Example of function not Fourier Invertible in $L^1$ It is well known that, if $1 < p \le 2$, then, for every $f \in L^p$, 
$$ \int_{[-R,R]^n} e^{-2\pi i x \cdot y} \hat{f}(y) dy \rightarrow f(x) $$
As $R \rightarrow \infty$, in the $L^p$ sense. This is a corollary, for example, of the $L^p$ Theory for the Hilbert Transform. 
I have dug the internet, unfortunately in vain, in search of a counterexample for the $p=1$ case. A fellow at my local institute told me that this counterexample was due to Kolmogorov, but I still cannot find it anywhere - at least not the example I want. 
Any ideas of how can I build such a function, or even good references for such a construction? 
Thanks in advance. 
EDIT: To be noted that I'm not talking about Fourier Series. It is not a problem about periodic functions, it is about globally defined functions in $L^1$. The counterexample for Fourier Series is well known and due to Kolmogorov. 
 A: Well, I think this is a solution to the problem: 
1 - If there was convergence, then, by the Uniform Boundedness Principle, as $\sup_R \|S_R f \|_1 < \infty$, then the operators $S_R$ should be bounded in $L^1$. It can be shown that, conversely, it is enough that the latter condition holds for the convergence to hold.
2 - As the multiplier for $S_R$ is a dilation of the multiplier for $S_1$, then the norm $\|S_R\|_{1 \rightarrow 1} = \|S_1\|_{1 \rightarrow 1}$. 
3 - For it not to hold, then it should only be enough to check that there exists $f$ in $L^1$ such that $S_1 f \not \in L^1$. 
4 - As $f \in L^1 \Rightarrow \hat{f} \in C(\mathbb{R}^n) \rightarrow \chi_{C_1} \hat{f} \in L^1$, where $C_1 = [-1,1]^n$ is the unit cube. This implies immediately that $S_1 f = \mathcal{F}^{-1}(\chi_{C_1} \hat{f})$ is continuous. 
5 - Since $S_1 f$ is, by assumption, integrable, we must have also that $ \chi_{C_1} \hat{f} $ must be continuous. But, if we take $f = \frac{1}{(2\lambda)^n}\chi_{[-\lambda,\lambda]^n}$, we see that the Fourier Transform of $f$ is equal to 1 at zero. So, by choosing a suitable $\lambda$, we may assume that $\hat{f}$ is at least $\frac{1}{2}$ inside $C_1$. Then, $\chi_{C_1} \hat{f}$ cannot ever be a continuous function, contradicting what we have supposed. 
