A Fourier Transform Which Is Cartesian Separable We say that the Fourier transform of a complex-valued function $f\in L^{1}(\mathbb{R}^{n})$ is separable if there exist single-variable functions $g_{1},\ldots,g_{n}$ such that
$$g_{1}(\xi_{1})\cdots g_{n}(\xi_{n})=\widehat{f}(\xi), \text{   $\forall \xi=(\xi_{1},\ldots,\forall\xi_{n})\in\mathbb{R}^{n}$}$$
If $g_{1}(0)\cdots g_{n}(0)=\int_{\mathbb{R}^{n}}f(y)dy\neq 0$, then it is not hard to verify from Fubini's theorem that
$$g_{j}(\xi_{j})=\widehat{f_{j}}(\xi_{j}), \forall \xi_{j}\in\mathbb{R}$$
where
$$f_{j}(y_{j})=\dfrac{1}{g_{1}(0)\cdots g_{j-1}(0)g_{j+1}(0)\cdots g_{n}(0)}\int_{\mathbb{R}^{n-1}}f(y)dy_{1}\cdots dy_{j-1}dy_{j+1}\cdots dy_{n},$$
for $j=1,\ldots,n$. Whence by the injectivity of the Fourier transform, $f_{1}\cdots f_{n}=f$ a.e. 
What happens, though, if we drop the assumption that $\int_{\mathbb{R}^{n}}f\neq 0$? If $\widehat{f}\in L^{1}(\mathbb{R}^{n})$, then it follows from Fubini's theorem that each $g_{j}\in L^{1}(\mathbb{R})$, whence $f=g_{j}^{\vee}\cdots g_{n}^{\vee}$ a.e. (i.e. the product of the inverse Fourier transforms). This seems like an unnecessarily strong assumption.

Question If the Fourier transform of $f\in L^{1}(\mathbb{R}^{n})$ is separable, is $f$ almost everywhere equal to a separable function?

I tried reducing the problem to the case $f\neq 0$ by writing in terms of positive and negative parts ($f=f_{+}-f_{-}$), but I couldn't get anywhere with this approach.
 A: For convenience, we consider the case $n=2$; the general case is completely analogous. Suppose $f\neq 0\in L^{1}(\mathbb{R}^{2})$. Then $\widehat{f}$ is not identically zero, whence there exists a point $(\xi_{1},\xi_{2})\in\mathbb{R}^{2}$ (an open ball, in fact, by the continuity of $\widehat{f}$) such that $\widehat{f}(\xi_{1},\xi_{2})=g_{1}(\xi_{1})g_{2}(\xi_{2})\neq 0$. We can therefore write
$$g_{1}(\eta_{1})=\dfrac{\widehat{f}(\eta_{1},\xi_{2})}{g_{2}(\xi_{2})}=\dfrac{1}{g_{2}(\xi_{2})}\int_{\mathbb{R}}\left(\int_{\mathbb{R}}f(x_{1},x_{2})e^{-2\pi i\xi_{2}x_{2}}dx_{2}\right)e^{-2\pi i\eta_{1}x_{1}}dx_{1}, \quad \forall \eta_{1}\in\mathbb{R}$$
and
$$g_{2}(\eta_{2})=\dfrac{\widehat{f}(\xi_{1},\eta_{2})}{g_{1}(\xi_{1})}=\dfrac{1}{g_{1}(\xi_{1})}\int_{\mathbb{R}}\left(\int_{\mathbb{R}}f(x_{1},x_{2})e^{-2\pi i\xi_{1}x_{1}}dx_{1}\right)e^{-2\pi i\eta_{2}x_{1}}dx_{2},\quad \forall \eta_{2}\in\mathbb{R}$$
Now define single-variable functions $f_{1},f_{2}$ by
$$f_{1}(x_{1})=\dfrac{1}{g_{2}(\xi_{1})}\int_{\mathbb{R}}f(x_{1},x_{2})e^{-2\pi i\xi_{2}x_{2}}dx_{2}, \quad  f_{2}(x_{2})=\dfrac{1}{g_{1}(\xi_{1})}\int_{\mathbb{R}}f(x_{1},x_{2})e^{-2\pi i \xi_{1}x_{1}}dx_{1}$$
By Fubini-Tonelli, $f_{1}$ and $f_{2}$ are measurable functions on $\mathbb{R}$ which are finite a.e. Moreover, since $f_{1}$ is dominated a.e. by $g_{2}(\xi_{2})^{-1}\int_{\mathbb{R}}\left|f\right|dx_{1}$ and analogously for $f_{2}$, we conclude that $f_{1},f_{2}\in L^{1}(\mathbb{R})$; whence, $f_{1}\cdot f_{2}\in L^{1}(\mathbb{R}^{2})$. By using Fubini's theorem again, it is easy to see that $\widehat{f_{1}}=g_{1}$, $\widehat{f_{2}}=g_{2}$, and $\widehat{f_{1}f_{2}}=g_{1}g_{2}$. From the inejctivity of the Fourier transform, we conclude that $f=f_{1}f_{2}$ a.e.
