The Riemann-Liouville integral is defined by
$$
I^\alpha f(x)=\frac{1}{\Gamma(\alpha)} \int_a^x f(t)(x-t)^{\alpha-1} d t
$$
where $\Gamma$ is the gamma function and $a$ is an arbitrary but fixed base point. Take $a = 0$ and $\alpha = 1/2$. Therefore we look at:
$$
I^x_{\frac{1}{2}} := \frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} d t
$$
Suppose $I^x_{\frac{1}{2}} = 0$ for all $x$. Can we then conclude $f=0$ a.e.? 

My approach so far has been to take the Fourier transform and use the convolution theorem. I cannot conclude because I do not know if $f \in L^2(\mathbb{R})$. Otherwise, I could conclude just by using the fact that the Fourier transform is an isometry between $L^2$ spaces.