Integración y una función en aumento.

Question

Deje $g_{1},g_{2}\in\mathcal{R}([a,b])$ (integrable a Riemann) tal que $$\int_{a}^{x}{g_{2}(t)dt}\leq\int_{a}^{x}{g_{1}(t)dt}\phantom{a}\text{for each}\phantom{a}x\in [a,b]$$ and $$\int_{a}^{b}{g_{1}(t)dt}=\int_{a}^{b}{g_{2}(t)dt}$$ Show that if $ f: [a, b] \ rightarrow {\ mathbb {R}}% PS

¿Alguien tiene alguna sugerencia?

Answer 1

Considere $$G(x) =\int_{a} ^{x} (g_1(t)-g_2(t))\,dt$$ which is continuous, of bounded variation, and non-negative on $ [a, b]$ further $ G (a) = G (b) = 0$. Now we are given that $ f$ is increasing on $ [a, b]$ and hence $ G$ is Riemann Stieltjes integrable with respect to $ f$ and $ f$ is Riemann Stieltjes integrable with respect to $ G$ so that $$\int_{a}^{b} f(x) \, dG(x) +\int_{a} ^{b} G(x) \, df(x) =G(b) f(b) - G(a) f(a) =0$$ The second integral on left side is non-negative and hence the first integral $$\int_{a} ^{b} f(x)\, dG(x) \leq 0$$ The result in question is established if one notices that $% #% PS