[Edited] I am not particularly looking for an answer from a reputable source as the bounty says. Just any clear explanation would suffice.
This question is induced by a remark in Folland's Real Analysis where Lebesgue-Stieltjes measure is discussed. In the chapter, it says one can show that the domain of a Lebesgue-Stieltjes measure $\bar{\mu_f}$ (which is the completion of the Borel measure $\mu_f$ generated by an increasing right-continuous function $f$) is always strictly larger than the Borel sigma algebra on $\mathbb{R}$. I wanted to know how to show this.
More generally, a natural question is that in the Carathéodory extension where we extend a premeasure $\mu$ from a measure-theoretic semi-ring $\mathcal{S}$ to a $\mu$-measurable algebra $\mathcal{M}_{\mu}$, I wanted to know when we could conclude that $\sigma (\mathcal{S})$, the sigma algebra generated by $\mathcal{S}$, is strictly smaller than $\mathcal{M}_{\mu}$.
Here are some (hopefully correct) facts that I know of:
- If the premeasure $\mu$ is sigma-finite, its extension to $\mathcal{M}_{\mu}$ is the completion.
- $\sigma(\mathcal{S})$ can be described by using transfinite induction indexed by the first uncountable ordinal.
- If the cardinality of $\mathcal{S}$ is between that of the natural numbers and the reals, then the cardinality of $\sigma(\mathcal{S})$ is that of the reals.
- [Added] Cantor set is uncountable but of Lebesgue measure zero, so the set of Lebesgue measurable sets has the same cardinality as $2^{\mathbb{R}}$. So at least it shows the set of Lebesgue measurable sets is strictly larger than Borel sets on $\mathbb{R}$.
I would appreciate any reference or help.