The Fourier transform of $x \mapsto |x|^{-1/3}e^{-x^{2}}$ is not in $L^{1}$. Okay previously my lecturer showed that this is so by proving in the following way:
Proof by contradiction. Suppose the transform is in $L^{1}$. Then as $f \in L^1$, we may use Fourier Inversion Formula thereby getting $$ f(x) = \int_{\mathbb{R}} \widehat f(t) e^{2\pi i xt} dt. $$ Therefore, $f$ is bounded almost everywhere and therefore globally.
Is this proof correct? I think according to my previous post, it seems not. If this proof is incorrect, can I modify? Or change it entirely and if then how?

The argument works here as $f$ can be represented by a continuous function (this can be seen by the inversion transform formula). So if $|f(x)|\leq M$ for almost every $x$, say for $x\in \Bbb R\setminus N$, where $\lambda(N)=0$, then $\lambda$ has an empty interior, so $\Bbb R\setminus N$ is dense in $\Bbb R$. We conclude by continuity.
This proves that if an integrable function $f$ has a Fourier transform which is integrable, then $f$ is bounded. 
For the particular case in the OP, note that $f$ is not (essentially) bounded.
