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Fubini's theorem

Fubini's theorem is a cornerstone of mathematical analysis and measure theory, providing rigorous justification for evaluating multiple integrals via iterated integrals and interchanging their order under appropriate conditions on the integrand.^[1] Named after the Italian mathematician Guido Fubini, who established its general form in 1907, the theorem addresses the equality between a double integral over a product space and the corresponding iterated integrals. Specifically, for σ-finite measure spaces (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu) with product measure \mu \times \nu on the product σ-algebra, if f: X \times Y \to \mathbb{R} is integrable (i.e., \int_{X \times Y} |f| , d(\mu \times \nu) < \infty), then

\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y).

^[2] A related result, often considered a precursor or variant known as Tonelli's theorem, extends this to nonnegative measurable functions f \geq 0, where the iterated integrals always equal the double integral (possibly infinite), without requiring absolute integrability.^[2] In the classical setting of multivariable calculus with Riemann integrals, the theorem applies to continuous functions on compact rectangular domains, where the double integral over R = [a,b] \times [c,d] equals both possible iterated integrals.^[1] Fubini's original proof in his paper "Sugli integrali multipli" addressed Lebesgue integrals over bounded measurable sets, building on earlier work by Henri Lebesgue in 1904 and special cases known to Leonhard Euler in the 18th century for continuous functions.^[3] The theorem's significance lies in its foundational role for computations in higher-dimensional analysis, probability theory (e.g., joint distributions), and applications in physics and engineering involving volume, mass, or expectation calculations.^[4]

History

Origins and Development

The concept of interchanging the order of integration in multiple integrals has roots in the 18th century, where Leonhard Euler considered special cases for continuous functions over simple domains, effectively using early forms of what would become Riemann integrals. These ideas were informal and lacked the rigorous foundation provided by later developments in analysis.^[1] In the early 20th century, the creation of measure theory by Henri Lebesgue laid the groundwork for generalizing multiple integrals. Lebesgue's 1902 doctoral thesis Intégrale, longueur, aire introduced the Lebesgue measure and integral, enabling treatment of more general functions and sets. By 1904, Lebesgue had extended these ideas to multiple integrals, addressing issues of measurability and integrability that foreshadowed Fubini's theorem.^[2]

Key Contributors and Milestones

Guido Fubini, an Italian mathematician born in 1879, made a foundational contribution to multiple integration by proving a theorem allowing the evaluation of double Lebesgue integrals over bounded measurable sets via iterated integrals for integrable functions. This result appeared in his 1907 paper "Sugli integrali multipli," published in the Rendiconti dell'Accademia dei Lincei (series 5, volume 16, pages 608–614).^[5] Building on Fubini's work, Leonida Tonelli, another prominent Italian mathematician (1885–1946), extended the theorem in 1909 to apply to non-negative measurable functions, removing the absolute integrability restriction and broadening its applicability in analysis. Tonelli's extension was detailed in his paper "Sull'integrazione per parti," published in the Rendiconti dell'Accademia dei Lincei (series 5, volume 18, pages 246–253).^[5] The development of Fubini's theorem was deeply influenced by the earlier foundations of measure theory laid by Henri Lebesgue in his 1902 doctoral thesis Intégrale, longueur, aire, which introduced the Lebesgue integral and measure, enabling rigorous treatment of integration over more general spaces. This work, published in the Annali di Matematica Pura e Applicata (series 3, volume 7, pages 1–163), provided the theoretical groundwork for later generalizations of multiple integrals.^[6] In the 1930s, further refinements emerged within functional analysis and abstract measure theory, notably in Stanisław Saks's 1937 monograph Theory of the Integral, which extended Fubini's theorem to abstract product measure spaces and clarified conditions for its validity in broader settings (see sections 8 and 9, pages 76–82).^[7] These advancements solidified the theorem's role in modern analysis. Key milestones in the theorem's history include Fubini's initial 1907 formulation for Lebesgue integrable functions over bounded sets, Tonelli's 1909 extension to non-negative functions, and its incorporation into standard measure theory texts during the 1930s, exemplified by Saks's comprehensive treatment.^[7]

