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The Beta density has finite positive Lebesgue integral

Statement

Assume Countable Choice. For every α,β>0, the nonnegative function p↦pα−1(1−p)β−1 is Lebesgue measurable on (0,1) and ∫(0,1)pα−1(1−p)β−1 dλ1(p)=B(α,β)=Γ(α)Γ(β)Γ(α+β)∈(0,∞).

Facts & Assumptions

Given: Countable Choice and positive real numbers α,β.

[A1]

Countable Choice is the axiom assumed by the bounded-Riemann-to-Lebesgue integral theorem. The Axiom of Countable Choice (ACω).

[F1]

The real Beta function is the improper integral of pα−1(1−p)β−1 on (0,1). Euler's real Beta integral.

[F2]

For positive α,β, B(α,β)=Γ(α)Γ(β)/Γ(α+β). The real Beta--Gamma identity.

[F3]

The Euler Gamma integral converges to a finite real value for every positive argument. Euler's Gamma integral converges exactly for positive real parameters.

[F4]

A bounded Riemann-integrable function on a closed bounded interval is Lebesgue measurable and has the same Lebesgue and Riemann integrals under Countable Choice. A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral.

[F5]

Continuous functions on a closed bounded interval are bounded and Riemann integrable. A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion.

[F6]

Increasing limits of nonnegative measurable functions have integrals equal to the increasing limit of their integrals. Monotone convergence for the integral.

[F7]

The improper Beta integral converges for positive parameters. Euler's Beta integral converges exactly for two positive parameters.

[F8]

Positive real powers are continuous on (0,∞). Continuity and derivatives of positive-base real powers.

[F10]

For s>0, Γ(s) is the improper integral ∫0∞ts−1e−t dt, with separate improper ends at zero and infinity. The real Gamma function by Euler's integral.

[F12]

The logarithm is strictly increasing on (0,∞) and log⁡1=0. Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm.

[F13]

The exponential is continuous and strictly increasing on R, and exp⁡(x)>0 for every real x. The exponential function is strictly increasing, The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x).

[F14]

Proper Riemann integration preserves nonnegativity and pointwise order; if an integrable function is bounded below by c on [a,b], its integral is at least c(b−a). If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a).

[F15]

Splitting a convergent improper integral at a nonsingular finite point changes its value by the corresponding proper integral and preserves convergence of the remaining tail. Improper convergence is independent of finite truncations and split points.

Proof

Given: Countable Choice and α,β>0.

1.1A1F4F5F8F9construct

By [F8], f(p)=pα−1(1−p)β−1 is continuous and hence Borel and Lebesgue measurable on (0,1) with the measure in [F9]. For each integer m≥3, one has 1/m<1−1/m, so its restriction to [1/m,1−1/m] is continuous, hence bounded and Riemann integrable by [F5], and [F4] identifies its Lebesgue integral with its Riemann integral. The indicators of these intervals increase pointwise to one, so the truncated functions increase to f.

1.2givenF5F8F11F12F13F14algebra

Fix s>0 and put hs(t)=ts−1e−t for t>0. For 1≤t≤2, [F12] gives 0≤log⁡t≤log⁡2, so (s−1)log⁡t≥min⁡{0,(s−1)log⁡2}. By [F11] and [F13], hs(t)≥cs:=exp⁡(min⁡{0,(s−1)log⁡2})exp⁡(−2)>0. The function hs is continuous on [1,2] by [F8] and [F13], hence bounded and Riemann integrable by [F5]. The lower integral bound [F14] therefore gives ∫12hs(t) dt≥cs(2−1)=cs>0.

2.1step 1.1F1F6F7

By [F6], the Lebesgue integral of f is the limit of the truncated integrals from step 1.1. The same truncations define the improper integral in [F1], which converges by [F7]; therefore the limit is B(α,β).

2.2F3F10F11F13F14F15step 1.2

By [F3] and [F10], Γ(s) is finite. The integrand hs is positive on (0,∞) by [F11] and [F13]. Every compact truncation integral on either side of [1,2] is nonnegative by [F14], and its finite improper limit is nonnegative as well. Splitting at 1 and 2 by [F15] gives Γ(s)=∫01hs+∫12hs+∫2∞hs≥∫12hs≥cs>0. Thus 0<Γ(s)<∞ for every s>0, with strict positivity proved here rather than inferred from convergence alone.

3.1step 2.1step 2.2F2algebra∎

Apply step 2.2 with s=α,β,α+β. The identity in [F2] expresses B(α,β) as a quotient of finite positive numbers, hence 0<B(α,β)<∞. Together with step 2.1 this proves the displayed Lebesgue integral identity and its endpoint bounds.

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