Given $f$ increasing on $[a,b]$, $g(x)\in R(\alpha)$ on $[a,b]$, $\alpha \in C([a,b])$ and $\alpha \in BV([a,b])$ $$ \beta(x)=\int_a^xg(z)d\alpha(z) \text{ on [a,b]} $$
Why is the additional assumption $f$ continuous and what theorem if any is needed to show the we can substitute $d\beta$ in the following? $$ \int_a^bfd\beta= \int_a^bfgd\alpha $$
I thought that by the definition of $\beta$ it is differentiable and continuous on $[a,b]$ and I just did a direct substitution $d\beta=g(x)d\alpha(x)$. With an unused assumption I am concerned I overlooked something.
In general, for any increasing $\beta$ on $[a,b]$, $$ \beta(b) - \beta(a) \geq \int_a^b \beta'(t) dt $$ $\beta'$ exists a.e. since $\beta$ is increasing, so $\beta'$ is Riemann integrable. The above turns into equality when $\beta$ is absolutely continuous.