Suppose that $f, g: [0,1] \rightarrow \mathbb{R}$ are Riemann integrable and $f(x) \geq g(x)\ \forall x \in [0,1]$. Show that $\int_0^1 f \geq \int_0^1 g$.
So this should be relatively simple. We have $f$ is Riemann integrable if $L(f) = U(f)$. Where $L(f) = sup\{L(f,P)\ |\ \text{P a partition}\}$, $U(f) = inf\{U(f,P)\ |\ \text{P a partition}\}$.
Now we have that $f(x) \geq g(x)\ \forall x \in [0,1]$. Because this holds, we have $L(f) \geq L(g)$. Since $\int_0^1$ is the common value $L(f) = U(f)$. We have $U(f) \geq U(g)$ as well. And so $\int_0^1 f \geq \int_0^1 g$.
Not too versed in proving stuff of this sort so would appreciate help/clarification.
There's nothing wrong with it, but for someone at your stage, I would expect some explanation of why $f(x) \ge g(x)$ implies $L(f) \ge L(g)$, especially since this is the main part of the proof.