I think it is true for $\mathbb R$ with usual metric. How about others? How to prove it?
Motivation:
I got this idea when I was reading a proof for Lebesgue's Criterion of Riemann Integrability, here it is:
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Please note the red marker: "[$[a,b] - \mathop {\cup}^{n}_{j = 1} I_j$] is a finite union of closed intervals". I think [$[a,b] - \mathop {\cup}^{n}_{j = 1} I_j$] is compact here. I've no idea how to prove the equivalence.
P.S. here is the definition of Lebesgue outer measure of Carothers' Real Analysis:
I think I've fully understood this question. Yes, Przemysław Scherwentke's answer here is correct. However, [$[a,b] - \mathop {\cup}^{n}_{j = 1} I_j$] is more complicated than the general title coz $I_j$ and [a,b] both are intervals here(distinct by open and closed) that means they all have interiors that are not empty. Since {$j$} is finite, that means if {$I_j$} are disjoint each other, it will be trivial to show a finite union((n+1) elements) of closed intervals; Otherwise, suppose {$I_j$} have been rearranged from small to large by values of left endpoints. Without loss of generality, there exist $I_k$ and $I_m$ are not disjoint that imply the interval between left endpoints of $I_k$ and $I_m$ will be covered by $I_k$. Apparently, a finite union(less than (n+1) elements) of closed intervals are trivial again. Indeed, {$I_j$} is a cover for all points whose oscillation equal or greater than $1/k$, so it should be chosen as disjoint each other as possible.
In particular, One example of nowhere continuous function is the indicator function of the rational numbers on $[0,1]$, also known as the Dirichlet function on $[0,1]$. It is not Riemann Integrable and neither is indicator function of the irrational numbers. Actually, Lebesgue Outer measure of discontinuous points of Dirichlet function on $[0,1]$ is $1$ which means m*(${ω_f≥1/k}$) $> 0$ when $k≥N$ where ∃ $N∈ \mathbb N^+$. So reverse implication of the proof can not work on the Dirichlet function on $[0,1]$.