Derivative of Fourier transform: $F[f]'=F[-ixf(x)]$

You know, thanks to Riemann-Lebesgue theorem, that $\,\hat{f}$ and $\,\hat{g}$ are continuos and vanishing at $\infty$, where $g=\widehat{-ixf(x)}$. I prove that $$\hat{f}(y)-\hat{f}(0)=\int_0^y g(t)dt,$$ the result will follow from fundamental calculus theorem.

\begin{align*} \int_0^y g(t)dt&=\int_0^y\int_{\mathbb{R}}-ixe^{-ixt}dt\, f(x)dx=\\ &=\int_{\mathbb{R}} \left.e^{-ixt}\right|_{0}^y\,\, f(x)dx=\int_{\mathbb{R}}f(x)\left[e^{-ixy}-e^{ix0}\right]dx=\hat{f}(y)-\hat{f}(0). \end{align*}

The hypothesis $xf(x)\in L^1$, in addiction with $|e^{i \theta}|=1$ for all $\theta\in\mathbb{R}$, is used to apply Fubini's theorem.