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单词 ProofOfFubinisTheoremForTheLebesgueIntegral
释义

proof of Fubini’s theorem for the Lebesgue integral


Let μx and μy be measuresMathworldPlanetmath on X and Y respectively, let μ be the product measureMathworldPlanetmath μxμy, and let f(x,y) be μ-integrable on AX×Y. Then

Af(x,y)𝑑μ=X(Axf(x,y)𝑑μy)𝑑μx=Y(Ayf(x,y)𝑑μx)𝑑μy

where

Ax={y(x,y)A},Ay={x(x,y)A}

Proof: Assume for now that f(x,y)0. Consider the set

U=X×Y×

equipped with the measure

μu=μxμyμ1=μμ1=μxλ

where μ1 is ordinary Lebesgue measureMathworldPlanetmath and λ=μyμ1. Also consider the set WU defined by

W={(x,y,z)(x,y)A,0zf(x,y)}

Then

μu(W)=Af(x,y)𝑑μ

And

μu(W)=Xλ(Wx)𝑑μx

where

Wx={(y,z)(x,y,z)W}

However, we also have that

λ(Wx)=Axf(x,y)𝑑μy

Combining the last three equations gives us Fubini’s theorem. To remove the restrictionPlanetmathPlanetmath that f(x,y) be nonnegative, write f as

f(x,y)=f+(x,y)-f-(x,y)

where

f+(x,y)=|f(x,y)|+f(x,y)2,f-(x,y)=|f(x,y)|-f(x,y)2

are both nonnegative.

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更新时间:2025/5/4 15:59:58