Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)
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Uniqueness of a law from its characteristic function

Statement

Assume AC. Two Borel probability laws on R with equal characteristic functions are equal. In particular a real random variable has a real-valued characteristic function if and only if its law is symmetric under xx.

Facts & Assumptions

Given: The hypotheses and conventions in the statement.

[F1]

Probability and Fourier conventions correspond by an invertible frequency change. Characteristic function fourier stieltjes convention.

[F2]

Under AC finite complex Borel measures with equal transforms are equal. Uniqueness of finite Borel measures from their Fourier transforms.

[F3]

AC supplies the choices in the Fourier uniqueness proof. The Axiom of Choice.

[F4]

Reflection conjugates a characteristic function. Basic properties of characteristic functions.

[F5]

The reflected law has characteristic function phi(-t). Characteristic functions under affine maps and independent sums.

Proof

technique · direct
1.1

Let μ,ν be the two laws. For every real ξ, F1 gives μ^(ξ)=φμ(2πξ)=φν(2πξ)=ν^(ξ). Each positive probability law, regarded as a complex measure, has total variation one: every measurable partition has sum of absolute masses equal to its total mass. Thus the finite-variation hypotheses of Fourier uniqueness hold.

F1
2.1

Apply F2 in dimension one to obtain μ=ν. AC is inherited from that proof: it supplies the Hahn/Jordan and Radon–Nikodym selections used in Gaussian smoothing and covers its regularity argument. No inversion result from this page is used.

step 1.1F2F3
3.1

For the final equivalence, F5 with a=1,b=0 and F4 give φX(t)=φX(t)=φX(t). If φX is real-valued, the two characteristic functions agree and step 2.1 proves symmetry of the law. Conversely, symmetry means the two laws, hence their defining integrals, agree; the displayed identity then forces φX(t)=φX(t), so every value is real.

step 2.1F4F5

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