Deriving existence of classical Fourier transform via the space of temperate distributions

Question

If for some measurable function $f:{\bf R}^n\rightarrow{\bf R}$ the functional $h\mapsto\int fh$ is in ${\scr S}'$ (space of temperate distributions) and there exists some measurable $g$ such that the Fourier transform of $h\mapsto\int fh$ in ${\scr S}'$ equals $h\mapsto\int gh$, can we conclude that the Fourier transform of $f$ also exists in the classical sense, i.e. $\zeta\mapsto\int f(x)e^{-ix\eta}dx$ is defined almost everywhere, and that this functional equals $g$ almost everywhere?