Let's suppose that we have two functions $f(x)$ and $g(x)$ which are both bordered on $[a,b]$. We know that $g(x)$ is integrable on $[a,b]$ and $g(x) \geq 0$ for whichever $x \in [a,b]$. Also we know that $f(x) \cdot g(x)$ is integrable. This is the beginning of the proof of the mean value theorem for the Riemann integrals.
My first question is, knowing just what I've said above, doesn't it mean indirectly that $f(x)$ must be also continuous or having Lebesgue discontinuity points? If not, then I'd like someone to show me an example of $f(x)$ a function which is not continuous on $[a,b]$, and $g(x)$ and $f(x) \cdot g(x)$ being integrable functions on $[a,b]$.
Anyways, here is what I understood from the mean value theorem. We have the area below the graph of the function $g(x)$ on $[a,b]$. Also we have the area below the graph of the function $g(x) \cdot f(x)$ on $[a,b]$. Obviously the area below $f(x) \cdot g(x)$ is bigger than the area below the graph of $g(x)$. We suppose that the number we need to multiply with the area below the graph of $g(x)$ so that to equal the area below the graph of $f(x) \cdot g(x)$ is between the minimum and the maximum of $f(x)$ on $[a,b]$. And from that assumption there begins the proof. After some calculations it is proved that what we supposed was true. Is my approach correct?
Then here comes another question, at the beginning of the proof shouldn't it be necessary to specify that it must be $f(x) \geq 0$ on $[a,b]$? If not, then it is possible to prove the mean value theorem for the Riemann integrals knowing the conditions above and $f(x) < 0$ on $[a,b]$?
And my last question is the next. Generally if we know that $g(x)$ is integrable on $[a,b]$, without being strictly $g(x) \geq 0$ on $[a,b]$, $f(x)$ bordered on $[a,b]$ and also we know that $f(x) \cdot g(x)$ is integrable on $[a,b]$, doesn't it mean indirectly that $f(x)$ is continuous or has Lebesgue discontinuity points on $[a,b]$?
As for your last question, consider: $$f(x)=\left\{\begin{array}{ll} 1 & x\in\mathbb{Q}\cap[0,1]\\ 0 & \text{else} \end{array}\right.,\ g(x)=0$$ Then, it is clear that $f\cdot g$ is integrable, as well as $g$, but $f$ is not Riemann integrable.
That is not true, if $f(x)\leq1$ on $[a,b]$. For instance, take $f(x)=g(x)=x$ and $[a,b]=[0,1]$. Then $$f(x)\cdot g(x)=x^2\leq x=g(x),\ \forall x\in[0,1]$$
Except for that part of your thought, your idea of the proof is, intuitively, correct. Note that, multiplying two values in real numbers - which is not the case in natural numbers - does not neccessarily mean that we take larger values as a result.
As for $f(x)$ being non-negative, as you stated, the number we are seeking is between $m=\inf\limits_{x\in[a,b]}f(x)$ and $M=\sup\limits_{x\in[a,b]}f(x)$ and $m,M$ may be arbitrary real numbers, so, no need for non-negativeness.