If $g(x)$ is a continuous nonnegative function on $[a,b]$ and if $\int_{a}^b g(x)dx = 0$ then $g$ is identically $0$ on $[a,b]$
Proof:
Otherwise, since $g(x)$ is continuous, there is a nonempty interval $(c,d) \subseteq [a,b]$ and $\alpha$ satisfying $g(x) \ge \alpha/2$ for $x \in (c,d)$
Then $\int_{a}^b g(x)dx \ge \int_{c}^d g(x)dx \ge \alpha/2*(d-c) > 0$
contradicting $\int_{a}^b g(x)dx = 0$
My question is why do we need $\alpha/2$ instead of just $\alpha$
I think because there is a part missing in the proof, i.e. the definition of $\alpha$.
The proof should start with "Assume $f$ is non-zero, then there exists a point $x_0\in [a,b]$ such that $f(x_0)=:\alpha > 0$''
Now because the function might reach $\alpha$ at a single point and be strictly smaller in the neighbourhood you use $\alpha/2$.
Indeed in this case due to continuity you can find the $(c,d)$ interval in which the function is at least $\alpha/2$, otherwise the interval in which the function is $\geq \alpha$ might be a single point.
The proof starts with
"Suppose not, then $\exists y \in \mathbb{R}: g(y)= \alpha$, now since $g$ is continuous at $y,\forall \epsilon>0, \exists \delta >0:|y-x|<\delta \implies |g(y)-g(x)|<\epsilon$
Take $\epsilon=\alpha/2,$ then by continuity of $g$ at $y$, we have $\delta$ interval, say $y-\delta=c: y+\delta=d$...
and you can proceed the proof as you gave...
and draw pictures to see why we chose $\alpha/2$,
i.e take a function which satisfies the hypothesis as well as it has a maximum at $\alpha$ which is attained only once.(i.e$\exists !z \in [a,b]: g(z)=\alpha$)
In this case, We can't find an interval $(c,d)$ such that given any point $x \in (c,d), g(x)\geq \alpha$, we need such a interval in which the image of all the points is greater than a fixed positive real number to conclude the proof easily, so we are making the easiest choice $\alpha/2$, and proceeding... do you understood now...?