Foundations

Product Measure Spaces

In measure theory, the product measure space arises from two given measure spaces (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu). The underlying set is the Cartesian product X \times Y, and the product \sigma-algebra \mathcal{A} \otimes \mathcal{B} is the smallest \sigma-algebra on X \times Y that contains all measurable rectangles of the form E \times F, where E \in \mathcal{A} and F \in \mathcal{B}. This construction ensures that the product \sigma-algebra captures the joint measurability structure derived from the individual spaces.^[8] The product measure \mu \otimes \nu on \mathcal{A} \otimes \mathcal{B} is defined initially on the collection of measurable rectangles by setting (\mu \otimes \nu)(E \times F) = \mu(E) \nu(F) for E \in \mathcal{A} and F \in \mathcal{B}. The rectangles form a semi-ring, and the set function defined on them is a premeasure. Under the assumption that \mu and \nu are \sigma-finite, Carathéodory's extension theorem guarantees a unique extension of this premeasure to a measure on the product \sigma-algebra. Without \sigma-finiteness, such uniqueness may fail, as multiple extensions could exist.^[9] A measure on a space is \sigma-finite if the space can be expressed as a countable union of measurable sets each having finite measure; thus, both X = \bigcup_{n=1}^\infty X_n with \mu(X_n) < \infty for each n, and similarly for Y. This condition is crucial for the well-definedness of the product measure, as it allows the outer measure to coincide with the extension on the product \sigma-algebra and ensures additivity over disjoint unions covering the space. For a measurable rectangle E \times F, the integral of its indicator function with respect to the product measure satisfies

\int_{X \times Y} 1_{E \times F} \, d(\mu \otimes \nu) = (\mu \otimes \nu)(E \times F) = \mu(E) \nu(F).

This equality follows directly from the definition of the measure on rectangles and the linearity of the integral for simple functions.^[2]

Measurable Functions and Integrability

In the context of a product measure space (X \times Y, \mathcal{A} \otimes \mathcal{B}, \mu \otimes \nu), where (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu) are \sigma-finite measure spaces, a function f: X \times Y \to \mathbb{R} is measurable if the preimage f^{-1}(B) belongs to the product \sigma-algebra \mathcal{A} \otimes \mathcal{B} for every Borel set B \subseteq \mathbb{R}.^[8] This ensures that the function respects the measurable structure of the product space, allowing integration to be well-defined.^[10] Measurable functions on the product space can be approximated by simple functions, which are finite linear combinations of characteristic functions of measurable sets in \mathcal{A} \otimes \mathcal{B}. For a non-negative measurable function f: X \times Y \to [0, \infty), there exists a sequence of non-negative simple functions \{\phi_n\} such that $0 \leq \phi_1 \leq \phi_2 \leq \cdots \leq f and \phi_n(x,y) \to f(x,y) pointwise for all (x,y) \in X \times Y.^[11] If f is bounded on a measurable subset E \subseteq X \times Y, the convergence is uniform on E.^[12] This approximation property facilitates the extension of the integral from simple functions to general measurable functions via limits. Absolute integrability on the product space requires that f be measurable and that the integral of its absolute value be finite: \int_{X \times Y} |f| \, d(\mu \otimes \nu) < \infty. This condition defines the space L^1(\mu \otimes \nu) of integrable functions on the product, equipped with the norm \|f\|_{L^1} = \int_{X \times Y} |f| \, d(\mu \otimes \nu).^[8] For such functions, the product integral of |f| equals the iterated integral \int_X \left( \int_Y |f(x,y)| \, d\nu(y) \right) d\mu(x), and likewise for the reverse order.^[8] For non-negative measurable functions f: X \times Y \to [0, \infty], integrability is extended to allow values in the extended non-negative reals, defined as \int_{X \times Y} f \, d(\mu \otimes \nu) = \sup \left\{ \int_{X \times Y} \phi \, d(\mu \otimes \nu) : 0 \leq \phi \leq f, \, \phi \text{ simple} \right\}, which may be infinite.^[13] This framework supports the application of Tonelli's theorem by ensuring the integral is well-defined even when finite integrability fails.^[13]

Main Theorems

Fubini's Theorem for Integrable Functions

Fubini's theorem provides a foundational result in measure theory for interchanging the order of integration in double integrals over product spaces, specifically when the integrand is absolutely integrable. Consider two σ-finite measure spaces (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu), with the product measure space (X \times Y, \mathcal{A} \otimes \mathcal{B}, \mu \otimes \nu). A measurable function f: X \times Y \to \overline{\mathbb{R}} belongs to L^1(\mu \otimes \nu) if \int_{X \times Y} |f| \, d(\mu \otimes \nu) < \infty, establishing absolute integrability.^[14] Under these conditions, the slice functions f_x: Y \to \overline{\mathbb{R}} defined by f_x(y) = f(x, y) for each fixed x \in X are \nu-integrable for \mu-almost every x, meaning f_x \in L^1(\nu) and \int_Y |f_x| \, d\nu < \infty for \mu-a.e. x. Similarly, the slices f^y: X \to \overline{\mathbb{R}} given by f^y(x) = f(x, y) lie in L^1(\mu) for \nu-a.e. y. The theorem asserts that the double integral equals both iterated integrals:

\int_{X \times Y} f \, d(\mu \otimes \nu) = \int_X \left( \int_Y f(x, y) \, d\nu(y) \right) d\mu(x) = \int_Y \left( \int_X f(x, y) \, d\mu(x) \right) d\nu(y).

