# Are all measures on $\mathbb{R}$ Lebesgue-Stieltjes measures on all measureable sets?

(Inspiration 1: Are all measures Lebesgue-Stieltjes measures?)

(Inspiration 2: Bass' real analysis textbook, exercise 4.1:

Let $$\mu$$ be a measure on the Borel $$\sigma$$-algebra on $$\mathbb{R}$$ such that $$\mu(K) < \infty$$ whenever $$K$$ is compact, define $$\alpha(x) = \mu((0,x])$$ if $$x\geq 0$$ and $$\alpha(x) = -\mu((x,0])$$ if $$x < 0$$. Show that $$\mu$$ is the Lebesgue-Stieltjes measure corresponding to $$\alpha$$.

)

My question, in a nutshell, is whether the above statement in the exercise can be extended to all $$\mu^*$$-measurable sets, where $$\mu^*$$ is the Lebesgue-Stieltjes (hereafter, L-S) measure corresponding to $$\alpha$$. The motivation for this question follows:

When solving the exercise, I have used Carathéodory extension theorem, after proving that $$\mu((a,b]) = \ell((a,b]) := \alpha(b) - \alpha(a)$$, to show that $$\mu$$ and $$\mu^*$$ are both extensions of $$\ell$$ onto the $$\sigma$$-algebra generated by the algebra generated all intervals of the form $$(a,b]$$, which clearly contains the Borel $$\sigma$$-algebra.

I was wondering if the argument could somehow be modified to show that $$\mu$$ coincides with $$\mu^*$$ for all $$\mu^*$$-measurable sets, thereby showing that $$\mu$$ is either exactly the L-S measure or an extension of it.

If it were false, then I would suspect that there is a measure that coincides with the L-S measure on all Borel sets but does not coincide for at least one non-Borel, yet L-S-measurable set. I cannot think of what consequences this would imply, but my beginner's intuition rings an alarm here.

Just for clarity, here is my exact question:

Let $$(\mathbb{R}, \mathcal{A}, \mu)$$ be a measure space. Let $$\alpha$$ be defined as above, $$\mu^*$$ be the L-S measure corresponding to $$\alpha$$, and $$\mathcal{A}'$$ be the collection of $$\mu^*$$-measurable sets. If $$\mathcal{A}' \subseteq \mathcal{A}$$, then does $$\mu|_{\mathcal{A}'}=\mu^*$$ hold?

Yes, it is true. The main idea is to use Caratheodory extension theorem. But we need to check $$\ell$$-finiteness. (We can check this easily because the union of $$K_i$$ is finite since $$\mu(K)$$ is finite if $$K$$ is compact.) Also, we can verify that $$\mu=\ell$$ at finite union of collections of the form $$(a, b]$$. Finally, the measure is unique and it is a L-S measure, since $$\mu$$ is measure on Borel sigma algebra, $$\mu=m$$ as desired.
• Thank you for the answer! I would presume $\ell$-finiteness corresponds to $\sigma$-finiteness in Bass' vocabulary, that is, there is a countable cover of $\mathbb{R}$ with finite measures. I still do not understand how this answer incorporates non-Borel L-S-measurable sets, as the Carathéodory extension thm. only deals with the $\sigma$-algebra generated by the $(a,b]$-type intervals, which then becomes the Borel $\sigma$-algebra. Could you please elaborate?