Textbooks always take some care when defining the fourier transform on $L^2$. This suggests that the naive definition fails, i.e. that there is some square-integrable $f$ for which
$$\int_{-\infty}^\infty f(x)\exp(i\omega x)dx$$
does not converge for almost every $\omega$. What is an example of such an $f$?
Note that if we choose $f(x):=x^{-1}\mathbf{1}_{[1,\infty )}(x)$ then $$ \int_{\Bbb R }|f(x)e^{-i\omega x}|\,\mathrm d x=\int_{[1,\infty )}\frac1{x}\,\mathrm d x=\infty $$ so $f\notin L^1(\Bbb R )$ and the Fourier transform can not be defined in the obvious way. However $$ \int_{\Bbb R }|f(x)|^2\,\mathrm d x=\int_{[1,\infty )}\frac1{x^2}\,\mathrm d x=1 $$ so $f\in L^2(\Bbb R )$. Indeed we have that $L^2(\Bbb R )\not\subset L^1(\Bbb R )$ so $f(x)e^{-i\omega x}$ is not integrable in general for any chosen $L^2$ function.