Fundamental Theorem of Calculus. For the Riemann–Stieltjes integral. I am looking of a fundemental theorem of calculus for the Riemann-Stieljes integral.
The Fundamental theorem of calculus : https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus ,
links differentiation to Riemann integration. 
One can also define the Riemann-Stieljes integral : https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral#Formal_definition. 
I.e (for $f,g:\mathbb{R}\to\mathbb{R}$) consider $$I(t)=\int_0^t f(s)dg(s) $$
If $g$ is a differentiable function then we can write
$$I(t)=\int_0^t f(s)g'(s)ds $$
and the fundemental theorem of calculus tells us that 
$$I'(t)=f(t)g'(t). $$
$\textbf{Question}$ : what if $g$ is not differentiable, say $g$ is only monotonically increasing and non-negative. Then what can we say about $I'(t)$ ? Can we bound it from above? 
In the above let us assume the function $f$ is a continuous and non-negative, and   Riemann-Stiejles integrable.
 A: Assume only that $f$ is Riemann-Stieltjes integrable with respect to $g$ and $g$ is monotonically increasing on $[a,b]$.  Under these assumptions $f$ is almost everywhere continuous and $g$ is almost everywhere differentiable. We can then say that $I$ is differentiable at any point $t \in [a,b]$ where simultaneously $f$ is continuous and $g$ is differentiable, and $I'(t) = f(t) g'(t)$.
Proof:
Since a Riemann-Stieltjes integrable function must be bounded there exists finite $m,M$ such that $m = \inf_{s \in [t,u]}f(s) $ and $M = \sup_{s \in [t,u]}f(s) $, and since $g$ is increasing,
$$m[g(u)-g(t)] \leqslant I(u) - I(t) =\int_t^uf \, dg \leqslant M[g(u) - g(t)]$$
Thus,
$$m \frac{g(u) - g(t)}{u-t} \leqslant \frac{I(u)-I(t)}{u-t} \leqslant M\frac{g(u) - g(t)}{u-t}$$
Taking the limit as $u \to t$ and applying the squeeze theorem we get $I'(t) = f(t) g'(t)$. Note that $m,M \to f(t)$ as $u \to t$ follows from the continuity of $f$ at $t$.
This result holds more generally if $g$ is of bounded variation.
