Fourier transform turns L1 convolution into multiplication

Apply [F7] to the real and imaginary components of $f$ and $g$, changing infinite values on their null sets to zero, to obtain finite Borel representatives. Euclidean Lebesgue spaces are sigma-finite (the boxes $[-k,k{]}^n$ have finite volume and cover them). F2 gives product measurability. Translation invariance and Tonelli give $\iint \mathrm{\mid }f(x-y)g(y)\mathrm{\mid }dxdy=\int \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\mid }g(y)\mathrm{\mid }dy=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1<\mathrm{\infty }$. Thus the exponential-weighted integrand is also absolutely integrable at every fixed frequency, with this same bound.

Fubini now permits exchanging the integrals in the transform of the convolution representative. The inner $x$ integral is the translation transform from F5, giving $\widehat{f\ast g}(\xi )=\int g(y)e^{-2\pi iy\cdot \xi }\widehat{f}(\xi )dy=\widehat{f}(\xi )\widehat{g}(\xi )$. The convolution is defined arbitrarily on its null exceptional set; F6 makes that choice irrelevant at every frequency. This proves the asserted equality, including when either input is zero.