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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29
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For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function

Statement

Let (X,A,μ) be a measure space, let f:XC be measurable, and let 0<p<. Then Xfpdμ=p0tp1μ({f>t})dt=p0tp1Af(t)dt, where either side may be +.

Facts & Assumptions

Given: A measurable function f:XC and a real number 0<p<.

[L1]

The distribution function is Af(t)=μ({f>t}). (The distribution function of absolute value)

[L2]

The derivative of tp is ptp1 on (0,), and the fundamental theorem of calculus recovers ap by integrating that derivative. (Continuity and derivatives of positive-base real powers, The second fundamental theorem: if G is differentiable on [a,b] with G=f and f is integrable, then abf=G(b)G(a))

[L3]

Every nonnegative measurable function admits increasing simple approximations. (Every nonnegative measurable function admits an explicit increasing sequence of simple approximations)

[L4]

Monotone convergence passes increasing limits through the integral. (Monotone convergence for the integral)

Proof

technique · direct
1.1

Fix a0. By [L2], ap=0aptp1dt=0ptp11{t<a}dt.

L2
2.1

Let s=j=1maj1Ej be a nonnegative simple function, with the sets Ej pairwise disjoint and the coefficients aj0. Then Xspdμ=j=1majpμ(Ej). Using step 1.1 for each coefficient and exchanging the resulting finite sum with the real integral gives Xspdμ=p0tp1μ({s>t})dt.

step 1.1algebra
3.1

By [L3], choose simple functions snf. Then snpfp, and for each t0 one has 1{sn>t}1{f>t}, hence μ({sn>t})μ({f>t})=Af(t) by [L1]. Applying [L4] first on X and then on (0,) to the identities from step 2.1 yields Xfpdμ=p0tp1μ({f>t})dt=p0tp1Af(t)dt.

L1L3L4step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources