Show that if $f(x)=\int_{0}^x f(t)dt$, then $f=0$
differentiating both sides gives $f'(x) = f(x)$ which has solutions $f(x) = Ae^x$. But the original statement gives that $f(0) = \int_0^0f(t)dt = 0$ so we have $Ae^0 = 0$ so $A = 0$ and the function is constantly zero.
If $f$ is continuous, the the argument suggested by Tom Oldfield gives the answer. If $f$ i only locally integrable, you can apply Gronwall's inequality. Choosing $\alpha =0$, you get $f \leq 0$. Applying the same reasoning to $-f$ you get $f \geq 0$. Hence $f=0$ identically.
PS: I think that a more direct proof can be found.