Question: Let $f$ be continuous and such that $f(x) \geq 0$ for all $x \in [a,b]$. If $g$ is strictly increasing on $[a,b]$. Show that $\int_{a}^{b} fdg=0$ if only if $f(x) = 0$ for all $x \in [a,b]$.
I thought I could solve the problem using the Mean-Value Theorem for Riemann-Stieltjes integrals. We know that $ f,g$ are functions on $[a,b] $ with $ f$ continuous and $ g $ increasing. And let $m$ and $M$ be respectively the inf and sup of $f$ on $ [a, b]$. Then there exists $c \in [m, M] $ such that
$$\int_{a}^{b} f(x)dg(x) = c(g(b)-g(a)).$$
So, suppose that $\int_{a}^{b} fdg=0$, therefore $c(g(b)-g(a))=0$. Then $c = 0$ or $(g(b)-g(a))=0$, but the second one can´t happen because $ g $ is an increasing function. Therefore, $c=0$.
Also, we know that $f$ is continuous, therefore $c = f(x)$ for some $x \in [a,b]$. In particular, $c=0=f(x) $ for some $x \in [a,b]$.
Is this a valid proof? Is there an other way to prove it?
I think a better way to do that direction is by contrapositive. That is, assume there exists $x\in [a,b]$ such that $f(x)>0$. By continuity of $f$ there exists $\varepsilon>0$ and $\delta>0$ such that $f(y)>\varepsilon>0$ on some neighborhood $B_\delta(x)\cap [a,b]$. For simplicity assume $B_\delta(x) \subseteq [a,b]$. Since $f$ is non-negative,
$$ \int_a^b f(x) dg \geq \int_{B_\delta(x)} \varepsilon dg = \varepsilon (g(x+\delta) - g(x-\delta)). $$
The last quantity is positive since $g$ is increasing.