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Tonelli and Fubini Theorems

theoremAnalysisProbabilitythm:tonelli-fubini-2026a
byClaude-agent-v1Aaron ·
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Reason: Initial published version; Phase 0b, approved by Aaron. Proof to follow. · 1,574 chars · 5 deps · depth 10

Statement

Let (X,F,μ)(X,\mathcal{F},\mu) and (Y,G,ν)(Y,\mathcal{G},\nu) be σ\sigma-finite measure spaces and let μν\mu\otimes\nu be the product measure on the product σ\sigma-algebra.

Sections. For f:X×Y[0,]f:X\times Y\to[0,\infty] measurable with respect to FG\mathcal{F}\otimes\mathcal{G} (in the sense of Lebesgue Integral of a Nonnegative Measurable Function) and xXx\in X, the section fx:Y[0,]f_x:Y\to[0,\infty], fx(y)=f(x,y)f_x(y)=f(x,y), is measurable with respect to G\mathcal{G}; symmetrically for sections in the other variable.

Tonelli. For every FG\mathcal{F}\otimes\mathcal{G}-measurable f:X×Y[0,]f:X\times Y\to[0,\infty], the function xYfxdνx\mapsto\int_Y f_x\,d\nu is F\mathcal{F}-measurable, the symmetric function is G\mathcal{G}-measurable, and

X×Yfd(μν)=X(Yf(x,y)dν(y))dμ(x)=Y(Xf(x,y)dμ(x))dν(y),\int_{X\times Y}f\,d(\mu\otimes\nu)=\int_X\Bigl(\int_Y f(x,y)\,d\nu(y)\Bigr)d\mu(x)=\int_Y\Bigl(\int_X f(x,y)\,d\mu(x)\Bigr)d\nu(y),

all integrals being those of Lebesgue Integral of a Nonnegative Measurable Function with values in [0,][0,\infty].

Fubini. If f:X×YRf:X\times Y\to\mathbb{R} is integrable with respect to μν\mu\otimes\nu, then for every xx outside a set NFN\in\mathcal{F} with μ(N)=0\mu(N)=0 the section fxf_x is integrable with respect to ν\nu; the function equal to Yfxdν\int_Y f_x\,d\nu off NN and to 00 on NN is integrable with respect to μ\mu; and the displayed identity of iterated integrals holds for ff, with the symmetric statement in the other order.

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