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Mathematics · 术语表

什么是 integrable function?

Definition 10.9 University Mathematics — Year 3 · 第 10 章 — The Lebesgue Integral

A measurable f ⁣:X→Rf \colon X \to \R (or C\C) is integrable if ∫∣f∣  ⁣dμ<∞\int\abs f\,\dd\mu < \infty; then ∫f=∫f+−∫f−\int f = \int f^+ - \int f^- (positive and negative parts; real and imaginary parts in the complex case). The integral is linear on integrable functions (decompose and recombine positive parts; the complex case reduces to the real one) and satisfies ∣∫f∣≤∫∣f∣\abs{\int f} \leq \int\abs f (real case: ±∫f=∫(±f)≤∫∣f∣\pm\int f = \int(\pm f) \leq \int\abs f; complex case: multiply by a unimodular constant to make the integral real). A property holds almost everywhere (a.e.) if it fails only on a μ\mu-null set; modifying ff on a null set changes no integral (the difference is dominated by ∞⋅1N\infty\cdot\mathbf 1_N, of integral 00).

示例

Example 10.13

1Q\mathbf 1_\Q is nowhere continuous: not Riemann-integrable — but Lebesgue-trivial: ∫1Q  ⁣dλ=λ(Q)=0\int\mathbf 1_\Q\,\dd\lambda = \lambda(\Q) = 0. Thomae’s function ( 1q\frac1q at rationals pq\frac pq, 00 elsewhere) is continuous exactly at the irrationals: Riemann-integrable with integral 00. And improper Riemann integrals are a different notion: ∫0∞sin⁡xx  ⁣dx\int_0^\infty\frac{\sin x}x\,\dd x converges as a limit of ∫0A\int_0^A (the weekend problem computes it =π2= \frac\pi2), but sin⁡xx∉L1((0, +∞))\frac{\sin x}x \notin L^1(\intoo0{+\infty}): the absolute integral diverges like the harmonic series (Exercise 10.6). Lebesgue’s theory trades conditional convergence for robust limit theorems.

Example 10.16 (The Gamma function)

For t>0t > 0 let

Γ(t)=∫0+∞xt−1e−x  ⁣dx.\Gamma(t) = \int_0^{+\infty} x^{t-1}\eu^{-x}\,\dd x .

The integral converges: near 00, xt−1x^{t-1} is integrable (t>0t > 0); at infinity, xt−1e−x≤Ce−x/2x^{t-1}\eu^{-x} \leq C\eu^{-x/2}. Integration by parts (on [ε,A][\varepsilon, A], then limits via MCT) gives the functional equation Γ(t+1)=t Γ(t)\Gamma(t + 1) = t\,\Gamma(t), whence Γ(n+1)=n!\Gamma(n+1) = n!: the factorial interpolated. On every [a, b]⊆(0, +∞)\intcc ab \subseteq \intoo0{+\infty}, ∂t(xt−1e−x)=ln⁡x⋅xt−1e−x\partial_t\bigl(x^{t-1}\eu^{-x}\bigr) = \ln x\cdot x^{t-1}\eu^{-x} is dominated by ∣ln⁡x∣(xa−1+xb−1)e−x\abs{\ln x}(x^{a-1} + x^{b-1})\eu^{-x}, integrable: Γ\Gamma is C1\mathcal C^1, and by induction C∞\mathcal C^\infty, with Γ(k)(t)=∫0∞(ln⁡x)kxt−1e−x ⁣dx\Gamma^{(k)}(t) = \int_0^\infty(\ln x)^kx^{t-1}\eu^{-x}\dd x. The value Γ(12)=π\Gamma(\frac12) = \sqrt\pi is the Gaussian integral in disguise (Problem 10.1).

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