We have $\mu$, final non-negative measure on Borel sets on $[a,b]$. Define $\alpha (x) = \mu ([a,x])$. Assume that $\alpha$ is continuous at the interval$[a,b].
$f$ is continuous at the interval $[a,b]$. Show that $\int_{[a,b]}f(x)d\mu (x) = \int_a^b f(x)d\alpha (x)$.
Integral on the right side is Stielts integral.
I started with $\mu ([a,x]) - \phi([a,x]) = V(x) - V(a) - [V(x) - V(a) - (\alpha(x) - \alpha(a))] = \alpha (x) - \alpha(a) = \int_a^bf(x)d\alpha(x) $
And I know that $\int_a^b f(x)d\mu (x)$ generalizes $\int_a^bf(x)d\alpha(x)$