Borel algebra always strictly smaller than its completion? [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.
 A: I'm going to stick with the Lebesgue-Stieltjes part of the question. Hopefully Folland had something simpler in mind, but here's one way to show that the domain of the completion has cardinality $2^{\mathbb{R}}$.
Note: I'm not much of a set theorist, so in my mind the only uncountable subsets of $\mathbb{R}$ have size $|\mathbb{R}|$. If someone wants to go over this argument with a more discerning eye in this regard and make some comments, I'd love to know if some things should be changed.
The main idea, as already noted in the original question and the comments, will be to find an uncountable Borel set $G$ with $\mu_{f}(G) = 0$. By completeness, every one of its $2^{\mathbb{R}}$-many subsets will be in the domain of $\overline{\mu}_{f}$.
A good reference for what I'm about to do is Giovanni Leoni's A First Course in Sobolev Spaces (1st edition). All the theorems and propositions I state will be from that book.
The first step will be to decompose $f$ into two increasing functions:
$$f = f_{AC} + f_{S},$$
where $f_{AC}$ is absolutely continuous and $f_{S}$ is differentiable with derivative $0$ $\mathcal{L}^{1}$-a.e (see Theorem 3.73).
With this decomposition, we have
$$\mu_{f} = \mu_{f_{AC}} + \mu_{f_{S}}.$$
Thus, our goal is to find an uncountable Borel set $S$ with measure $0$ with respect to each of these two measures.
We will start with $f_{S}$. By (the proof of) Theorem 3.72, the set $E = \{x : f_{S}'(x) = 0\}$ satisfies $\mathcal{L}^{1}(\mathbb{R} \backslash E) = 0$ and $\mathcal{L}^{1}(f_{S}(E)) = 0$. Note that $E$ is Borel.
Since $f_{S}$ is increasing, it has only countably many discontinuities. Thus the set
$$F = E \backslash \{x : f_{S} \text{ is discontinuous at } x\}$$
is Borel and satisfies $\mathcal{L}^{1}(\mathbb{R} \backslash F) = 0$, $\mathcal{L}^{1}(f_{S}(F)) = 0$, and $f_{S}$ is continuous on $F$. By Proposition 5.9,
$$\mu_{f_{S}}(F) = \mathcal{L}^{1}(f_{S}(F)) = 0.$$
Now we have to deal with $f_{AC}$. By the absolute continuity of $f_{AC}$, we know that $\mu_{f_{AC}}$ is absolutely continuous with respect to the Lebesgue measure. Since $F$ is an uncountable Borel set, it contains a nonempty perfect set $G \subseteq F$ of Lebesgue measure $0$ (as per bof's answer to this question), which means that $\mu_{f_{AC}}(G) = 0$.
All told, we have
$$\mu_{f}(G) = \mu_{f_{AC}}(G) + \mu_{f_{S}}(G) = 0,$$
which completes the proof.
