I would like to show that the (positive) measure $n$ on $\Bbb R \setminus \{0\}$ defined by $n(E) = \displaystyle \int_E \frac{dx}{|x|}$ is outer regular ($dx$ being the usual Lebesgue measure).
— In particular, I have to show that for any fixed Borel set $E$, I have $$\forall \epsilon>0 \quad \exists O \supset E \text{ open,} \quad 0 ≤ n(O)-n(E)<\epsilon$$
If $n(E)=\infty$, then this is trivial. If $n(E)<\infty$, then $f(x) = 1/x$ is $L^1(E,m)$ and by regularity of the Lebesgue measure $m$, we have : $$\exists O \supset E \text{ open,} \quad m(O)-m(E)<\epsilon/ \|f\|_{L^1(E,m)}$$
— But I don't know how to conclude. The minus sign in $-n(E)$ bothers me, and even if I had a function $f$ in $L^1(\Bbb R \setminus \{0\})$ (with $f \geq 0$), I wouldn't know how to do it.
Could I say $$n(O)-n(E) = n(O \setminus E) = \int_{O \setminus E} \frac{dx}{|x|} \leq m(O \setminus E) \|f\|_{L^1(O,m)} < \epsilon$$ if I suppose that $f \in L^1(O,m)$ and I take $m(O)-m(E)<\epsilon/ \|f\|_{L^1(O,m)}$ ?
Thank you in advance for any help!