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Let $(X, \beta, \mu)$ be a measure space, and $(X, \tau)$ be a Hausdorff topological space such that:

  1. $\mathcal{B}(\tau)\subset\beta$; where $\mathcal{B} (\tau)$ is the Borel set generated by $\tau$.
  2. There is a sequence $$ (K_n)_{n\in \mathbb{N}} \in X $$ of compact sets such that $X$ is equal to the union of all those sets in the sequence.
  3. The measure $\mu(K)$ is finite, for all $K$ compact that is in $X$.

I want to prove that, if $\mu$ is regular, then for each $A\in\beta$, and for all $\epsilon>0$, there exists an open set $U \supset A$ such that: $$\mu (U-A) \leq \epsilon$$.

To solve the problem, do we need full regularity or just outer regularity? And why does the space have to be specifically Hausdorff? Can the problem still be solved with a stronger/weaker separation axiom?
In general, what restrictions given in the problem statement can we remove/add while still having a similar solution?

Edit: The following is a relevant post about regular measures.
Is every finite Borel measure on a locally compact Hausdorff, $\sigma$-compact and separable space automatically regular?

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    $\begingroup$ I'm not sure whether this is a research-level question; Math.SE might be more suitable. But either way, can you state precisely what you mean by "$\mu$ is regular"? Different authors use different definitions, and for some, the statement you want to prove is literally part of the definition of being regular. $\endgroup$
    – Nate Eldredge
    Commented Jan 21, 2021 at 17:01
  • $\begingroup$ @NateEldredge , thanks for the comment. Regarding the level of the question, I thought it was of a level high enough for MO; however, I'll be more cautious and try MathSE instead next time. Re: regularity, similarly to this post, I define regularity as having both inner regularity and outer regularity. $\endgroup$
    – evaristegd
    Commented Jan 21, 2021 at 17:08
  • $\begingroup$ I see, so you know only that it's inner and outer regular on Borel sets, but you want to know whether the outer regularity also holds on sets $A \in \beta$ which are not necessarily Borel. I don't think it is true. $\endgroup$
    – Nate Eldredge
    Commented Jan 21, 2021 at 17:17
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    $\begingroup$ In particular, see math.stackexchange.com/questions/209532/…. If $A \subset \mathbb{R}$ is a set with zero inner Lebesgue measure and nonzero outer Lebesgue measure, there is an extension $\mu'$ of Lebesgue measure to $\beta = \sigma(\mathcal{B} \cup \{A\})$ for which $\mu(A)=0$. But for any open set $U \supset A$ we have $\mu'(U) = m(U) \ge m^*(A) > 0$. $\endgroup$
    – Nate Eldredge
    Commented Jan 21, 2021 at 17:22

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