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Uniqueness of the L1 Fourier transform
Statement
Assume countable choice. If and , then almost everywhere. Equality of the transforms almost everywhere already suffices.
Facts & Assumptions
Given: , the stated inputs and The Axiom of Countable Choice ().
The transform is linear and continuous as a function of frequency (The L1 transform is bounded and uniformly continuous).
A function and its integrable transform obey inversion almost everywhere (L1 Fourier inversion with an integrable transform).
Proof
Set . By linearity, . If equality was given only almost everywhere, continuity still implies this everywhere: a nonzero value would remain bounded away from zero on an open ball, which contains a positive-volume box and cannot be null. Thus the transform of h is the zero integrable function.
F2 applies to h, since both h and its zero transform are integrable. It gives almost everywhere, so as classes. Countable choice is inherited from inversion (and the Euclidean measure interface in the optional almost-everywhere hypothesis).