Product Measures, Fubini–Tonelli, and Change of Variables

For two sigma-finite measure spaces, Tonelli’s theorem permits either integration order for a nonnegative product-measurable integrand; the common value may be $+\infty$. Fubini’s theorem permits the same reordering for a real- or complex-valued integrand when its absolute value is integrable over the product space; its sections are then integrable for almost every outer variable. For a $C^1$ diffeomorphism $\Phi:U\to V$ between open subsets of $\mathbb R^n$, Lebesgue measure transforms by the absolute Jacobian $|\det D\Phi|$.

These statements do not authorize rearranging conditionally convergent integrals, replacing global injectivity by a nonzero local determinant, or treating a formal continuum symbol $\mathcal D\phi$ as an infinite product of Lebesgue measures. The QFT application below stays at fixed finite regulator dimension and uses an ordinary matrix determinant.

Required background. Lebesgue Integration and Convergence Theorems provides the integral, monotone convergence, and absolute-integrability criteria used in the proofs.

Product measures · Tonelli · Fubini · Change of variables · Field-variable change · Exercises

Product sigma-algebras and product measures

Let

$$(X,\Sigma,\mu), \qquad (Y,\mathcal T,\nu)$$

be sigma-finite measure spaces. A measurable rectangle is a set $A\times B$ with $A\in\Sigma$ and $B\in\mathcal T$. The product sigma-algebra is

$$\Sigma\otimes\mathcal T = \sigma\!\left( \{A\times B:A\in\Sigma,\ B\in\mathcal T\} \right).$$

It is the smallest sigma-algebra on $X\times Y$ containing all measurable rectangles; see Axler 2020, Definitions 5.2, 5.3, and 5.7 and Theorems 5.6 and 5.9, pp. 117–119, PDF. It need not be the full power set, and a set does not become product-measurable merely because each of its one-variable sections happens to be measurable (Axler 2020, Exercise 5A.3, p. 128, PDF).

Product-measure theorem. There is a unique measure $\mu\otimes\nu$ on $\Sigma\otimes\mathcal T$ such that

$$(\mu\otimes\nu)(A\times B) = \mu(A)\nu(B)$$

for all $A\in\Sigma$ and $B\in\mathcal T$, with $0\cdot\infty=0$. Sigma-finiteness is weaker than finiteness of either total measure; its role is to provide countable covers by finite-measure pieces.

Product-measure construction: proof sketch

Finite disjoint unions of measurable rectangles form an algebra. On that algebra, assign

$$\pi\!\left( \bigsqcup_{j=1}^{m}A_j\times B_j \right) = \sum_{j=1}^{m}\mu(A_j)\nu(B_j).$$

Refining two rectangle decompositions by all pairwise intersections shows that this value is representation-independent. Countable additivity makes $\pi$ a premeasure, and the Carathéodory extension theorem extends it to $\Sigma\otimes\mathcal T$. Sigma-finite covers of $X$ and $Y$ produce a countable cover of $X\times Y$ by rectangles of finite product measure; the sigma-finite uniqueness theorem then makes the extension unique.

This is a proof sketch. An equivalent section-based construction, the rectangle rule, and sigma-finite uniqueness are developed in Axler 2020, Definition 5.25, Example 5.26, Theorem 5.27, and Exercise 5A.10, pp. 126–128, PDF. The Carathéodory construction just sketched is treated structurally in Folland 1999, §2.5, pp. 64–70.

Sections

For $E\subseteq X\times Y$, define

$$E_x=\{y\in Y:(x,y)\in E\}, \qquad E^y=\{x\in X:(x,y)\in E\}.$$

For a function $f:X\times Y\to\mathbb C$, define

$$f_x(y)=f(x,y), \qquad f^y(x)=f(x,y).$$

If $E\in\Sigma\otimes\mathcal T$, then $E_x\in\mathcal T$ for every $x$ and $E^y\in\Sigma$ for every $y$. If $f$ is product-measurable, every $f_x$ and $f^y$ is measurable. The proof is direct: the collections of sets whose sections are measurable are sigma-algebras containing all measurable rectangles.