^[14]^[4] This equality of the double integral and the iterated integrals holds due to the absolute integrability condition, which guarantees the integrability of the slices and justifies the interchange. The σ-finiteness of \mu and \nu ensures the product measure is σ-finite, allowing the theorem to apply to standard spaces such as Euclidean spaces equipped with Lebesgue measure. For signed functions, the result follows from applying the non-negative case (Tonelli's theorem) to the positive and negative parts after establishing integrability of |f|.^[15] The implications of Fubini's theorem are profound in analysis, enabling the reduction of multiple integrals to sequential single integrals, which simplifies computations in areas like partial differential equations and probability theory. It underscores the robustness of the Lebesgue integral over the Riemann integral by providing conditions under which order interchange is valid without additional assumptions on continuity or boundedness.^[14]

Tonelli's Theorem for Non-Negative Functions

Tonelli's theorem addresses the interchange of iterated integrals for non-negative measurable functions on product measure spaces, without the requirement of finite integrability that is necessary in Fubini's theorem. Let (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu) be \sigma-finite measure spaces, and let f: X \times Y \to [0, \infty] be a measurable function with respect to the product \sigma-algebra \mathcal{A} \otimes \mathcal{B}. Then,

\int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y) = \int_{X \times Y} f \, d(\mu \otimes \nu),

where all integrals are understood in the extended non-negative sense and may equal +\infty.^[16] The \sigma-finiteness condition on the measures \mu and \nu ensures that the product measure \mu \otimes \nu is well-defined and that the theorem holds; without it, counterexamples exist where the equality fails. The function f must be non-negative and measurable, allowing the integrals to potentially diverge while still equating the iterated and double integrals. Unlike Fubini's theorem, which requires f to be absolutely integrable (i.e., \int |f| \, d(\mu \otimes \nu) < \infty) to guarantee finite equal integrals, Tonelli's theorem permits infinite values but does not demand absolute integrability a priori.^[16] The proof of Tonelli's theorem relies implicitly on the monotone convergence theorem, which justifies passing limits through integrals when approximating the non-negative measurable f by an increasing sequence of simple functions. This approach establishes the equality for simple functions first and then extends it to general non-negative measurable functions via monotone limits.^[17] Tonelli's theorem, introduced by Leonida Tonelli in 1909, forms the foundation for handling broader classes of functions in multiple integrals. For signed integrable functions, Tonelli's theorem applies after decomposing f = f^+ - f^- into non-negative parts, yielding finite iterated integrals under the absolute integrability condition.^[16]

Fubini-Tonelli Theorem

The Fubini-Tonelli theorem unifies the results of Fubini's and Tonelli's theorems, providing conditions under which the order of integration can be interchanged for double integrals over product measure spaces. It applies to measurable functions that are either non-negative or integrable, assuming the underlying measure spaces are σ-finite to ensure the product measure is well-defined and the integrals are comparable. This theorem is fundamental in measure theory for computing multiple integrals by iteration, avoiding direct evaluation of the double integral.^[18] For a non-negative measurable function f: X \times Y \to [0, \infty], where (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu) are σ-finite measure spaces equipped with the product σ-algebra \mathcal{A} \otimes \mathcal{B} and product measure \mu \times \nu, the theorem asserts that the double integral equals both iterated integrals whenever they are defined:

\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y).

Here, the inner integrals may be infinite, but the equality holds by Tonelli's theorem for non-negative functions, which guarantees the measurability of the iterated integral functions and the equality of the integrals without requiring finiteness. The σ-finiteness of \mu and \nu is essential, as it allows the product space to be covered by countably many finite-measure rectangles, facilitating the extension of the product measure and ensuring no pathologies arise in the integration process.^[19]^[17] For a signed measurable function f that is integrable over X \times Y, meaning \int_{X \times Y} |f| \, d(\mu \times \nu) < \infty so f \in L^1(\mu \times \nu), the Fubini-Tonelli theorem extends the result by decomposition. Write f = f^+ - f^-, where f^+ = \max(f, 0) and f^- = \max(-f, 0) are non-negative, and both are integrable since |f| = f^+ + f^- is integrable. Applying the non-negative case to f^+ and f^- yields finite iterated integrals for each, and thus:

\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y).

The inner integrals now exist as finite Lebesgue integrals almost everywhere, and the equality in both orders follows directly from the non-negative result applied to the decomposition. This integrability condition ensures that the iterated integrals are equal and finite, distinguishing it from the non-negative case where infinity is permitted.^[20]

