If $f(x) \leq g(x)$ for all $x \in [a,b]$ then $\int_a^b f \leq \int_a^b g.$
What I did i followed Bartle's method but I cannot keep up on what He meant.
This is the uniqueness so that we'll be familiar:
S as "sum" and I'll just P' for P partitions
Referring to this:
$\left|S(f;P')-\int_a^bf\right|<\frac{\varepsilon}{2}$ and $\left|S(g;P')-\int_a^bg\right|<\frac{\varepsilon}{2}$ (1)
This was Bartle's method:
To prove the inequality, we note that the triangle inequality be applied to (1)
thus,
$\int_a^b f - \frac{\varepsilon}{2} < S(f;P')$ and $S(g;P')<\int_a^b g + \frac{\varepsilon}{2} $
If we use the fact that $S(f;P')\leq S(g;P')$ , we have
$\int_a^b f \leq \int_a^b g + \varepsilon$
But, since $\varepsilon$ is an arbitrary number we conclude that $\int_a^b f \leq \int_a^b g$
Please help me to understand this, how it turned out to this it's simplified and there were no further details