Can someone tell me if is this proof correct? I don't know if it is correct because I did it myself, so..
If $f\in R$ on $[a,b]$ and $g$ is a monotonous function on $[a,b],$ then there exist $\epsilon \in [a,b]$ such that $$\int_a^bfg=g(a)\int_a^{\epsilon}f+g(b)\int_{\epsilon}^bf.$$
Proof:
As $f\in R$ on $[a,b]$ then f is bounded and continuous almost everywhere.
Let $\alpha(x)=\int_a^xf(t)dt$. By the fundamental theorem of calculus, $\alpha'(x)=f(x).$ So we have $d\alpha(x)=fd(x)$.
Now, by the second mean value theorem for riemann-stieltjes integrals*, we have $$\int_a^bfg=\int_a^bgd(\alpha)=g(a)\int_a^{\epsilon}f+g(b)\int_{\epsilon}^bf$$
This is the $*2nd$ MVT for riemann-stieltjes integrals:
Let $f$ be increasing on $[a,b]$, $g$ continuous on $[a,b]$. Then there exist $c\in [a,b]:\int_a^bfdg(x)=f(a)\int_a^cdg(x)+f(b)\int_c^bdg(x)$.