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.

Rewriting this in terms of $h(x) = g(x)-f(x)$, we have $h(x)>0$ for all $x$, and want to know whether $\int_a^bh(x)dx = 0$ or $\int_a^bh(x)dx < 0$ is possible. This rewriting is a common trick, and makes it a bit easier conceptually because there is now only one function to consider, and $0$ is a nice value to compare functions to. — Arthur, Mar 25 '17 at 16:58
$\mathscr R[a,b]$ is defined as the set of all Riemann integrable functions over $[a,b]$? — AlexR, Mar 25 '17 at 17:00
Possible duplicate of Let $f:[a,b]\to\mathbb R$ be Riemann integrable and $f>0$. Prove that $\int_a^bf>0$. (Without Measure theory) — AlexR, Mar 25 '17 at 17:09
@Itay4 In that case, we have a duplicate. In essence: $\mathscr R[a,b] \ni f-g>0$ implies, that $f-g$ has at least one point of continuity. The integral around the continuous point must be $>0$ in some area, the rest of the function must have integral $\ge 0$ and thus the entire integral is $>0$. — AlexR, Mar 25 '17 at 17:11

score 2 · Accepted Answer · answered Mar 25 '17 at 17:06

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.

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

1 Answers1