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Lebesgue measure on the unit interval supplies Gaussian moment data

Statement

Assume Countable Choice. For every integer M≥1, Lebesgue measure on the relative Borel space [0,1] is Gaussian quadrature moment data of order 2M in the sense of Positive-measure Gaussian quadrature moment data.

Facts & Assumptions

Given: An integer M≥1 and Lebesgue measure λ on [0,1].

[F1]

A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included: λ([0,1])=1, and every nonempty relative interval neighborhood in [0,1] has positive measure equal to its length.

[F2]

Integrable real and complex functions, and their integrals and Monotonicity and nonnegative homogeneity of the nonnegative integral: if g is Borel measurable and 0≤g≤1 on [0,1], then g is integrable and 0≤∫g dλ≤1 because λ([0,1])=1.

[F3]

Positive-measure Gaussian quadrature moment data: order-2M data require a finite positive measure whose relative support contains at least M+1 points and whose absolute moments through degree 2M are finite.

[F4]

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function and The Borel sigma-algebra of a topological space: polynomial and absolute-value functions are continuous on [0,1]. Continuity makes preimages of open subsets of R relatively open in [0,1], so preimages of Borel sets belong to the relative Borel sigma-algebra generated by those opens.

[A1]

Countable Choice, required by [F1] for the Lebesgue interval-measure theorem.

Proof

1.1A1F1F3construct

The measure is finite and positive because [F1] under [A1] gives λ([0,1])=1. Every x∈[0,1] belongs to the relative support: any relative open neighborhood of x contains a relative interval of positive length, including when x=0 or x=1, so its measure is positive by [F1]. Thus the support contains the M+1 distinct points 0,1/M,2/M,…,1, as required in [F3].

2.1F1F2F3F4step 1.1algebra∎

For each integer 0≤k≤2M, [F4] makes x↦∣x∣k continuous and Borel measurable on [0,1], and it satisfies 0≤∣x∣k≤1 there (with ∣x∣0=1). Thus [F2] applies and its integral is at most one. These finite moments and step 1.1 verify every condition in [F3], proving the assertion.

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