Mean Value Theorm on Wikipedia why $c \in (a,b)$? From here

By IVT (as in the proof given on the wiki page) we can only deduce $c \in [a,b]$, where is it implied that $c \in (a,b)$?  
P.S. How do I scale the image down in size? 
EDIT: Here are my thoughts. I claim that if equality holds in their proof of the equation, 
$$ mI \le \int_a^b f(x) g(x) \, dx \le M I . $$ 
then we are reduced to either $f(c)=m$ for some $c \in (a,b)$ or $I=0$. Wlog, 
$$ mI = \int_a^b f g \, dx \Rightarrow \int_a^b (m-f)g \, dx $$ 
Hence, we have $(m-f(x))g(x) = 0$ a.e. on $(a,b)$. If $m \not= f(x)$ for all $x \in (a,b)$, then we must have $g(x)=0$ a.e., hence $I=0$. In this case, $c$ can be arbitrary. 
 A: Look at the following proof:
Let $F(x):= \int_{a}^x f(t) dt$ for $x \in [a,b]$. Since $f$ is continuous, $F$ is differentiable and $F'=f$ on $[a,b]$. Since $\int_{a}^b f(t) dt=F(b)-F(a)$, there is, by the Mean value Theorem, a number $c \in (a,b)$ such that
$\int_{a}^b f(t) dt=F'(c)(b-a) =f(c)(b-a).$
A: There is another non-trivial possibility where $\int_a^b g(x) \, dx \neq 0$ and $c$ can take the value $a$ or $b$ but not every value in $(a,b)$.
Consider the (continuous) function $f(x) = \left(1 + \frac{\sin x}{2} \right)^{-1}$ and the (integrable, positive) function $g(x) =1 + \frac{\sin x}{2} $ defined on the interval $[0,2 \pi]$.
We have
$$\int_0^{2\pi} f(x) g(x) \, dx = \int_0^{2\pi} g(x) \, dx =2\pi.$$
Hence,
$$\int_0^{2\pi} f(x) g(x) \, dx = f(c)\int_0^{2\pi} g(x) \, dx, $$
for exactly three points $c = 0, \pi, 2\pi$ where $f(c) = 1$.
From this example we see that it would not be incorrect to state as the conclusion of the mean value theorem that there exists at least one $c \in [a,b]$ such that
$$\int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx.$$
A: Assuming $a<b.$  You have shown that if $f(c)=m$ then $c$ can be any member of $[a,b].$
Similarly, if $f(c)=M,$ show  that $c$ can be any member of $[a,b].$
If $f(c) \not \in \{m,M\}$  then there exist $c_1,c_2$ in $[a,b]$ with $f(c_1)=m<f(c)<M=f(c_2).$ ..... So by def'n of $m$ and $M$, since $f$ is continuous there exists $ c'$ strictly between  $c_1$ and $c_2$ with $f(c')=f(c).$ Now $c'\in (a,b).$
