if $f>g$, can it be that $\int^{b}_{a}f =\int^{b}_{a}g$ or $\int^{b}_{a}f >\int^{b}_{a}g \ ?$

Question

I've just started studying integrals, and I've been thinking about the following problem:

Let $f,g \in {\mathscr R[a,b]} \ s.t. f>g, \forall x \in [a,b].$

Can it be that $\int^{b}_{a}f =\int^{b}_{a}g$ or $\int^{b}_{a}f >\int^{b}_{a}g \ ?$

I'm having hard time getting intuition, but I think it's possible.

Any help appreciated.

Answer 1

It must be that $\int_a^b f > \int_a^b g$. Indeed, $f>g$ means $f-g > 0$, so it suffices to show that if a Riemann integrable $f$ satisfies $f>0$, then $\int_a^b f > 0$. Are you aware of the fact that if $f$ is Riemann integrable, then it must have at least one point of continuity? (In fact, almost every point is a point of continuity). Once you know this, since $f>0$ at the point of continuity, you can lower bound $f > \epsilon$ on some ball around the point of continuity.