Existence of closed subset of positive measure

Question

Suppose that $\psi(A)>0$ where $\psi$ is some measure and $A$ is an open set. Is it true that we can always find a closed set such that $B \subseteq A$ and $\psi(B)>0$?

This is a follow up to the previous question:

closed subset that has the same size

Answer 1

If $\psi$ is a Radon measure, then $\psi(A)=\sup\{\psi(K): K\subseteq A, K~\text{is compact}\}$.

In Hausdorff space, compact set is closed.

Answer 2

I assume you are taking $\psi = \mathcal{L}$, the Lebesgue measure.

I would say no for a general topology, for instance if we equip $\mathbb{R}$ with the topology induced by the closed sets as follows: $C$ closed if and only if $C$ is finite of $C=\mathbb{R}$ or $\emptyset$ (check that these are closed), then clearly $(0,1)$ is open but every closed subset is finite and hence has measure zero.

On the other side, if we take the standard topology then yes. Let $A, \psi(A)>0$ be open, so that $A\neq \emptyset$. Since a basis is made of balls, then there is $x\in A, r>0$ such that $B_r(x)\subset A$, so $\overline{B_{r/2}(x)}\subset A$ and $\psi(\overline{B_{r/2}(x)})>0$.