The stronger section-measure lemma says that

$$x\longmapsto\nu(E_x), \qquad y\longmapsto\mu(E^y)$$

are measurable (Axler 2020, Theorem 5.20, pp. 124–125, PDF). Applying the indicator-function case of Tonelli then gives

$$\begin{aligned} (\mu\otimes\nu)(E) &= \int_X\nu(E_x)\,\mathrm d\mu(x) \\ &= \int_Y\mu(E^y)\,\mathrm d\nu(y). \end{aligned}$$

One proof first restricts to finite factor measures. There, the sets for which the formulas hold form a monotone class containing the algebra of finite unions of rectangles. A monotone-class theorem extends the formulas to the product sigma-algebra. Increasing sigma-finite exhaustions of $X$ and $Y$, followed by monotone convergence, remove the finite-measure restriction. See Axler 2020, Definition 5.25 and Theorem 5.28, especially formula 5.29, pp. 126 and 129–130, PDF.

A triangle checked in both directions

Let $\lambda$ be Lebesgue measure on $[0,1]$ and

$$E=\{(x,y)\in[0,1]^2:0\leq y\leq x\leq1\}.$$

For fixed $x$, the section $E_x=[0,x]$ has length $x$. For fixed $y$, the section $E^y=[y,1]$ has length $1-y$. Hence

$$\begin{aligned} (\lambda\otimes\lambda)(E) &= \int_0^1x\,\mathrm dx \\ &= \int_0^1(1-y)\,\mathrm dy = \frac12. \end{aligned}$$

The two section functions are different; their integrals agree because they measure the same product-measurable set.

Euclidean products and completion

For Euclidean Borel sigma-algebras,

$$\mathcal B(\mathbb R^m)\otimes\mathcal B(\mathbb R^n) = \mathcal B(\mathbb R^{m+n}).$$

The product of the Borel restrictions of $\lambda^m$ and $\lambda^n$ is the Borel restriction of $\lambda^{m+n}$ (Axler 2020, Theorem 5.39, Definition 5.40, and the following discussion, pp. 138–140, PDF). Completing that product gives ordinary Lebesgue measure on $\mathbb R^{m+n}$ (Tao 2011, Example 1.7.13, p. 199, PDF) and supports the notation

$$\mathrm d^{m+n}(x,y) = \mathrm d^m x\,\mathrm d^n y.$$

Completion changes one point of wording. Even if both factor measures are complete, their product on $\Sigma\otimes\mathcal T$ need not be complete. For example, if $N\subset\mathbb R$ is non-Lebesgue-measurable, then

$$\{0\}\times N \subset \{0\}\times\mathbb R$$

is a subset of a product-null set. It is measurable in the completed product, but its section at $x=0$ is $N$. Thus for completed-product representatives, section measurability and the resulting integral formulas are asserted almost everywhere, after choosing a product-measurable representative; they need not hold at every exceptional point. See Tao 2011, Example 1.7.13, p. 199, and Theorem 1.7.18 and Corollary 1.7.19, pp. 202–203, PDF.

Tonelli’s theorem for nonnegative integrands

Tonelli’s theorem. Let $(X,\Sigma,\mu)$ and $(Y,\mathcal T,\nu)$ be sigma-finite measure spaces. If

$$f:X\times Y\longrightarrow[0,\infty]$$

is $\Sigma\otimes\mathcal T$-measurable, then the functions

$$x\longmapsto\int_Yf(x,y)\,\mathrm d\nu(y), \qquad y\longmapsto\int_Xf(x,y)\,\mathrm d\mu(x)$$

are measurable, and

$$\begin{aligned} \int_{X\times Y}f\,\mathrm d(\mu\otimes\nu) &= \int_X \left( \int_Yf(x,y)\,\mathrm d\nu(y) \right) \mathrm d\mu(x) \\ &= \int_Y \left( \int_Xf(x,y)\,\mathrm d\mu(x) \right) \mathrm d\nu(y). \end{aligned}$$

All three values lie in $[0,\infty]$ and may equal $+\infty$. Tonelli needs no prior integrability assumption and does not prove that the answer is finite. The exact statement and proof appear in Axler 2020, Theorem 5.28, pp. 129–130, PDF.

