Without getting too much into details, we are trying to prove that if $f=f_1+f_2$ on $[a,b]$ and $f_1,f_2$ are Riemann-Stiljetes integrable, then $f$ is Riemann-Stiljetes integrable and $$\int_{a}^{b} fd\alpha =\int_{a}^{b} f_1 d\alpha+ \int_{a}^{b} f_2 d\alpha$$
I understand everything in Rudin's proof until (and including) line (21): $$\int_{a}^{b} fd\alpha \leq \int_{a}^{b} f_1 d\alpha+ \int_{a}^{b} f_2 d\alpha$$ but I don't understand the step that comes after it: "If we replace $f_1$, $f_2$ in (21) by $-f_1$, $-f_2$, the inequality is reveresed and equality is proved."
I understand that if we show that $\int fd\alpha \geq \int f_1 d\alpha+ \int f_2 d\alpha$, then equality is proved, but I don't understand how exactly we are suppose to deduce that from (21) and what does it have to do with $-f_1$, $-f_2$.
Maybe this is a very simple argument, but I don't understand it and so I would really appreciate if someone could explain it in details.
It has already been proved that, if $f_1$ and $f_2$ are Riemann-Stieltjes integrable, then$$\int_a^b(f_1+f_2)\,\mathrm d\alpha\leqslant\int_a^bf_1\,\mathrm d\alpha+\int_a^bf_2\,\mathrm d\alpha.$$Applying this to the functions $-f_1$ and $-f_2$, you get that\begin{align}-\int_a^b(f_1+f_2)\,\mathrm d\alpha&=\int_a^b(-f_1-f_2)\,\mathrm d\alpha\\&\leqslant\int_a^b-f_1\,\mathrm d\alpha+\int_a^b-f_2\,\mathrm d\alpha\\&=-\left(\int_a^bf_1\,\mathrm d\alpha+\int_a^bf_2\,\mathrm d\alpha\right),\end{align}and therefore$$\int_a^b(f_1+f_2)\,\mathrm d\alpha\geqslant\int_a^bf_1\,\mathrm d\alpha+\int_a^bf_2\,\mathrm d\alpha.$$