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The Jacobi weight defines a finite positive measure with full support

Statement

Assume Countable Choice. Let α,β>−1 and define wα,β(t)={(1−t)α(1+t)β,−1<t<1,0,t=±1. Let λ be Lebesgue measure restricted to the Borel subsets of [−1,1], and set μα,β(E)=∫Ewα,β dλ. Then μα,β is a finite positive Borel measure, its relative support is all of [−1,1], and every moment ∫∣t∣k dμα,β is finite for k∈N0.

Facts & Assumptions

Given: Countable Choice and real parameters α,β>−1.

[A1]

The Axiom of Countable Choice (ACω): the Countable Choice axiom is assumed.

[F1]

A power above exponent minus one is Lebesgue integrable at zero: for every q>−1, x↦xq on (0,1], extended by zero at zero, is nonnegative and Lebesgue integrable with integral 1/(q+1).

[F4]

The measure with density f relative to μ: a nonnegative measurable density w defines the measure w dλ by E↦∫Ew dλ.

[F6]

Monotonicity and nonnegative homogeneity of the nonnegative integral: the nonnegative integral is monotone and positively homogeneous.

[F8]

Monotone convergence for the integral and Closure properties of measurable functions used by the integral: increasing nonnegative measurable cutoffs have integrals increasing to the integral of their limit.

Proof

1.1A1F2F3givenalgebra

The density is nonnegative and continuous on (−1,1) by [F3]. For every real c, its strict superlevel is all of [−1,1] when c<0, and the relatively open set {t∈(−1,1):wα,β(t)>c} when c≥0; thus the endpoint extension is Borel, and [F2] makes it Lebesgue measurable. Put Mα=max⁡(1,2α) and Mβ=max⁡(1,2β). On (−1,0] the factor (1−t)α is at most Mα, and on [0,1) the factor (1+t)β is at most Mβ.

2.1A1F1F6F7F8F9step 1.1constructalgebra

To justify the endpoint integral comparisons, set wn=wα,βχ[−1+1/n,1−1/n] for n≥2. Its superlevels are Borel by step 1.1. It is bounded and continuous except possibly at the two cutoff points; isolating these in intervals of arbitrarily small total length and using uniform continuity on the remaining compact pieces proves the Darboux criterion [F9]. The same argument applies to each half-interval restriction and to the corresponding power cutoffs. Consequently [F7] and the Riemann substitutions u=1+t and u=1−t give ∫−10wn dλ≤Mα∫1/n1uβ dλ≤Mα/(β+1) and ∫01wn dλ≤Mβ∫1/n1uα dλ≤Mβ/(α+1), using [F1] and [F6] for the final bounds. Since wn↑wα,β pointwise, including at the endpoints, [F8] gives ∫[−1,1]wα,β dλ≤Mα/(β+1)+Mβ/(α+1)<∞.

3.1A1F2F4F5F6step 1.1step 2.1choose

By [F4], μα,β is a positive Borel measure, and step 2.1 makes its total mass finite. If U⊆[−1,1] is nonempty and relatively open, choose a nondegenerate closed interval [c,d]⊂U∩(−1,1). The density is continuous and strictly positive on [c,d], so [F5] gives a minimum m>0 there. By [F2] and monotonicity [F6], μα,β(U)≥∫[c,d]wα,β dλ≥m(d−c)>0. Therefore every point is in the relative support.

4.1F6step 2.1algebra∎

For every k∈N0, ∣t∣k≤1 on [−1,1], so [F6] gives ∫∣t∣k dμα,β≤μα,β([−1,1])<∞ by step 2.1. This proves all moment claims.

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