Proof

For an indicator $f=\mathbf1_E$, the theorem is exactly the section-measure lemma:

$$\int_Y\mathbf1_E(x,y)\,\mathrm d\nu(y) = \nu(E_x).$$

Finite nonnegative linear combinations of indicators give the result for nonnegative simple functions. Now choose product-measurable simple functions

$$0\leq s_1\leq s_2\leq\cdots\uparrow f.$$

For every fixed $x$, monotone convergence on $Y$ gives

$$\int_Ys_k(x,y)\,\mathrm d\nu(y) \uparrow \int_Yf(x,y)\,\mathrm d\nu(y).$$

The left side is a sequence of measurable functions of $x$, so its limit is measurable. Applying monotone convergence once more, now on $X$, yields

$$\begin{aligned} \int_X \left( \int_Yf(x,y)\,\mathrm d\nu(y) \right) \mathrm d\mu(x) &= \lim_{k\to\infty} \int_X \left( \int_Ys_k(x,y)\,\mathrm d\nu(y) \right) \mathrm d\mu(x) \\ &= \lim_{k\to\infty} \int_{X\times Y}s_k\,\mathrm d(\mu\otimes\nu) \\ &= \int_{X\times Y}f\,\mathrm d(\mu\otimes\nu). \end{aligned}$$

Interchanging $X$ and $Y$ gives the other order. This completes the proof.

Why sigma-finiteness appears

Let $X=Y=[0,1]$ with their Borel sigma-algebras. Put counting measure $\#$ on $X$ and Lebesgue measure $\lambda$ on $Y$. Counting measure on an uncountable set is not sigma-finite: its finite-measure sets are finite, and a countable union of finite sets cannot cover $[0,1]$.

For the diagonal

$$\Delta=\{(x,y)\in[0,1]^2:x=y\},$$

each $Y$-section is a Lebesgue-null singleton, whereas each $X$-section has counting measure one. Consequently,

$$\int_X \left( \int_Y\mathbf1_\Delta(x,y)\,\mathrm d\lambda(y) \right) \mathrm d\#(x) =0,$$

but

$$\int_Y \left( \int_X\mathbf1_\Delta(x,y)\,\mathrm d\#(x) \right) \mathrm d\lambda(y) =1.$$

The two nonnegative iterated integrals disagree. Thus the clean symmetric Tonelli theorem above cannot simply discard sigma-finiteness. Specialized extensions outside the sigma-finite setting require additional hypotheses; the example does not claim that every such extension fails. See Axler 2020, Example 5.30, p. 131, PDF and Tao 2011, Exercise 1.7.22, pp. 203–204, author-hosted preliminary PDF.

Fubini’s theorem for absolutely integrable integrands

Fubini’s theorem. Let the two factor spaces be sigma-finite, and let

$$f:X\times Y\longrightarrow\mathbb C$$

be product-measurable. If

$$\int_{X\times Y}|f|\,\mathrm d(\mu\otimes\nu)<\infty,$$

then

$$f_x\in L^1(\nu) \quad\text{for }\mu\text{-almost every }x,$$

and

$$f^y\in L^1(\mu) \quad\text{for }\nu\text{-almost every }y.$$

Define each inner integral to be zero on its exceptional null set. The resulting functions of the outer variable are measurable and integrable, and

$$\begin{aligned} \int_{X\times Y}f\,\mathrm d(\mu\otimes\nu) &= \int_X \left( \int_Yf(x,y)\,\mathrm d\nu(y) \right) \mathrm d\mu(x) \\ &= \int_Y \left( \int_Xf(x,y)\,\mathrm d\mu(x) \right) \mathrm d\nu(y). \end{aligned}$$

All three integrals are finite. This complex-valued form includes the real case. See Axler 2020, Theorem 5.32, pp. 132–133, PDF and Tao 2011, Theorem 1.7.21, pp. 204–205, author-hosted preliminary PDF.

