Integrability of function and its Fourier transform implies differentiabilty Is the the following  true: "Assume $f\in L^1[0,1]$ and $\hat{f}\in L^1(\mathbb{R})$, then $f$ is differentiable a.e" 
 A: Theorem (Bernstein). If $f:\mathbb R\to\mathbb R$ is compactly supported and $C^\alpha$ continuous for some $\alpha>1/2$, then $\hat f\in L^1(\mathbb R)$. 
This theorem gives a negative answer to your question, since an appropriate version of the Weierstrass function satisfies the above conditions. 
The proof is the same as the proof of the corresponding result for Fourier series. For the sake of completeness, I modified the proof and included it below.  
Proof. For $h\in\mathbb R$  write $f_h(t)=f(t-h)$. The Fourier transform of $f_h-f$ is 
$$(e^{-i \xi h}-1)\hat f(\xi)$$
The $L^2$ norm of $f_h-f$ is $O(h^\alpha)$. By the Parseval identity 
$$\int_{\mathbb R}|e^{-i \xi h}-1|^2 |\hat f(\xi)|^2\,d\xi = O(h^{2\alpha})$$
We want to bound $ |e^{-i\xi h}-1|$ from below. There is no uniform bound for all $\xi$, but if we focus on some dyadic range $D_k=\{\xi: 2^k\le |\xi|< 2^{k+1}\}$, then choosing $h=\frac{2\pi}{3}\cdot 2^{-k}$ gives good result: $\frac{2\pi}{3}\le |\xi h|< \frac{4\pi}{3}$, which keeps $e^{-i\xi h}$ far away from $1$. Thus,
$$ \int_{D_k} |\widehat{f}(\xi)|^2\,d\xi    = O(2^{-2\alpha k}) $$
By the Cauchy-Schwarz inequality, 
$$ \int_{D_k} |\widehat{f}(\xi)| \,d\xi  = O(2^{k/2})\, O(2^{-\alpha k})= O(2^{k(1/2-\alpha)}) $$ 
and the sum over $k$ converges.
