Translation, modulation, linear dilation and reflection laws

The maps $y\mapsto y+a$ and $x\mapsto Ax$ are $C^1$ diffeomorphisms of the open set ${\mathbb{R}}^n$, with determinants $1$ and $\det A\mathrm{\ne }0$. Applying F2 to $\mathrm{\mid }f\mathrm{\mid }$ proves $\mathrm{\parallel }{\tau }_{a}f{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }f\circ A{\mathrm{\parallel }}_1=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$; unit modulus gives $\mathrm{\parallel }{M}_{b}f{\mathrm{\parallel }}_1=\mathrm{\parallel }\bar{f}{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$. These operations preserve null equivalence, by the same substitution applied to indicators of null sets (or directly by its null-set proof). Countable choice is precisely the assumption inherited from this Lebesgue substitution interface.

Substituting $x=y+a$ in the absolutely convergent translation integral gives $\int f(y)e^{-2\pi i(y+a)\cdot \xi }dy=e^{-2\pi ia\cdot \xi }\widehat{f}(\xi )$. Combining exponential factors in the modulation integral gives $e^{-2\pi ix\cdot (\xi -b)}$, hence the modulation formula.

Substituting $y=Ax$ gives $x\cdot \xi =y\cdot A^{-T}\xi$ and $dx=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}dy$, proving the dilation formula. For $A=-I$ its determinant has absolute value one, proving reflection even when orientation is reversed. Finally conjugate the componentwise integral for $\widehat{f}(-\xi )$: conjugation commutes with its real and imaginary integrals and sends $e^{2\pi ix\cdot \xi }$ to $e^{-2\pi ix\cdot \xi }$. This proves the last identity. All equalities are pointwise because F1 gives absolute convergence at each frequency.