Show that the Lebesgue Stieltjes measure corresponding to $\alpha(x) = \mu((0,x])$ is $\mu$.

Question

This is exercise 4.1 from Bass: Let $\mu$ be a measure on the Borel $\sigma$-algebra fo $R$ such that $\mu(K) < \infty$ whenever $K$ is compact, define $\alpha(x) = \mu((0,x])$ if $x \ge 0$ and $\alpha(x) = -\mu((x,0])$ if $x < 0$. Show that $\mu$ is the Lebesgue-Stieltjes measure corresponding to $\alpha$.

I've been trying to tackle this problem using the $\pi-\lambda$ theorem. I've gotten to the point of proving that the set of half open intervals is a $\pi$ system, and that since the measures are both $\sigma$-finite, $\mu$ and $m^*$ are the same on the Borel $\sigma$-algebra, which is generated by the half open intervals. But, I'm stuck here, because I think that I need to now show that $\mu$ and $m^*$ are the same on the larger class of Lebesgue-measurable sets, and I'm not sure how to do that.

I've thought about using Caretheodory's extension theorem for this - my main issue being that I don't know how to apply this because I don't know any generating sets for the Lebesgue $\sigma$-algebra...

Answer 1

I don't see why you think you need to show they are the same for all Lebesgue measurable sets. Consider the example $\alpha$ being the Cantor function https://en.wikipedia.org/wiki/Cantor_function. Then the Cantor set has $\mu$ measure equal to $1$, but Lebesgue measure zero. Then you should be able to find a subset $E$ of the Cantor set which is non-measurable with respect to $\mu$, but clearly $E$ is Lebesgue measurable.