Measure on Borel $\sigma$-algebra of $\mathbb{R}$ is Lebesgue-Stieltjes measure 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 $x \ge 0$ and $\alpha(x) = -\mu((x, 0])$ if $x < 0$. How do I see that $\mu$ is the Lebesgue-Stieltjes measure corresponding to $\alpha$?

Let $\mu_\alpha$ the Lebesgue_Stieljes measure corresponding to $\alpha$. 
The function $x \mapsto \alpha(x)$ is right-continuous, non-decreasing.
For all $ a<b$, $\mu_\alpha((a,b)) :=\alpha(b)-\alpha(a) = \mu( (0,b) )-\mu((0,a))=\mu(a,b)$. Thus, $\mu_g = \mu$.
