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The shape-rate Gamma density is normalized

Statement

Assume Countable Choice. For a,b>0, ∫(0,∞)baΓ(a)xa−1e−bx dλ1(x)=1.

Facts & Assumptions

Given: Countable Choice and positive real parameters a,b.

[A1]

Countable Choice is the axiom required by the bounded-Riemann-to-Lebesgue integral and Lebesgue change-of-variables suppliers. The Axiom of Countable Choice (ACω).

[F1]

Γ(a)=∫0∞ta−1e−t dt as a mixed improper integral. The real Gamma function by Euler's integral.

[F2]

This Euler integral is finite for a>0 by convergence and strictly positive by the positive-integrand observation in the Gamma definition. Euler's Gamma integral converges exactly for positive real parameters The real Gamma function by Euler's integral.

[F3]

Bounded Riemann-integrable functions on compact intervals have the same Lebesgue integral. A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral.

[F4]

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

[F5]

Monotone convergence holds for nonnegative measurable functions. Monotone convergence for the integral.

[F6]

For a C1 diffeomorphism between open sets, Lebesgue change of variables holds for nonnegative measurable functions under Countable Choice. A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions.

[F7]

Proof

Given: Countable Choice and a,b>0.

1.1A1F3F4F7F8construct

By [F7], g(t)=ta−1e−t is continuous and hence Borel and Lebesgue measurable on (0,∞) with the measure in [F8]. For each integer m≥2, its restriction to [1/m,m] is continuous, hence bounded Riemann integrable by [F4], and [F3] identifies its Riemann integral with its Lebesgue integral. The truncated functions g 1[1/m,m] increase to g.

2.1step 1.1F1F2F5

Monotone convergence [F5] identifies ∫g dλ1 with the limit of those truncated integrals. By the mixed-improper-integral definition [F1] and convergence [F2], this limit is Γ(a), so g has finite positive Lebesgue integral.

3.1A1F2F6step 2.1algebra∎

Apply [F6] to the diffeomorphism t=bx of (0,∞). Since dt=b dx and x=t/b, it gives ∫0∞xa−1e−bx dx=b−a∫0∞ta−1e−t dt=b−aΓ(a). Multiplication by ba/Γ(a) is valid by [F2] and yields one.

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