Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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L1 Fourier inversion with an integrable transform

Statement

Assume countable choice. If f∈L1(Rn;C) and f^∈L1, then g(x)=∫f^(ξ)e2πix⋅ξ dξ is bounded and continuous, equals f almost everywhere, and equals the specified value at every Lebesgue point of f.

Facts & Assumptions

Given: n≥1, f,f^∈L1 and The Axiom of Countable Choice (ACω).

[F1]

Gaussian Fourier means recover each Lebesgue value (Gaussian Fourier summability at Lebesgue points).

[F2]

An L1 transform is bounded and uniformly continuous (The L1 transform is bounded and uniformly continuous).

[F3]

Dominated convergence permits passage under an integral with a single integrable majorant (Dominated convergence).

[F4]

Almost every point of a locally integrable function is a Lebesgue point under countable choice (Almost every point is a Lebesgue point of a locally integrable function).

Proof

1.1F2F3given

Since f^∈L1, F2 applied to it shows that g(x)=F(f^)(−x) is bounded and uniformly continuous. For fixed x and the explicit sequence tm=1/(m+1), m∈N, the damped integrands converge pointwise to f^(ξ)e2πix⋅ξ and have majorant ∣f^∣. F3 therefore gives Stmf(x)→g(x).

2.1F1F4step 1.1∎

At any Lebesgue point with value a, F1 gives the same sequence limit a, so g(x)=a. Since f∈L1 is locally integrable, F4 gives such points with a=f(x) outside a null set (for complex inputs apply the real conclusion to both components and bound the complex oscillation by their sum). Hence g=f almost everywhere. In particular g represents the original L1 class. Countable choice is precisely inherited from F1 and F4; no statement about all values of an arbitrary representative follows.

Depends on

Dependency tree · 2 levels

Sources