If $f$ is continuous, nonnegative on $[a, b]$, show that $\int_{a}^{b} f(x) d(x) = 0$ iff $f(x) = 0$
Using continuity of $f$ at $x = x_0$, you can find an open interval $(x_0-\delta,x_0+\delta)$ such that $|f(x)-f(x_0)| < \dfrac{f(x_0)}{2} \to f(x) > \dfrac{f(x_0)}{2} \to \displaystyle \int_{a}^b f(x)dx \geq \displaystyle \int_{x_0-\delta}^{x_0+\delta} f(x)dx > \displaystyle \int_{x_0-\delta}^{x_0+\delta} \dfrac{f(x_0)}{2}dx = \delta\cdot f(x_0) > 0$