# Proving that every set in the Borel $\sigma$-algebra on $\mathbb{R}$ is Lebesgue-Stieltjes Measurable

I am reading through Richard Bass's Real Analysis and on page 25 we have the following proposition, using $\alpha$ is a Stieltjes function and $m^*$ is the Lebesgue-Stieltjes outer measure on $\ell((a,b])=\alpha(b)-\alpha(a)$:

Proposition 4.7 Every set in the Borel $\sigma$-algebra on $\mathbb{R}$ is $m^*$-measurable.

The proof begins with:

It suffices to show that every interval $J$ of the form $(c, d]$ is $m^∗$-measurable. Let $E$ be any set with $m^∗(E) < \infty$; we need to show $$m^∗(E) \ge m^∗(E \cap J) + m^∗(E \cap J^c).$$ Choose $I_1, I_2, \ldots,$ each of the form $(a_i, b_i]$, such that $E \subset \cup_i I_i$ and $$m^∗(E) \ge \sum_i [\alpha(b_i) − \alpha(a_i)] − \varepsilon.\qquad(*)$$

Then via a series of calculations the proof concludes:

Thus $$m^∗(E \cap J) + m^∗(E \cap J^c) \le \sum_i m^∗(I_i) \le m^∗(E) + \varepsilon.$$ Since $\varepsilon$ is arbitrary, this proves $(*)$.

My problem with this proof is the existence of $I_1,I_2,\ldots$ that satisfy $(*)$. It seems to me that if $\varepsilon$ is sufficiently small that it may become the case that $(*)$ no longer holds for any $I_1,I_2,\ldots$.

Why is it valid to claim that we can have $I_1,I_2,\ldots$ satisfying $(*)$ for any $\varepsilon$? Or am I misunderstanding something else in this proof?

This is really just a consequence of the definition of $m^*(E)$ the author gives on the page right before the theorem (all taken from the link you provided): it is defined as the inifimum over all such coverings by all such intervals. If you then assume the logical juxtaposition of (*). in other words that there is an $\epsilon > 0$ such that
$$d:=m^∗(E) + \varepsilon \lt \sum_i [\alpha(b_i) − \alpha(a_i)] \quad (**)$$
for all such coverings, then certainly the infimum $I$ satisfies
$$m^∗(E) \lt d \le I,$$
in contradiction to the definition of $m^∗(E)$. Hence, (**) is wrong, and (*) is true.