Extensions and Variations

For Complete Measure Spaces

A measure space (X, \mathcal{M}, \mu) is complete if every subset of a set of \mu-measure zero is measurable and has measure zero. The completion of an incomplete measure space (X, \mathcal{M}, \mu), denoted (X, \mathcal{M}^*, \mu^*), extends the \sigma-algebra \mathcal{M}^* to include all subsets of \mu-null sets, with \mu^* defined by \mu^*(E) = \inf \{ \mu(A) : A \in \mathcal{M}, E \subset A \} for E \subset X, ensuring \mu^* agrees with \mu on \mathcal{M} and assigns measure zero to all null sets and their subsets. This construction incorporates null sets more robustly, enlarging the class of measurable sets without altering the measure on the original \sigma-algebra.^[21] For complete product measure spaces, Fubini's theorem extends to the completed product \mu^* \otimes \nu^* on the completed product \sigma-algebra \mathcal{M}^* \otimes \mathcal{N}^*, where (Y, \mathcal{N}, \nu) is another measure space with completion \nu^*. Specifically, if (X, \mathcal{M}^*, \mu^*) and (Y, \mathcal{N}^*, \nu^*) are \sigma-finite complete measure spaces and f: X \times Y \to [-\infty, \infty] is measurable with respect to \mathcal{M}^* \otimes \mathcal{N}^* and \int_{X \times Y} |f| \, d(\mu^* \otimes \nu^*) < \infty, then f_x(y) = f(x,y) is \mathcal{N}^*-measurable and integrable for \mu^*-almost every x \in X, f^y(x) = f(x,y) is \mathcal{M}^*-measurable and integrable for \nu^*-almost every y \in Y, and the iterated integrals equal the double integral: \int_X \left( \int_Y f(x,y) \, d\nu^*(y) \right) d\mu^*(x) = \int_{X \times Y} f \, d(\mu^* \otimes \nu^*) = \int_Y \left( \int_X f(x,y) \, d\mu^*(x) \right) d\nu^*(y). A similar statement holds for non-negative measurable functions via Tonelli's theorem, replacing integrability with the requirement that at least one iterated integral is finite.^[22] This extension differs from the standard Fubini-Tonelli theorem for incomplete spaces by permitting integration over a broader class of functions that are measurable only with respect to the completed product \sigma-algebra, which may include sets not in the original \mathcal{M} \otimes \mathcal{N}. However, it necessitates caution with cross-sections (slices), as these may fail to be measurable with respect to the original \sigma-algebras \mathcal{M} or \mathcal{N}, though they remain measurable almost everywhere with respect to the completions \mu^* and \nu^*. The completed product measure \mu^* \otimes \nu^* is generally larger than the completion of the incomplete product \mu \otimes \nu, ensuring the theorem applies directly without additional modifications.^[21] An illustrative example arises in the context of Lebesgue measure on \mathbb{R}, which is complete. The product Lebesgue measure \lambda^2 on \mathbb{R}^2 has completion \lambda^{*2} = \lambda^* \otimes \lambda^*, resolving measurability issues for sets like the product of a Vitali non-measurable set V \subset [0,1] with a singleton \{0\}, which has outer measure zero in the product but is not measurable in the incomplete product \sigma-algebra; under the completion, such sets become measurable with measure zero, allowing Fubini's theorem to apply to indicator functions of these sets without violating integrability conditions.^[22]

Generalizations to Multiple Integrals

The generalization of the Fubini-Tonelli theorem to multiple integrals extends the two-dimensional case through iterative application to n-fold product measures. Consider σ-finite measure spaces (X_i, \mathcal{A}_i, \mu_i) for i = 1, \dots, n. The n-fold product measure \mu = \mu_1 \otimes \cdots \otimes \mu_n is defined on the product space X = X_1 \times \cdots \times X_n equipped with the product \sigma-algebra \mathcal{A} = \mathcal{A}_1 \otimes \cdots \otimes \mathcal{A}_n.^[8] For a non-negative measurable function f: X \to [0, \infty] or an integrable function f \in L^1(\mu), the multiple integral equals the iterated integral in any order:

\int_X f \, d\mu = \int_{X_1} \left( \cdots \int_{X_{n-1}} \left( \int_{X_n} f(x_1, \dots, x_n) \, d\mu_n(x_n) \right) d\mu_{n-1}(x_{n-1}) \cdots \right) d\mu_1(x_1),

and all such iterated integrals coincide. This holds under the condition that all measures \mu_i are \sigma-finite.^[23] These generalizations find key applications in probability theory, where they enable the computation of expectations for functions of several random variables by iterating over joint densities on product probability spaces. In partial differential equations, they support the evaluation of higher-dimensional multiple integrals, such as those appearing in convolution representations of solutions.^[24]^[25]

