For non-negative functions, are Riemann-Stieltjes and Lebesgue integrals equivalent?

Question

For functions $f : \Bbb{R} \mapsto \Bbb{R^+ \cup 0}$ which are non-negative everywhere, does existence of the are the Riemann-Stieltjes integral imply existence of the and the Lebesgue integral (and is the converse true? If the integrals exist, are they always equal?

I suspect the answer to the existence issue is that existence of the Lebesgue integral does not imply that the Riemann integral exists, but I can't find a counterexample function. If the probelm is restricted to proper integrals, I suspect the equivalence does hold.

Answer 1

If $f$ is Lebesgue-integrable, and $h=f$ everywhere except on a set of Lebesgue measure zero, then $h$ is also Lebesgue-integrable (and $\int h = \int f$). This property certainly doesn't hold for the Riemann-Stieltjes integral.

For instance, the characteristic function of the rationals on $[0,1]$ is Lebesgue-integrable (with integral $0$), but not Riemann-integrable. In fact, almost all functions that are Lebesgue-integrable fail to be Riemann(-Stieltjes)-integrable.

Answer 2

You can not really compare the two, because Lebesgue integral $$\int f \;d\mu$$ only depends on $f$, and on the other hand the R-S integral $$\int f\; dg$$ depends on both $f$ and $g$.

For example, if we take $g$ to be a constant function, then the R-S integral $$\int f \;dg = 0$$ for any function $f$, even when $f = \chi_{A}$ where $A$ is a non-Lebesgue measurable set, here clearly $f$ is not Lebesgue integrable because it is not Lebesgue measurable.