Sets of Divergence for Fourier Partial Integals It is a consequence of Carleson's theorem together with a transference argument that (see Section 4.3.5 in L Grafakos, Classical Fourier Analysis for proof) that the Fourier partial integrals of a function $f\in L^{p}(\mathbb{R})$, where $1<p\leq 2$, converge pointwise almost everywhere (a.e.) to $f$:
$$\lim_{R\rightarrow\infty}\int_{\left|\xi\right|\leq R}\widehat{f}(\xi)e^{2\pi i\xi x}\mathrm{d}\xi=f(x)\quad \text{a.e.} \tag{1}$$
Kahane and Katznelson showed that given any subset of the torus $E$ of measure zero, there exists a continuous periodic complex-valued function $f\in C(\mathbb{T})$ such that the Fourier partial sums of $f$ diverge at every $x\in E$ (and possibly at other points too). My question is whether there is an analogoue of Kahane's and Katznelson's result for continuous, $p$-integrable functions on the real line. A cursory search has been unable to find one.

Question. Given a subset $E\subset\mathbb{R}$ of measure zero, does there exist a continuous function $f\in L^{p}$ with locally integrable (distributional) Fourier transform $\widehat{f}$, such that
$$S_{R}f(x):=\int_{\left|\xi\right|\leq R}\widehat{f}(\xi)e^{2\pi i\xi x}\mathrm{d}\xi \tag{2}$$
diverges for all $x\in E$?

I think this question makes sense as its written. I know that, unlike in the setting of the torus, the Fourier transform of a function $f\in L^{p}(\mathbb{R})$ isn't necessarily a function, if $p>2$. But I think I have taken care of that issue by requiring the Fourier transform to be locally integrable so the integral in (2) makes sense.
My motivation for this question stems from the problem of inverting the Fourier transform $\widehat{f}$ using the cutoff function $\chi_{[-R,R]}$ (see this question), when $f\in L^{1}(\mathbb{R})$ and $\widehat{f}$ is only assumed to be locally integrable. How badly can the equality in (2) fail if $\widehat{f}\notin L^{1}(\mathbb{R})$?

Edit 1. I incorrectly stated Kahane and Katznelson's result in the original question. They only showed that given a measure zero subset $E\subset\mathbb{T}$, the partial Fourier sums diverge on $E$ and possibly at other points as well. It's not necessarily the case that $S_{N}f(x)\rightarrow f(x)$ for all $x\notin E$. In fact, characterizing the sets $E$ for which $S_{N}f$ diverges on $E$ and converges on $E^{c}$ appears to be an open problem.
Also, it cannot be the case that $S_{N}f(x)\rightarrow a\in\mathbb{C}$, where $a\neq f(x)$. The Fejer means $\sigma_{N}f(x)$ of the partial sums converge to $f(x)$ at every point of continuity of $f$, whence everywhere. So if the partial sums converge, then it must be to $f(x)$.
 A: Of course the answer must be yes. You can actually use Kahane's example on the torus to construct an example on the line, using a somewhat standard technique as in Replicating Kolmogorov's Counterexample for Fourier Series in Context of Fourier Transforms .
Ah, you were the guy in that other thread; I can fudge a few details. And again I'm going to set $2\pi=1$. (Also $\pi=1$.)
Say $E\subset\mathbb R$ has measure $0$. Define $$E'=E+2\pi\mathbb Z=\{t+2\pi n:t\in E,n\in\mathbb Z\}.$$
So $m(E')=0$, $E\subset E'$, and $E'$ is $2\pi$-periodic.
So Kahane tells us there is a continuous $2\pi$-periodic function $g$ such that $S_N(x)$ diverges for all $x\in E'$, where $S_N$ is the $N$-th partial sum of the Fourier series for $g$. 
Now say $\phi$ is a Schwarz function with no zero on $\mathbb R$, such that $\psi=\hat\phi$ has support in $(-1/2,1/2)$. Let $f=\phi g$.
Then $f$ is continuous and lies in every $L^p$ space. Exactly as in that other thread, $\phi(x)S_N(x)$ is a partial integral for $f(x)$, so $f$ is bad at every point of $E'$.

Details: How we know we can get $\phi$ as above. Start with a non-zero function $\psi_0\in C^\infty_c$. The Fourier transform of $\psi_0$ is an entire function, so there is some horizontal line in $\mathbb C$ where it has no zero. Considering $\psi_t(x)=e^{tx}\psi_0(x)$ shifts that horizontal line down to the $x$-axis.
Or do this to get $\phi>0$: Start with a Schwarz function $\phi_0$ such that $\hat\phi_0$ has compact support. So $\phi_0$ is the restriction of an entire function to the real axis, hence has only countably many zeroes. Let $\Phi=|\phi_0|^2$. Then $\Phi$ is a Schwarz function and $\hat \Phi$ still has compact support. Let $\phi=\Phi+\tau_x\Phi$ for suitable $x$ (where $\tau_x$ denotes translation).
