Is it possible to abstract a Riemann integral into a "higher" integral with measure?

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I'm not very comfortable with more generalised integrals such as the Lebesgue integral yet, but I'm working through some material to achieve that goal.

I have a question which stems simply from curiosity. Suppose I have a Riemann integral $$I=\int_a^bf(x)g(x)dx.$$ Is it possible to recast this form of Riemann integral into a more general integral such as the Lebesgue integral, in which the function $g(x)$ is absorbed into the measure?, e.g. does there exist a measure $\alpha$ such that $$I=\int_a^b f(x)d\alpha,$$ and if so what form would it take? If so, how would this relate to the equivalent Riemann-type sum (clearly it wouldn't actually be a Riemann sum)

In my notes from a long time ago, I recall seeing a theorem concerning the Riemann-Stieltjes integral, which states (probably very roughly!)

Let $f\in RS(\alpha)$ and suppose $\alpha$ has a continuous derivative on $X= [a,b]$. Then $$\int_X fd\alpha = \int_a^b f(x)\alpha'(x)dx,$$ and the Riemann integral exists,

which seems vaguely related to my question.