TheoremBase

Tonelli's and Fubini's Theorems

For σ-finite measure spaces, the integral of a nonnegative product-measurable function against the product measure equals both iterated integrals (Tonelli); for an integrable function the sections are integrable outside a null set, and the iterated integral, set to zero on that null set, is integrable with the same integral (Fubini), in either order.

Statement

In the setting of Product Measures, Lebesgue Measure on ℝⁿ and Image Measures: Notation, let (X,F,μ)(X,\mathcal{F},\mu) and (Y,G,ν)(Y,\mathcal{G},\nu) be measure spaces with μ\mu and ν\nu σ\sigma-finite.

Tonelli's theorem. Let f∈M+(X×Y,F⊗G)f\in\mathcal{M}^{+}(X\times Y,\mathcal{F}\otimes\mathcal{G}). Then fx∈M+(Y,G)f_{x}\in\mathcal{M}^{+}(Y,\mathcal{G}) and fy∈M+(X,F)f^{y}\in\mathcal{M}^{+}(X,\mathcal{F}) for all x∈Xx\in X and y∈Yy\in Y, the functions x↦∫Yfx dνx\mapsto\int_{Y}f_{x}\,d\nu and y↦∫Xfy dμy\mapsto\int_{X}f^{y}\,d\mu lie in M+(X,F)\mathcal{M}^{+}(X,\mathcal{F}) and M+(Y,G)\mathcal{M}^{+}(Y,\mathcal{G}), and

∫X×Yf d(μ⊗ν)=∫X(∫Yfx dν)dμ(x)=∫Y(∫Xfy dμ)dν(y).\int_{X\times Y}f\,d(\mu\otimes\nu)=\int_{X}\Bigl(\int_{Y}f_{x}\,d\nu\Bigr)d\mu(x)=\int_{Y}\Bigl(\int_{X}f^{y}\,d\mu\Bigr)d\nu(y).

Fubini's theorem. Let f∈L1(μ⊗ν)f\in\mathcal{L}^{1}(\mu\otimes\nu). The set

N={x∈X:fx∉L1(ν)}N=\{x\in X:f_{x}\notin\mathcal{L}^{1}(\nu)\}

lies in F\mathcal{F} and μ(N)=0\mu(N)=0; the function I:X→RI:X\to\mathbb{R} with I(x)=∫Yfx dνI(x)=\int_{Y}f_{x}\,d\nu for x∈X∖Nx\in X\setminus N and I(x)=0I(x)=0 for x∈Nx\in N lies in L1(μ)\mathcal{L}^{1}(\mu); and

∫X×Yf d(μ⊗ν)=∫XI dμ.\int_{X\times Y}f\,d(\mu\otimes\nu)=\int_{X}I\,d\mu.

Fubini's theorem in the other order. Let f∈L1(μ⊗ν)f\in\mathcal{L}^{1}(\mu\otimes\nu). The set

M={y∈Y:fy∉L1(μ)}M=\{y\in Y:f^{y}\notin\mathcal{L}^{1}(\mu)\}

lies in G\mathcal{G} and ν(M)=0\nu(M)=0; the function J:Y→RJ:Y\to\mathbb{R} with J(y)=∫Xfy dμJ(y)=\int_{X}f^{y}\,d\mu for y∈Y∖My\in Y\setminus M and J(y)=0J(y)=0 for y∈My\in M lies in L1(ν)\mathcal{L}^{1}(\nu); and

∫X×Yf d(μ⊗ν)=∫YJ dν.\int_{X\times Y}f\,d(\mu\otimes\nu)=\int_{Y}J\,d\nu.

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