Let $g_{1},g_{2}\in\mathcal{R}([a,b])$ (Riemann-integrable) such that $$\int_{a}^{x}{g_{2}(t)dt}\leq\int_{a}^{x}{g_{1}(t)dt}\phantom{a}\text{for each}\phantom{a}x\in [a,b]$$ and $$\int_{a}^{b}{g_{1}(t)dt}=\int_{a}^{b}{g_{2}(t)dt}$$ Show that if $f: [a,b]\rightarrow{\mathbb{R}}$ is increasing, then $$\int_{a}^{b}{f(t)g_{1}(t)dt}\leq\int_{a}^{b}{f(t)g_{2}(t)dt}$$
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Consider $$G(x) =\int_{a} ^{x} (g_1(t)-g_2(t))\,dt$$ 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 $$\int_{a} ^{b} f(x) \, dG(x) =\int_{a} ^{b} f(x) (g_1(x)-g_2(x))\,dx$$