Theorem 33.4 ii: Ross, Elementary Analysis 2nd Edition We are given that:
"If $g$ is a continuous nonnegative function on $[a,b]$ and if $\int_{a}^{b}g=0$, then $g$ is identically $0$ on $[a,b]$."
First of, the proof is done by contradiction. I am unsure if the contradiction is the following: 
"If $g$ is a continuous.... then there exists some $x$ in $[a,b]$ such that $g\ne0$." Is this right?
The proof goes as follows in Ross's book:
"Otherwise, since $g$ is continuous, there is a nonempty interval $(c,d) \subseteq [a,b]$ and $\alpha >0$ satisfying $g(x) \geq \alpha/2$ for $x \in (c,d)$. Then 
$$\int_{a}^{b} g \geq \int_{c}^{d} g \geq \frac{\alpha}{2}(d-c)$$
contradicting $\int_{a}^{b}g=0$.
The logic and structure makes sense to me. Other than my first question above, the other thing I cannot understand about this proof is how this is justified:
"Otherwise, since $g$ is continuous, there is a nonempty interval $(c,d) \subseteq [a,b]$ and $\alpha >0$ satisfying $g(x) \geq \alpha/2 $ for $x \in (c,d)$."
How does this follow? I was thinking of using the intermediate value theorem: 
Since $g$ is continuous, there is some $\delta$ positive such that for $\epsilon=\frac{\alpha}{2}$ and $x \in (c,d)$:
$$|g(x)-g(x_0)|< \frac{\alpha}{2}$$
but I'm having trouble obtaining $g(x)> \alpha /2$ 
Someone suggested using the reverse triangle inequality then:
$$g(x)> g(x_0) - \alpha/2 > \alpha/2$$
but this part doesn't seem to hold:
$g(x_0) - \alpha/2 > \alpha/2$ because of the negative sign...
 A: For the first question, the desired result is $g \equiv 0$, so we assume the opposite ($g$ is positive at some point, i.e. $g(x_0) = \alpha > 0$).
Note that $g$ is nonnegative, so $|g| = g$.  Make $g(x)$ the subject of the inequality.
$$|g(x_0)| - |g(x)| \le ||g(x)| - |g(x_0)|| = |g(x)-g(x_0)|< \frac{\alpha}{2}$$
$$g(x) > g(x_0)-\frac{\alpha}{2} = \alpha - \frac{\alpha}{2} = \frac{\alpha}{2}$$
A: Firstly,  the contradiction that is arrived at is that $\int_a^bg\gt0$.
Note that you are assuming the contrary of$g=0$ on $[a,b]$.  Since $g$ is nonnegative, $\exists \alpha\gt0$ such that $g\gt\alpha $ on some interval $[c,d]\subset[a,b]$.  This follows from the assumption that $g$ is positive somewhere (or not identically zero) together with continuity of $g$...
A: The contradiction is :
If $g$ is not identically $0$ on $[a,b]$, then there exists a point $x_0 \in [a,b]$ such that $g(x_0)>0$
Now you use the continuity of $g$ at $x=x_0$
Let $\epsilon = g(x_0)/2$
There exists a $\delta >0$ such that if $ x\in [a,b]$ and $|x-x_0|<\delta ,$ then $|g(x)-g(x_0)|<\epsilon.$ 
$$   |g(x)-g(x_0)|<\epsilon \implies g(x_0)-\epsilon< g(x)<g(x_0)+\epsilon$$ 
thus $$ g(x)>g(x_0)-\epsilon = g(x_0)/2 $$ on a neighborhood of $x_0$
The rest of the proof goes as you have it.
