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Fourier transform of a finite complex Borel measure
Statement
Assume countable choice. Let be a complex Borel measure on , , with finite total variation. Then is a bounded uniformly continuous function and .
Facts & Assumptions
Given: The measure of A complex measure is a finite-valued countably additive set function, finite variation, The Axiom of Countable Choice (), and the simple-limit integral of Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|).
The integral modulus is bounded by its absolute integral against total variation (Integrals against signed or complex measures are bounded by total variation).
Dominated convergence holds for positive measures (Dominated convergence).
The exponential has modulus one on imaginary arguments and satisfies addition (, , and , , and the complex exponential extends the real exponential).
Proof
First repair the positive integral used to construct the complex-measure integral. Augment every finite disjoint nonnegative-simple display by its complement with coefficient . Pairwise intersections of two augmented displays partition the space and carry equal coefficients on nonempty cells, so finite additivity and prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants and the sets , , give monotone convergence; increasing simple approximations give nonnegative additivity and finite linearity after positive/negative and real/imaginary decomposition. Fatou follows by applying MCT to ; applying Fatou to proves dominated convergence when . For a canonical nonzero-level complex simple function, the triangle inequality and the definition of variation give . Passing to simple limits using this bound constructs the complex integral, makes it independent of the approximants, and preserves the same variation bound and finite linearity. Each exponential is bounded Borel with modulus one, so it is integrable against . The locally reconstructed complex integral exists and gives ; no Radon–Nikodym representation is needed.
By the addition law and the locally proved variation bound, , independently of . Under the stated countable choice the sequential Euclidean limit criterion applies; the locally proved dominated convergence, with majorant on this finite measure space, makes this bound tend to zero as tends to zero. Hence the transform is uniformly continuous. Countable choice also covers the near-maximizing partition selections in the total-variation additivity proof. Finite variation was assumed; no general finiteness theorem, RN or Hahn decomposition is used.
Depends on
- A complex measure is a finite-valued countably additive set function
- Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|)
- Integrals against signed or complex measures are bounded by total variation
- Dominated convergence
- $\exp(x+iy)=e^x(\cos y+i\sin y)$, $|\exp(x+iy)|=e^x$, and $e^{i\pi}+1=0$
- $\exp(z+w)=\exp z\,\exp w$, and the complex exponential extends the real exponential
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
Used by
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Sources
- Gerald Teschl, Topics in Real and Functional Analysis (2017) (standard reference, not scraped)