Riemann-Stieltjes integral with respect to functions equal almost everywhere

Question

Let $f, g$ be Lebesgue integrable, real-valued functions on $[0, 1]$ with bounded variation, and let $\phi : [0, 1] \to \mathbb R$ be continuous. Assuming that $f = g$ almost everywhere, does one have $$\int_0^1 \phi \mathrm{d}f = \int_0^1 \phi \mathrm{d}g$$ ?

Answer 1

Any continuous function $\phi$ is RS-integrable with respect to any functions $f,g$ with bounded variation. This implies that $f,g$ are RS-integrable with respect to $\phi$ (integration by parts for RS integrals) and

$$\int_0^1\phi\,df- \int_0^1 \phi \, dg = \phi(1)[f(1)-g(1)]-\phi(0)[f(0)-g(0)] - \left[\int_0^1 f \, d\phi -\int_0^1 g \, d\phi \right] $$

Suppose $f(x) = g(x)$ for all $x \in [0,1)$ and $f(1) \neq g(1)$, then it is straightforward to show using RS-sums that when $ \phi$ is continuous, we have

$$\int_0^1 f \, d\phi =\int_0^1 g \, d\phi $$

Hence, if $\phi(1) \neq 0$, it follows that

$$\int_0^1\phi\,df- \int_0^1 \phi \, dg = \phi(1)[f(1)-g(1)]\neq 0$$