Proof

Apply Tonelli to the nonnegative measurable function $|f|$. The section norm

$$A(x)=\int_Y|f(x,y)|\,\mathrm d\nu(y)$$

is measurable and satisfies

$$\int_XA(x)\,\mathrm d\mu(x) = \int_{X\times Y}|f|\,\mathrm d(\mu\otimes\nu) < \infty.$$

An integrable nonnegative function is finite almost everywhere, so $f_x\in L^1(\nu)$ for almost every $x$. Moreover, on the good set,

$$\left| \int_Yf(x,y)\,\mathrm d\nu(y) \right| \leq A(x),$$

which proves integrability in the outer variable after the exceptional values are set to zero.

For real $f$, apply Tonelli separately to $f^+$ and $f^-$. Their product integrals and almost all section integrals are finite because both are bounded by $|f|$. Subtracting the two finite equalities proves Fubini in the first order. Repeating with $X$ and $Y$ exchanged proves the second order. For complex $f$, apply the real result to its real and imaginary parts. This completes the proof.

A practical test

Sometimes the product integral of $|f|$ is not known in advance, but one iterated absolute integral is. If, for example,

$$\int_X \left( \int_Y|f(x,y)|\,\mathrm d\nu(y) \right) \mathrm d\mu(x) < \infty,$$

then Tonelli applied to $|f|$ first proves $f\in L^1(\mu\otimes\nu)$. Fubini may then be applied to $f$. The same criterion works with the variables reversed.

Available information	Legal result	Conclusion
$f\geq0$, product-measurable	Tonelli	Either order; $+\infty$ allowed
$\int\lvert f\rvert\,\mathrm d(\mu\otimes\nu)<\infty$	Fubini	Finite equality; sections integrable a.e.
One iterated integral of $\lvert f\rvert$ is finite	Tonelli, then Fubini	Absolute product integrability and reordering
Only conditional or improper convergence	No general license	A regulator or separate argument is required
Factors are not sigma-finite	Standard symmetric theorem unavailable	Additional structure is required

Conditional summation can depend on order

Take $X=Y=\mathbb N$ with counting measure and define

$$a_{mn} = \mathbf1_{\{n=m\}} - \mathbf1_{\{n=m+1\}}.$$

For each fixed $m$, the row contains one $+1$ and one $-1$, so

$$\sum_{m=1}^{\infty} \left( \sum_{n=1}^{\infty}a_{mn} \right) =0.$$

The first column contains one $+1$, while every later column contains a $+1$ and a $-1$. Therefore

$$\sum_{n=1}^{\infty} \left( \sum_{m=1}^{\infty}a_{mn} \right) =1.$$

Every inner sum is finite, but

$$\sum_{m,n}|a_{mn}|=\infty.$$

The positive and negative parts both have infinite total mass, so the signed product integral is undefined as $\infty-\infty$. Mere existence of the two iterated sums does not supply Fubini’s absolute-integrability hypothesis. This discrete example is discussed in Tao 2011, Remark 0.0.3, pp. xiv–xv, author-hosted preliminary PDF.

“Almost every section” is sharp

On $\mathbb R^2$, let

$$f=\mathbf1_{\{0\}\times\mathbb R}.$$

The vertical line is product-null, so $f\in L^1(\mathbb R^2)$ and its integral is zero. For every $x\neq0$, the section $f_x$ vanishes. At $x=0$, however, $f_0(y)=1$ on all of $\mathbb R$ and is not integrable. Fubini promises integrable sections almost everywhere, not at every outer point.

Change of variables and transformed densities

A change of variables has two layers. The first is measure-theoretic and requires no derivative. If $\Phi:U\to V$ is measurable and $\mu$ is a measure on $U$, its pushforward satisfies

$$\int_Vh(y)\,\mathrm d(\Phi_\#\mu)(y) = \int_Uh(\Phi(x))\,\mathrm d\mu(x)$$

for nonnegative measurable $h$, and for every integrable real- or complex-valued $h$. This is the integration form of the pushforward definition.

