(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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – 이희원
    Sep 20 at 4:36

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