Let $\mu$ be a Lebesgue-Stieltjes Measure who's distribution function F has a positive continuous derivative on R. Show that $d \mu = F'd \lambda$
My thoughts are to show that $\forall E,$ $\mu(E) = \int _E F' d \lambda$. I was thinking to show it using a Borel set $[a,b)$ and then using the fundamental theorem of calculus? We have that if the function is Riemann integrable, it's Lebesgue integrable and the two integrals.
So it would look like: Consider any $[a,b) \in B(R)$, then:
$\mu ([a,b) = F(b) - F(a) = \int_{[a,b)} F' d \lambda$
with the first equality coming from the definition of a distribution function and the second equality combing from the fundamental theorem of Calculus.
This seems too simple. Am I missing something?