Theorem 0.2
Let \(f,g\) be Riemann integrable functions on \([a,b]\). Suppose that \(f(x) \leq g(x)\) for all \(x \in [a,b]\). Then $$ \begin{align*} \int_a^b f(x) \ dx \leq \int_a^b g(x) \ dx \end{align*}

Proof

For any partition \(\mathcal{P}\), we have

$$ \begin{align*} M_j(f) \leq M_j(g) \end{align*} $$

so

$$ \begin{align*} U(f, \mathcal{P}) \leq U(g, \mathcal{P}) \quad\quad\quad (1) \end{align*} $$

But \(f\) and \(g\) are Riemann integrable. This means that the lower integral is equal to the upper integral by definition

$$ \begin{align*} \int_a^b f(x)dx = (U) \int_a^b f(x)dx = (L) \int_a^b f(x)dx \quad\quad\quad (2) \end{align*} $$

This is true for both \(f\) and \(g\). Hence, by (1) and (2)

$$ \begin{align} \int_a^b f(x)dx \leq \int_a^b g(x)dx \end{align} $$

References