If $g$ is a continuous nonnegative function on $[a,b]$ and if $\int_{a}^{b} g = 0$, then $g$ is identically $0$ on $[a,b]$

Question

I'm reading Elementary Analysis by Ross and don't understand the usage of $\frac{\alpha}{2}$ in the proof of the theorem below. Why not just use $\alpha (d - c)$ in the third inequality?

Proof:

Otherwise, since $g$ is continuous, there is a nonempty interval $(c,d) \subseteq [a,b]$ and $\alpha > 0$ satisfying $g(x) \geq \frac{\alpha}{2}$ for $x \in (c,d)$. Then $\int_{a}^{b} g \geq \int_{c}^{d} g \geq \frac{\alpha}{2}(d - c) >0$.

This post doesn't answer my question because I would still wonder why let $\epsilon = \frac{\alpha}{2}$?

Answer 1

We have $g(x) \ge \frac{\alpha}2$ for all $x \in (c,d)$.

Integrating both sides, gives us the result.

We do not know that $g(x) \ge \alpha$. Hence you can't drop the $\frac12$.

We could have modified the proof by saying that $g(x) \ge \alpha$ directly, then you can just state $\int_c^d g(x)\, dx \ge \alpha (d-c) $ directly.