ such that for any $\epsilon >0$ and each $A\in\tau$, there is $K\in\mathcal K$ such that $K\subseteq A$ and $m(A-K)<\epsilon$.

An easy result: if $X$ is compact, then P1 $\implies$ P2, but not conversely.

Q) Is there is any counterexample for this? I was not able to think of any non Hausdorff topological measure space that has P2 but not P1.

I welcome any idea, comment, or even reference.

  • $\begingroup$ I would guess you'd have to work in a space that isn't locally compact. For instance, consider $\ell^\infty(\mathbb{R})$, and let $A$ be the unit ball; since closed balls in this space are not compact, P2 should probably not hold. Of course it depends on the measure you put on this space, and I'm not really sure where to go from there. $\endgroup$ – Chris Dec 12 '18 at 22:49

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