Is $f=0$ if the integral is zero If $f: [0,1]\to \mathbb R$ is continuous and for all $s\in [0,1]$ 
$$ \int_0^s f(t)dt = 0$$
does it then follows that $f=0$? I can show it for $f \ge 0$ but I am wondering if it is also true if $f$ not positive.

Assume $a:=f(x_0)\ne0$ for some $x_0\in(0,1)$. Then $\frac1a{f(x)}>\frac12$ on some open neighbourhood $x_0\in (x_0-r,x_0+r)\subset(0,1)$ because $f$ is continous. Hence
$$ 0=\frac1a\int_0^{x_0+r}f(t)\,\mathrm dt-\frac1a\int_0^{x_0-r}f(t)\,\mathrm dt=\int_{x_0-r}^{x_0+r}\frac1af(t)\,\mathrm dt\ge \int_{x_0-r}^{x_0+r}\frac12\,\mathrm dt=r>0.$$
Again by continuity, $f(0)=f(1)=0$, too.