Proofs

Outline for Lebesgue Integrals

The proof of the Fubini-Tonelli theorem in the Lebesgue integration setting proceeds by leveraging the approximation of measurable functions by simple functions and employing key convergence theorems from measure theory. In a product measure space (X \times Y, \mathcal{M} \otimes \mathcal{N}, \mu \times \nu), where \mu and \nu are \sigma-finite measures, the strategy begins with establishing the product measure via the Carathéodory extension theorem applied to the semi-ring of measurable rectangles A \times B with A \in \mathcal{M}, B \in \mathcal{N}, and measure \mu(A)\nu(B). This construction ensures the double integral is well-defined for non-negative measurable functions, with \sigma-finiteness guaranteeing countable additivity and avoiding issues in non-finite spaces.^[17]^[20] A foundational step is verifying the measurability of slices: for a measurable set E \subset X \times Y, the x-slice E_x = \{ y \in Y \mid (x,y) \in E \} is \mathcal{N}-measurable for \mu-almost every x \in X, and the measure of the slice \nu(E_x) is measurable in x. Similarly for y-slices. This relies on the \sigma-finiteness to approximate E by unions of rectangles and apply properties of the product \sigma-algebra. For indicator functions of rectangles, $1_{A \times B}, the equality holds explicitly:

\iint 1_{A \times B} \, d(\mu \times \nu) = \mu(A) \nu(B) = \int_X \nu(B) 1_A(x) \, d\mu(x) = \int_X \left( \int_Y 1_{A \times B}(x,y) \, d\nu(y) \right) d\mu(x),

with the iterated integral in the reverse order matching by symmetry. \sigma-finiteness ensures this extends to indicators of finite disjoint unions of rectangles via linearity and countable additivity.^[17]^[20] The result extends to simple functions, which are finite non-negative linear combinations \sum c_i 1_{R_i} of disjoint rectangles R_i = A_i \times B_i, by linearity of the integral, yielding equality between the double integral and both iterated integrals. For general non-negative measurable functions f \geq 0, approximate f by an increasing sequence of simple functions f_n \uparrow f pointwise; the monotone convergence theorem then implies

\iint f \, d(\mu \times \nu) = \lim_{n \to \infty} \iint f_n \, d(\mu \times \nu) = \lim_{n \to \infty} \int_X \int_Y f_n(x,y) \, d\nu(y) \, d\mu(x) = \int_X \int_Y f(x,y) \, d\nu(y) \, d\mu(x),

with the reverse iterated integral following analogously, provided the slices f_x(y) = f(x,y) are measurable for almost every x. This uses the monotone convergence theorem applied to the sequence of inner integrals \int_Y f_n(x,y) \, d\nu(y) \uparrow \int_Y f(x,y) \, d\nu(y).^[17]^[20] For integrable functions f with \iint |f| \, d(\mu \times \nu) < \infty, decompose f = f^+ - f^- into non-negative parts and apply Tonelli's theorem to each, ensuring the iterated integrals converge absolutely. Alternatively, approximate by simple functions dominated by an integrable g (e.g., |f|) and invoke the dominated convergence theorem to justify the limits, confirming

\iint f \, d(\mu \times \nu) = \int_X \int_Y f(x,y) \, d\nu(y) \, d\mu(x) = \int_Y \int_X f(x,y) \, d\mu(x) \, d\nu(y).

The role of \sigma-finiteness is pivotal throughout, as it facilitates the measurability of slices for indicators (via exhaustion by finite-measure sets) and enables the approximations without divergence. Unlike proofs for Riemann integrals on compact sets, which depend on uniform continuity, this measure-theoretic approach handles general \sigma-finite spaces.^[17]^[20]

Proof for Riemann Integrals

Fubini's theorem for Riemann integrals applies to bounded continuous functions defined on a compact rectangular domain in the plane. Specifically, let f: [a, b] \times [c, d] \to \mathbb{R} be continuous on the compact set [a, b] \times [c, d]. Continuity implies that f is bounded and uniformly continuous, ensuring that the double Riemann integral \iint_{[a,b] \times [c,d]} f(x,y) \, d(x,y) exists and is finite, where the integral is defined via upper and lower Darboux sums over partitions of the rectangle into subrectangles.^[26] The theorem states that this double integral equals the iterated Riemann integrals in either order:

\iint_{[a,b] \times [c,d]} f(x,y) \, d(x,y) = \int_a^b \left( \int_c^d f(x,y) \, dy \right) dx = \int_c^d \left( \int_a^b f(x,y) \, dx \right) dy.

For fixed x \in [a, b], the function y \mapsto f(x, y) is continuous on the compact interval [c, d], so the inner integral \int_c^d f(x,y) \, dy exists as a Riemann integral, yielding a function g(x) = \int_c^d f(x,y) \, dy that is continuous on [a, b] due to the uniform continuity of f. Thus, the outer integral \int_a^b g(x) \, dx exists. The argument for the reverse order is symmetric. To establish equality with the double integral, consider product partitions: partition [a, b] into subintervals of lengths \Delta x_i with points x_{i-1} < x_i for i = 1, \dots, m, and [c, d] into subintervals of lengths \Delta y_j with points y_{j-1} < y_j for j = 1, \dots, n. A Riemann sum for the double integral is

S = \sum_{i=1}^m \sum_{j=1}^n f(\xi_i, \eta_j) \Delta x_i \Delta y_j,

where \xi_i \in [x_{i-1}, x_i] and \eta_j \in [y_{j-1}, y_j]. This sum can be rewritten as

S = \sum_{i=1}^m \left( \sum_{j=1}^n f(\xi_i, \eta_j) \Delta y_j \right) \Delta x_i.

