(related to Proposition: Linearity and Monotony of the Riemann Integral)
It follows from the definition of Riemann integrable functions that they are bounded. By applying the rules of adding Riemann upper and lower integrals of bounded functions we get the of inequations \[\int_{a~*}^{b}f(x)dx + \int_{a~*}^{b}g(x)dx \le \int_{a~*}^{b}(f+g)(x)dx \le \int_{a}^{b~*}(f+g)(x)dx \le \int_{a}^{b~*}f(x)dx + \int_{a}^{b~*}g(x)dx.\] Because, by hypothesis \[\int_{a~*}^{b}f(x)dx =\int_{a}^{b~*}f(x)dx = \int_{a}^{b}f(x)dx\quad\quad\text{and}\quad\quad\int_{a~*}^{b}g(x)dx =\int_{a}^{b~*}g(x)dx = \int_{a}^{b}g(x)dx\] we get the result that the function \(f+g:[a,b]\mapsto\mathbb R\) is Riemann integrable and that \[\int_{a~*}^{b}(f+g)(x)dx = \int_{a}^{b~*}(f+g)(x)dx = \int_{a}^{b}(f+g)(x)dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx.\] Similarly, by applying the rules of scalar multiplication of Riemann upper and lower integrals, we get the equation chain \[\int_a^{b~*}(\lambda\cdot f)(x)dx=\lambda\cdot\int_a^{b~*}f(x)dx =\lambda\cdot\int_a^{b}f(x)dx=\lambda\cdot\int_{a~*}^{b}f(x)dx=\int_{a~*}^{b}(\lambda\cdot f)(x)dx.\] This means that the function \(\lambda\cdot f:[a,b]\mapsto\mathbb R\) is Riemann integrable and that \[\int_a^{b}(\lambda\cdot f)(x)dx =\lambda\cdot\int_a^{b}f(x)dx.\]
If \(f(x)\le g(x)\) for all \(x\in[a,b]\), then we have \(f(x)-g(x)\le 0\) for all \(x\in[a,b]\). It follows from the linearity rules that
\[\int_a^{b}(f-g)(x)dx=\int_a^{b}f(x)dx-\int_a^{b}g(x)dx\ge \int_a^{b}0dx=0.\] It follows \[\int_a^{b}f(x)dx\le\int_a^{b}g(x)dx.\]
