If $f \geq 0$ is continuous and $\int_{a}^{b} f(x) \, dx = 0$, then $f =0$

Solution 1:

It's sufficient to consider the inferior sum. Also, you should specify which partition is $P$ (I will denote it by $P_0$): It should be a partition such that one of the subintervals determined by $P_0$ is contained in $B_\delta (x_0)$, in this way, if this interval is $I$, your argument should be

$$ L(f,P_0) = \sum_{P_0([a,b])} m_i \Delta_i \geq \inf_{x \in I} |f(x)| \mathcal{V}(I)>0 $$

(here we used the fact that $f\geq 0$) and therefore

$$\int_a^b f(x)=\sup_{P}L(f,P)\geq L(f,P_0)>0$$

(Notice that you must put a more restricted condition to $\epsilon$ in order to conclude $\inf_{x \in I} |f(x)| \mathcal{V}(I)>0$)