# 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.

• The Cantor set is not merely uncountable. It has cardinal $|\Bbb R|.$ Can we show by transfinite recursion that $|dom(\mu_f)|\le |\Bbb R|$ and that there exists $S\in dom (\mu)$ with $|S|=|\Bbb R|$ and $\mu(S)=0?$ Feb 3 at 1:07
• @DanielWainfleet So you mean for any Lebesgue-Stieltjes measure, the Cantor set is measurable, so we can conclude the set of measurable set by the complete measure is strictly greater than the set of Borel sets? I guess this does not apply to Caratheodory consturction in general? Feb 3 at 12:03
• What I meant was to ask to whether there must exist an uncountable closed set $S$ with $\mu(S)=0$.... As every closed uncountable subset of $\Bbb R$ has cardinal $|\Bbb R|$. Feb 3 at 17:51
• @DanielWainfleet I see. $|dom(\mu_f)|=|\mathbb{R}|$ since it is just the Borel algebra. I do not have much idea how to exhibit such a set $S$ though; it should not be the same for all Lebesgue-Stieltjes measures since the Cantor set has measure $1$ with the Cantor measure. Feb 3 at 18:04

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.

• Thank you for your insight. I would very much like to see how the construction of the Cantor-like set will proceed. Feb 5 at 20:00
• will allow us to do a Cantor-like construction to get an uncountable Borel subset $G$ with $\mu(G)=0$. I suspect the details of this construction may become technical Just split the interval $I$ into, say, $8$ equal parts and chose two non-adjacent ones with total measure $\le \mu(I)/2$. Repeat for each chosen subinterval. Feb 7 at 11:47
• @fedja I'm not 100% convinced that something can't go horribly wrong. For instance, maybe $F$ is the complement of some Cantor space $C \subset \mathbb{R}$, and in paring down $F$ like this, we might have accidentally repeated the construction of $C$ (which should leave us with nothing). I am convinced that that's almost all of the main idea though, and you'd only have to be marginally careful if you had to be careful at all. In any case, I found an argument elsewhere on SE that seems slicker to me, so I stuck that in my answer. Feb 9 at 21:36
• @MichaelJesurum Thank you so much for providing the detail after it is chosen as the answer; I really appreciate it. Feb 10 at 0:19
• @MichaelJesurum I didn't make myself clear, indeed. I do not use the preliminary decomposition into $F$, etc., just start with the whole interval and the original finite measure and construct a Cantor set of measure $0$ (if the measure is just $\sigma$-finite, some extra care is needed, indeed, here you are completely right, but the OP requested merely the result for the Lebesgue-Stiltjes measures) Feb 12 at 0:28