Show that $\mu$ is the Lebesgue-Stieltjes measure corresponding to $\alpha$

1k Views Asked by At

I am preparing the quiz for the Lebesgue measure courses, and I am stuck with the problem of the book "Real analysis for graduate students"

The problem is following.

Let $\mu$ be a measure on the Borel-$\sigma$ algebra of $\mathbb{R}$ 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 $\alpha$

The only hint I got so far is that I have to use Caratheodory Extention Thorem. Could anybody help to solve that?

Thanks in advance.

1

There are 1 best solutions below

2
On

Going off user8268's comment, it's easy to see that the function $\sigma((a,b]):=\alpha(b)-\alpha(a)=\mu((a,b])$. The Caretheodory extension theorem says that you can uniquely extend this measure $\sigma$ to a Borel measure $\nu$ such that $\nu((a,b])=\mu((a,b])$. By uniqueness, $\mu$ is precisely the Lebesgue-Stieltjes measure of $\alpha$.