The inner sum approximates \int_c^d f(\xi_i, y) \, dy, and uniform continuity of f ensures that the approximation error is bounded uniformly in i by \epsilon \cdot (d - c) as the mesh norm of the partition of [c, d] approaches zero, for any \epsilon > 0. Thus, as the mesh norms of both partitions tend to zero, S converges to \int_a^b \left( \int_c^d f(x,y) \, dy \right) dx. The proof for the other iterated integral follows analogously by regrouping the sums.^[27] The historical development traces back to Paul du Bois-Reymond, who proved the result for continuous functions in 1882 by reducing the double integral to one-dimensional estimates via uniform continuity and bounds on partial sums. Guido Fubini's 1907 work extended the theorem to Lebesgue integrals, but his approach similarly reduced multiple integrals to iterated forms through limiting processes on measurable sets, influencing later Riemann treatments.^[28] This proof relies crucially on continuity and compactness, which guarantee uniform continuity and the existence of all integrals. However, the result fails for merely Riemann integrable (possibly discontinuous) functions, as counterexamples exist where the double integral is finite but one or both iterated integrals diverge or yield unequal values, highlighting the need for the more robust Lebesgue theory to handle broader classes of functions.^[26]

Counterexamples and Limitations

Failures in Non-σ-Finite Spaces

In non-σ-finite measure spaces, Tonelli's theorem fails to hold in general, as the double integral of a non-negative measurable function with respect to the product measure may differ from the iterated integrals. This occurs because the product measure is not uniquely determined without σ-finiteness, and the standard construction leads to discrepancies in how sets or functions are assigned measure. Specifically, sets that appear "thin" in slices can have positive or infinite product measure if they span the entire non-σ-finite space, violating the equality guaranteed by the theorem under σ-finite assumptions.^[29] A canonical counterexample uses an uncountable set with a measure that vanishes on countable subsets. Let X be an uncountable set, \mathcal{A} = 2^X the power set of X, and \mu: \mathcal{A} \to [0,\infty] the measure defined by

\mu(A) = \begin{cases} 0 & \text{if $A$ is countable}, \\ \infty & \text{if $A$ is uncountable}. \end{cases}

This defines a complete, σ-additive measure on (X, \mathcal{A}), but it is not σ-finite, as any countable collection of finite-measure sets (i.e., countable subsets) covers at most a countable portion of X. Consider the product measure space (X \times X, \mathcal{A} \otimes \mathcal{A}, \mu \times \mu), where the product σ-algebra \mathcal{A} \otimes \mathcal{A} is the power set $2^{X \times X} (generated by all rectangles A \times B with A, B \in \mathcal{A}), and the product measure satisfies (\mu \times \mu)(A \times B) = \mu(A) \mu(B) with the convention that $0 \cdot \infty = 0 and \infty \cdot 0 = 0. Now define the non-negative measurable function f: X \times X \to [0,\infty) as the indicator of the diagonal,

f(x,y) = \mathbf{1}_D(x,y) = \begin{cases} 1 & \text{if $x = y$}, \\ 0 & \text{otherwise}, \end{cases}

where

D = \{(x,x) \mid x \in X\}&#36;. The set

D$ is measurable in the product σ-algebra, as it belongs to the power set.^[4] For the iterated integrals, fix x \in X; the y-section is f_x(y) = \mathbf{1}_{\{x\}}(y), so

\int_X f(x,y) \, d\mu(y) = \mu(\{x\}) = 0,

since \{x\} is countable. Thus, the iterated integral is

\int_X \left( \int_X f(x,y) \, d\mu(y) \right) d\mu(x) = \int_X 0 \, d\mu(x) = 0.

By symmetry, the reverse iteration \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\mu(y) also equals 0. However, the double integral with respect to the product measure is

\int_{X \times X} f \, d(\mu \times \mu) = (\mu \times \mu)(D) = \infty.