A Jacobian appears only in the second layer, when the pushed-forward measure is compared with a chosen reference measure such as Lebesgue measure.

Euclidean change-of-variables theorem. Let $U,V\subseteq\mathbb R^n$ be open, and let

$$\Phi:U\longrightarrow V$$

be a $C^1$ diffeomorphism: it is bijective and both $\Phi$ and $\Phi^{-1}$ are continuously differentiable. Define

$$J_\Phi(x)=|\det D\Phi(x)|.$$

For every nonnegative measurable $h:V\to[0,\infty]$,

$$\int_Vh(y)\,\mathrm d^n y = \int_Uh(\Phi(x))J_\Phi(x)\,\mathrm d^n x$$

as an equality in $[0,\infty]$. The same formula holds for an integrable real- or complex-valued $h$, and the transformed integrand is then integrable. A precise $L^1$ statement appears in Dyatlov 2022, §10.1.3, Theorem 10.5, p. 108, PDF; the nonnegative and integrable forms are treated in Folland 1999, Theorem 2.47, p. 76.

For a Borel set $E\subseteq U$, the set form is

$$\lambda^n(\Phi(E)) = \int_EJ_\Phi(x)\,\mathrm d^n x.$$

The absolute value belongs to ordinary measure transformation. Oriented differential forms retain an orientation sign and obey a related but different theorem.

Affine maps: a complete check

Let

$$\Phi(x)=Ax+b, \qquad A\in GL(n,\mathbb R).$$

Translations preserve Lebesgue measure. Write a singular-value decomposition

$$A=Q_1DQ_2, \qquad D=\operatorname{diag}(\sigma_1,\ldots,\sigma_n), \qquad \sigma_j>0,$$

with $Q_1,Q_2$ orthogonal. Orthogonal maps preserve Euclidean volume. Successive one-dimensional substitutions for the diagonal map give

$$\mathrm d^n(Dx) = \left(\prod_{j=1}^n\sigma_j\right)\mathrm d^n x.$$

Because

$$\prod_{j=1}^n\sigma_j=|\det A|,$$

the volume-scaling formula holds for rectangles. To extend it without assuming that arbitrary measurable sets are finite unions of rectangles, define two Borel measures on the domain by

$$\alpha(E)=\lambda^n(AE+b), \qquad \beta(E)=|\det A|\,\lambda^n(E).$$

They agree on the $\pi$-system of half-open rectangles and are sigma-finite, so uniqueness of sigma-finite measures gives $\alpha=\beta$ on every Borel set. An invertible affine map and its inverse send null sets to null sets, so the same equality holds after Lebesgue completion. Taking $E=\Phi^{-1}(B)$ proves the substitution formula for indicators of arbitrary Lebesgue-measurable sets $B$. Finite linear combinations now give all nonnegative simple functions, and monotone convergence gives every nonnegative measurable $h$; positive and negative parts or real and imaginary parts give the integrable case. Thus

$$\int_{\mathbb R^n}h(y)\,\mathrm d^n y = \int_{\mathbb R^n}h(Ax+b)|\det A|\,\mathrm d^n x.$$

This proves the affine case and fixes both the determinant direction and its absolute value.

General diffeomorphisms: proof sketch

For a general $C^1$ diffeomorphism, the derivative gives the first-order approximation

$$\Phi(x+\delta x) = \Phi(x)+D\Phi(x)\,\delta x+o(\|\delta x\|).$$

On sufficiently small cells, the affine result therefore controls volume distortion by $|\det D\Phi(x)|$, with errors made uniform on compact subsets. A disjoint covering argument sums those local estimates; simple approximation treats nonnegative functions, and an increasing compact exhaustion handles noncompact open sets. The details needed to control the covering and limiting errors are not reproduced here; the cited Folland theorem gives a full proof.

