Lebesgue-Stieltjes measures on $\mathbb{R}$ are outer regular and inner regular by compact sets

Choose $\eta >0$ so small that $F((b+\eta {)}^{-})-F(b)<\epsilon$; then the open interval $U:=(a,b+\eta )$ contains $I$ and [L1] gives

${\mu }_F(U\setminus I)={\mu }_F((b,b+\eta ))<\epsilon \mathrm{.}$

Likewise choose $\delta >0$ with $F((a+\delta {)}^{-})-F(a)<\epsilon$; then the closed interval $K:=[a+\delta ,b]$ is compact by [L4], satisfies $K\subseteq I$, and [L1] gives

${\mu }_F(I\setminus K)={\mu }_F((a,a+\delta ))=F((a+\delta {)}^{-})-F(a)<\epsilon \mathrm{.}$

[L1, L4, choose]

$A\subseteq U,{\mu }_F((U\cap I_m)\setminus A)<\epsilon \mathrm{.}$

By step 1.1, every trace $H\cap I_m$ with $H\in \mathcal{H}$ lies in ${\mathcal{O}}_m$. Those traces form an algebra on $I_m$, because $\mathcal{H}$ is an algebra on $\mathbb{R}$ and traces preserve complements and finite unions. By [L5] together with the generator statement of [L3], they generate the Borel sigma-algebra of the subspace $I_m$. [step 1.1, L3, L5]

Then $U:={\bigcup }_n{U}_n$ is open, contains $A$, and

$(U\cap I_m)\setminus A\subseteq {\bigcup }_n((U_n\cap I_m)\setminus A_n),$

so ${\mu }_F((U\cap I_m)\setminus A)<\epsilon$. If instead $A_n\downarrow A$, then $A_0\subseteq I_m$ has finite measure by [L1], so [L2] gives ${\mu }_F(A_n\setminus A)\to 0$; choose $n$ with ${\mu }_F(A_n\setminus A)<\epsilon \mathrm{/}2$, then choose open $U_n\supseteq A_n$ with ${\mu }_F((U_n\cap I_m)\setminus A_n)<\epsilon \mathrm{/}2$. The open set $U_n$ contains $A$, and

${\mu }_F((U_n\cap I_m)\setminus A)\le {\mu }_F((U_n\cap I_m)\setminus A_n)+{\mu }_F(A_n\setminus A)<\epsilon \mathrm{.}$

[step 2.1, L1, L2]

Therefore [L3] gives that every Borel subset of $I_m$ lies in ${\mathcal{O}}_m$: every bounded Borel set is relatively outer regular inside a compact interval. [step 2.1, step 3.1, L3]

Choose open $U\supseteq I_m\setminus A$ with ${\mu }_F((U\cap I_m)\setminus (I_m\setminus A))<\epsilon$. Put $K:=I_m\setminus U$. Then $K$ is closed in the compact interval $I_m$, hence compact by [L4], and $K\subseteq A$. Moreover

$A\setminus K=A\cap U=(U\cap I_m)\setminus (I_m\setminus A),$

so ${\mu }_F(A\setminus K)<\epsilon$. Thus every bounded Borel set is also inner regular by compact sets. [step 4.1, L4]

Then $E_m\uparrow E$, so [L2] gives ${\mu }_F(E)={\sup }_m{\mu }_F(E_m)$. For each $m$, step 5.1 gives a compact $K_m\subseteq E_m\subseteq E$ with ${\mu }_F(E_m\setminus K_m)<1\mathrm{/}(m+1)$. Hence

$\sup \{{\mu }_F(K):K\subseteq E,K\text{ compact}\}\ge {\sup }_m{\mu }_F(K_m)={\sup }_m{\mu }_F(E_m)={\mu }_F(E),$

while the reverse inequality is monotonicity. [step 5.1, L2, algebra]

This is possible directly from [L2] on $E\cap I_m\uparrow E$. Step 4.1 gives an open $V\supseteq E\cap I_m$ with ${\mu }_F((V\cap I_m)\setminus E)<\epsilon \mathrm{/}3$. For each of the two tails $E\cap (m,\mathrm{\infty })$ and $E\cap (-\mathrm{\infty },-m)$, decompose into unit half-open strips and apply step 4.1 on each compact strip, choosing the errors summably below $\epsilon \mathrm{/}3$; enlarging each strip by a thin open shell whose ${\mu }_F$-measure is also chosen summably below $\epsilon \mathrm{/}3$ via [L1], the union of those stripwise open sets is open and contributes total excess below $2\epsilon \mathrm{/}3$. Taking the union with $V$ gives an open set $U\supseteq E$ with ${\mu }_F(U\setminus E)<\epsilon$. [step 4.1, step 6.1, L1, L2]

So $\inf \{{\mu }_F(U):E\subseteq U,U\text{ open}\}=+\mathrm{\infty }={\mu }_F(E)$. Combining this with step 7.1 gives the outer regularity equality in all cases. The inner regularity equality is step 6.1, and the finite-measure $\epsilon$-approximation statement is exactly the combination of steps 6.1 and 7.1. [step 6.1, step 7.1, algebra] ∎