Consider $f \in L^2(0,1)$ and $F = \int_0^x f(t) dt \in \mathcal{C}([0,1])$. Is it true that $f$ is the weak derivative of $F$, in the sense that given any $g \in \mathcal{C}_c^1([0,1])$, we have $\int Fg' = -\int fg$? This is obviously true if $f$ is continuous, but I'm having trouble proving the general case. Any help is appreciated. Thanks!
(Or, in case $f$ is not necessarily the weak derivative of $F$, is there some other way to show that the operator $f \mapsto \int_0^x f(t) dt$ is injective, i.e., if $\int_0^x f(t) dt = 0$ for some $f \in L^2(0,1)$, then $f = 0$ a.e.?)
Yes.
You actually need to verify that formula for $g\in C^1_c((0,1))$, not $C^1_c([0,1])$. This implies that $g(0)=0$, so $g(x)=\int_0^x g'(t)\,dt$. Insert that into $\int_0^1 f(x)g(x)\,dx$ and apply Fubini.
Edit: The OP says he had to make a small change to what I said. For the record, here's exactly what I had in mind:
Using the fact that $\int_t^1 f=\int_0^1 f-\int_0^t f$, we see $$\begin{aligned}\int_0^1 fg&=\int_0^1 f(x)\int_0^x g'(t)\,dtdx \\&=\int_0^1 g'(t)\int_t^1 f(x)\,dxdt \\&=\left(\int_0^1 g'\right)\left(\int_0^1 f\right)-\int_0^1g'(t)\int_0^t f(x)\,dxdt \\&=0-\int_0^1 g'F.\end{aligned}$$