This holds because any set E \subset X \times X of finite product measure must satisfy \mu(E_x) < \infty for \mu-almost every x \in X, implying E_x is countable (hence measure 0) for almost all x; the only such sets of positive measure are effectively empty in the uncountable directions. Since every section D_x = \{x\} is non-empty (though measure 0), D cannot have finite product measure and must have infinite measure under this construction.^[29] This counterexample illustrates that without σ-finiteness, the product measure assigns infinite measure to the diagonal despite the slices having measure zero, leading to iterated integrals of 0 but a double integral of \infty. Consequently, Tonelli's theorem requires the σ-finiteness condition to ensure that non-negative measurable functions have equal iterated and double integrals, preventing such pathologies in the product construction.^[4]

Failures for Non-Measurable or Non-Integrable Functions

Fubini's theorem relies on the function being measurable with respect to the product σ-algebra and absolutely integrable over the product measure space; violations of these conditions can lead to the double integral being undefined or the iterated integrals failing to exist or coinciding with it. A prominent counterexample for non-measurable functions involves the Vitali set, a non-Lebesgue measurable subset V of [0,1] constructed via the axiom of choice by selecting one representative from each coset of the rationals in the reals modulo 1. Define the function f: [0,1] × [0,1] → {0,1} by f(x,y) = 1 if x - y ∈ V and 0 otherwise. The set {(x,y) | f(x,y) = 1} is not product-measurable, as its projection onto either axis would imply V is measurable if the set were, leading to a contradiction with the properties of V. Consequently, the double integral ∬ f d(μ × ν) is undefined, where μ and ν are Lebesgue measures on [0,1]. However, the iterated integrals yield different values: one order computes to 0 and the other to 1, due to the disjoint countable covering of [0,1] by rational translates of V, which forces the vertical sections to have measure 0 almost everywhere while the horizontal sections have measure 1 almost everywhere in an outer measure sense. This discrepancy highlights that without product measurability, Fubini's theorem cannot guarantee equality between the double and iterated integrals.^[30] For non-integrable functions, consider f(x,y) = (x - y)^{-2} on (0,1) × (0,1), which is Borel measurable but fails absolute integrability over the product space, as ∬_{(0,1)^2} |f(x,y)| , dμ dν = ∞ due to the non-integrable singularity along the diagonal x = y, where the behavior resembles ∫ 1/z^2 dz near 0, which diverges. The double integral of f is thus undefined in the Lebesgue sense because the positive and negative parts (though f is positive here, the divergence prevents definition). The iterated integrals ∫_0^1 \left( ∫_0^1 f(x,y) , dy \right) dx and the reverse also diverge in the Lebesgue sense, as the inner integral ∫0^1 (x - y)^{-2} dy = \left[ \frac{1}{x - y} \right]{y=0}^{y=1} formally yields 1/(x-1) - 1/x = -(1/x + 1/(1-x)), but the improper nature causes divergence to +∞ from the singularity at y = x, preventing the inner function from being Lebesgue integrable for any x ∈ (0,1). Without the integrability condition, the theorem fails because the orders cannot be interchanged reliably, and conditional convergence does not hold in the Lebesgue framework—specifically, ∫_0^1 ∫_0^1 |f| , dy , dx = ∞, while naive computations might suggest finite values via symmetric limits or principal values, underscoring the need for absolute integrability to ensure the integrals exist and coincide.^[4]

Issues with Non-Maximal Product Measures

In measure theory, the standard construction of the product measure space for two measure spaces (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu) begins with the product \sigma-algebra \mathcal{A} \otimes \mathcal{B}, which is the smallest \sigma-algebra on X \times Y containing all measurable rectangles of the form A \times B where A \in \mathcal{A} and B \in \mathcal{B}. The product measure \mu \times \nu is then defined uniquely on this \sigma-algebra under \sigma-finiteness assumptions. However, this product \sigma-algebra is often referred to as non-maximal because it does not include all subsets of null sets under the product measure, unlike the completion \overline{\mu \times \nu} of the product measure, which enlarges the \sigma-algebra to include all subsets of null sets and is essential for applications like Lebesgue integration on \mathbb{R}^n. When Fubini's theorem is applied using only the non-maximal product \sigma-algebra \mathcal{A} \otimes \mathcal{B}, certain functions that are integrable with respect to the completed product measure may fail to be measurable, rendering the double integral undefined. For instance, consider the spaces ([0,1], \mathcal{B}([0,1]), m) and ([0,1], \mathcal{B}([0,1]), m), where \mathcal{B}([0,1]) is the Borel \sigma-algebra and m is Lebesgue measure (incomplete on Borel sets). Let N \subset [0,1] be a Borel null set (e.g., a fat Cantor set of measure zero), and let V \subset N be a non-Borel subset of N. The set E = V \times [0,1] \subset [0,1]^2 has product measure zero since m(V) = 0, and in the completed product space (corresponding to Lebesgue measure on [0,1]^2), E is measurable with m \times m (E) = 0. Thus, the characteristic function \chi_E is integrable over the completed space with integral zero. However, \chi_E is not measurable with respect to the product Borel \sigma-algebra \mathcal{B}([0,1]) \otimes \mathcal{B}([0,1]), because if it were, the projection onto the first coordinate would imply V is Borel, a contradiction. Consequently, the double integral \iint \chi_E \, dm \, dm is undefined in the non-maximal setting, even though the iterated integrals exist and equal zero: \int_0^1 \left( \int_0^1 \chi_E(x,y) \, dy \right) dm(x) = \int_V 1 \, dm(x) = 0 and similarly for the reverse order. This example illustrates the implication for Fubini's theorem: to ensure that measurable functions in the completed product space have measurable slices almost everywhere and allow the interchange of integrals for a broader class of integrands, the theorem must be formulated using the completed product \sigma-algebra \overline{\mathcal{A} \otimes \mathcal{B}}. Without completion, the sections of sets (or functions) measurable in the full space may not be measurable with respect to the original \sigma-algebras \mathcal{A} and \mathcal{B}, violating the hypotheses of Fubini even when the iterated integrals formally exist. In extensions to complete measure spaces, the completed product aligns the theorem with the product of the individual completions, preserving the equality of iterated and double integrals almost everywhere. In the context of probability theory, where measure spaces are often taken to be complete, a related issue arises with the "rectangle algebra"—the algebra generated by finite disjoint unions of measurable rectangles, which is coarser than the full product \sigma-algebra. Independent events generated by independent \sigma-algebras may not belong to this rectangle algebra, as they can require infinite operations to construct (e.g., limits of rectangle approximations for events like \{X = Y\} for independent random variables X, Y on atomless spaces). Consequently, integrals over such events cannot be computed using only rectangle-based approximations without passing to the full product \sigma-algebra, underscoring the need for the maximal structure in Fubini's application to joint distributions.

