Let $f$ be a continuous function on $\mathbb{R}$ such that the Fourier transform of $f$ exists almost everywhere. Is $f\in L^1 \cup L^2$.

Question

Problem: Let $f$ be a continuous function on $\mathbb{R}$ such that the Fourier transform of $f$ exist almost everywhere. Is $f\in L^1 \cup L^2$ ?

My opinion: We know that if $f\in L^1\cup L^2$, then the Fourier transform $\hat{f}$ exist a.e., but if for a continuous function $f$, $\hat{f}$ exist a.e., then can we say that $f\in L^1 \cup L^2$?

More precisely, what is the necessary condition to define the Fourier transformation in the classical sense (not the distributional sense)?