If $f$ is Riemann integrable and $f>0$ for $x\in[a,b]$ then $\int{f>0}$ I've seen the proof and it seems much less trivial than what I had in mind. 
Can anyone point out to me what is wrong about the following proof:
Let P be some partition of $[a,b]$ such that: 
$$ P= \begin{Bmatrix} a = x_0 < x_1 < x_2 < · · · < x_{n−1} < x_n = b \end{Bmatrix}$$
$$\forall  x\in[a,b]: f(x)>0 \Rightarrow 0<\min_{x\in[x_{i-1},x_i]}f(x)\Longrightarrow 0<\sum_{i=1}^{n} \min_{x\in[x_{i-1},x_i]}{f(x)} \cdot \Delta x_{i}$$
let $L(f, P)$ be the lower sum of partition P
$$0<\sum_{i=1}^{n} \min_{x\in[x_{i-1},x_i]}{f(x)} \cdot \Delta x_{i}=L(f,P)\le\sup_{P}{L(f,P')}=\int_{a}^{b}f(x)dx $$
hence
$$\int_{a}^{b}f(x)dx > 0$$
 A: To be clear: the statement is correct,but the proof you are demonstrating is not. There is a very, very slight point that contains a mistake. You are saying that since $f(x)>0$ for all $x$, it is $\min_{x\in[x_{i-1},x_i]}f(x)>0$. Why do you think this minimum exists? If it is a minimum, then fine, it will be equal to $f(\xi_i)$ for some $\xi_i\in[x_{i-1},x_i]$ therefore strictly positive. But this is not always the case! The following two examples illustrate why this is not always the case:
For example: There is a famous theorem that says $f$ is Riemann integrable if and only if the discontinuities of $f$ form a set of measure zero. Now we can have sets of measure zero that are very, very weird. For example the Cantor set. Denote by $C$ the classic cantor set and define a function $f:[0,1]\to(0,\infty)$ by $f(x)=1$ if $x\not\in C$ and we define $f$ however we want on $C$, but it must be strictly positive. We can define $f$ to be positive and satisfying for any $\delta>0$
$$\inf_{x\in[1/3-\delta,1/3+\delta]}f(x)=0$$
(Look at the wikipedia link to visualize the Cantor set. Can you think of a way to define such an $f$? remember, you can define $f(x)$ however you like on $C$ and you will still get a Riemann integrable function. The only restriction is that $f(x)>0$ for all $x$)
This function provides a counter example, since if we partition $[0,1]$ in subintervals, $1/3$ has to belong to a subinterval, therefore the infimum of $f$ in this subinterval will be $0$.
Edit: a much simpler counter example: set $f(x)=1$ everywhere on $[0,1]$ except  $\{0.5+1/n: n\geq3\}$. For these points, set $f(0.5+1/n)=1/n$. Then the same thing happens at $0.5$ (and of course $f$ is positive, Riemann integrable).