Applications

Iterated Integrals and Product Formulas

Fubini's theorem enables the evaluation of integrals over product measure spaces by expressing the double integral as an iterated integral. For an integrable function f: X \times Y \to \mathbb{R} on \sigma-finite measure spaces (X, \mathcal{A}, \mu) and (Y, \mathcal{B}, \nu), with product measure \mu \times \nu,

\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y).

This equality justifies computing the double integral by successive single-variable integrations, provided f is integrable with respect to the product measure.^[31] Tonelli's theorem extends this result to non-negative measurable functions, ensuring the iterated integrals coincide (possibly infinite) without requiring prior integrability. Specifically, for a non-negative measurable f,

\int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) = \int_{X \times Y} f \, d(\mu \times \nu) = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y).

This facilitates applications where functions may not be integrable but yield finite iterated integrals.^[31] A key application arises for product functions f(x,y) = g(x) h(y), where g: X \to [0,\infty) is \mathcal{A}-measurable and h: Y \to [0,\infty) is \mathcal{B}-measurable. Tonelli's theorem implies

\int_{X \times Y} g(x) h(y) \, d(\mu \times \nu) = \left( \int_X g \, d\mu \right) \left( \int_Y h \, d\nu \right).

The proof sketch proceeds by evaluating the iterated integral: the inner integral \int_Y g(x) h(y) \, d\nu(y) = g(x) \int_Y h \, d\nu, and the outer integral then factors as \left( \int_Y h \, d\nu \right) \int_X g \, d\mu, with equality in the reverse order by symmetry. For general integrable product functions (not necessarily positive), the result follows from Fubini's theorem applied to the positive and negative parts separately.^[32]^[31]

Computational Examples

Fubini's theorem facilitates the evaluation of double integrals by allowing the interchange of the order of integration, often simplifying computations that would be challenging as a single double integral. One classic example is the computation of the normalizing constant for the bivariate normal distribution. Consider the double integral over \mathbb{R}^2:

\iint_{\mathbb{R}^2} e^{-(x^2 + y^2)/2} \, dx \, dy.

By Fubini's theorem, since the integrand is positive and the integral is absolutely convergent, this equals

\int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \right) e^{-y^2/2} \, dy = \left( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \right)^2 = (\sqrt{2\pi})^2 = 2\pi.

The inner Gaussian integral is a standard result, and the squaring step leverages the separability enabled by Fubini. This evaluation is crucial in probability theory for normalizing the standard bivariate normal density.^[33] Another illustrative example is the double integral representation of the dilogarithm function at 1, which equals the Riemann zeta function at 2:

\int_0^1 \int_0^1 \frac{-\ln(1 - xy)}{xy} \, dx \, dy = \frac{\pi^2}{6}.

To compute this using Fubini's theorem, first note the series expansion -\ln(1 - xy) = \sum_{n=1}^\infty \frac{(xy)^n}{n}, so the integrand becomes \sum_{n=1}^\infty \frac{(xy)^{n-1}}{n}. Integrating term by term, justified by Fubini for the positive terms,

\int_0^1 \int_0^1 \sum_{n=1}^\infty \frac{(xy)^{n-1}}{n} \, dx \, dy = \sum_{n=1}^\infty \frac{1}{n} \left( \int_0^1 x^{n-1} \, dx \right) \left( \int_0^1 y^{n-1} \, dy \right) = \sum_{n=1}^\infty \frac{1}{n^2} = \zeta(2) = \frac{\pi^2}{6}.

This approach highlights how Fubini allows separation into iterated integrals, leading to the Basel problem solution.

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