Integration problem. Does $\left|\int_0^\infty f(x)dx\right|=0$ imply $f=0$ over $(0,\infty)$? Let $f$ be a continuous function over $(0,\infty)$. 
Does $\left|\int_0^\infty f(x)dx\right|=0$ imply $f(x)=0, \forall x\in (0,\infty)$? 
 A: Take $f(x)=\sin(x)$ if $x\in[0,2\pi]$ and $f(x)=0$ otherwise.
Then $f$ is continuous and $|\int_0^\infty f(x)dx|=0$ but $f \neq 0$.
A: No. One may consider
$$
f(x)=\frac1{1+x^2}-\frac \pi2 e^{-x}
$$ we have $f\neq 0$ and continuous over $(0,\infty)$:
$$
\left|\int_0^\infty f(x) dx\right|=\left|\int_0^\infty \frac1{1+x^2}dx-\frac \pi2\int_0^\infty e^{-x}dx \right|=\left|\frac\pi2-\frac \pi2\cdot 1\right|=0.
$$
A vast family of counter examples: $g$ continuous over $[0,\infty)$, sastifying $\displaystyle \int_0^\infty |g(x)|\:dx<\infty$, $\displaystyle \int_0^\infty g(x)\:dx=C$, then
$$
\left|\int_0^\infty \left(g(x)-C e^{-x}\right) dx\right|=0,
$$ $f=g- Ce^{-x}$ is continuous on $(0,\infty)$, in general $f\neq0$.
A: The question is slightly unclear. As it is $0$, there is no use in taking the absolute value of the whole expression; we gain no new information. Do you mean to write $\int_{\mathbb{R_{\geq 0}}}|f| = 0$? If so, the statement is true. $|f| \geq 0$ and so for any point $c$ where $f(c) \neq 0$ we have a neighbourhood where $|f| > 0$. One can easily see the integral in this neighbourhood is positive considering a lower sum. This implies $f(x)$ must be $0$ everywhere. 
If you mean to ask the question as it is written, then there are counterexamples. Forget the absolute value. We simply want a function, not everywhere $0$, such that $\int_{\mathbb{R_{\geq 0}}}f  = 0$. Indeed, one can have $f(x)$ with only countably many zeros. Consider 
$$f(x) = \frac{1}{n}\sin(2\pi x) , \  x \in [n-1,  n] \ \text{and} \ n \in \mathbb{N}$$