Densities under pushforward

Suppose a measure on $U$ has density

$$\mathrm d\mu(x)=\rho(x)\,\mathrm d^n x, \qquad \rho\geq0.$$

Apply change of variables to the pushforward identity. For every nonnegative test function $h$,

$$\begin{aligned} \int_Vh(y)\,\mathrm d(\Phi_\#\mu)(y) &= \int_Uh(\Phi(x))\rho(x)\,\mathrm d^n x \\ &= \int_V h(y)\, \rho(\Phi^{-1}(y)) |\det D\Phi^{-1}(y)| \,\mathrm d^n y. \end{aligned}$$

The Radon–Nikodym derivative is therefore

$$\frac{\mathrm d(\Phi_\#\mu)}{\mathrm d\lambda^n}(y) = \rho(\Phi^{-1}(y)) |\det D\Phi^{-1}(y)|.$$

This inverse determinant describes the density of the pushed-forward measure as a function of the new point $y$. By contrast, when the old variable is written as $y=\Phi(x)$ inside an integral, the substitution formula contains the forward factor $|\det D\Phi(x)|$. Stating the map direction prevents the two formulas from being confused.

Orientation and polar-coordinate checks

In one dimension, $\Phi(x)=-x$ has derivative $-1$ but $J_\Phi=1$. Reflection preserves positive Lebesgue measure. Omitting the absolute value would turn a positive integral into its negative.

For polar coordinates, take

$$\Phi(r,\theta) = (r\cos\theta,r\sin\theta)$$

on

$$(0,\infty)\times(0,2\pi).$$

This is a diffeomorphism onto the plane with the nonnegative horizontal ray removed. The omitted ray is Lebesgue-null, and

$$|\det D\Phi(r,\theta)|=r.$$

Thus, for nonnegative measurable or integrable $h$,

$$\int_{\mathbb R^2}h(y)\,\mathrm d^2y = \int_0^\infty \int_0^{2\pi} h(r\cos\theta,r\sin\theta)\, r\,\mathrm d\theta\,\mathrm dr.$$

The origin and angular seam cannot be hidden inside a claim that polar coordinates are one global diffeomorphism; they are harmless here because the deleted set is null.

Why a nonzero local determinant is not enough

Let

$$U=(-1,0)\cup(0,1), \qquad V=(0,1), \qquad \Phi(x)=x^2.$$

The derivative $2x$ is nonzero throughout $U$, but $\Phi$ is two-to-one. Applying one-dimensional substitution separately on the two branches gives

$$\int_Uh(x^2)\,2|x|\,\mathrm dx = 2\int_Vh(y)\,\mathrm dy.$$

The diffeomorphism formula would overcount the target if applied to the whole two-sheeted map. Noninjective area formulas include multiplicity or require a partition into injective branches.

A finite regulated field-variable change

The product and substitution theorems have a rigorous QFT-facing use before any continuum measure is introduced. Fix

$$N=n+m<\infty, \qquad q=(u,v)\in\mathbb R^n\times\mathbb R^m,$$

and let $S_E$ be a measurable real Euclidean action. The positive weight

$$e^{-S_E(u,v)}$$

is measurable on the finite-dimensional product space.

Reordering a finite split

For

$$Z_N = \int_{\mathbb R^{n+m}} e^{-S_E(u,v)} \,\mathrm d^n u\,\mathrm d^m v,$$

Tonelli permits either integration order:

$$\begin{aligned} Z_N &= \int_{\mathbb R^n} \left( \int_{\mathbb R^m} e^{-S_E(u,v)}\,\mathrm d^m v \right) \mathrm d^n u \\ &= \int_{\mathbb R^m} \left( \int_{\mathbb R^n} e^{-S_E(u,v)}\,\mathrm d^n u \right) \mathrm d^m v. \end{aligned}$$

This equality remains true if all three values are infinite. A normalized measure or expectation requires the separate condition

$$0<Z_N<\infty.$$

For a complex observable $\mathcal O$, Fubini requires

$$\int_{\mathbb R^{n+m}} |\mathcal O(u,v)|e^{-S_E(u,v)} \,\mathrm d^n u\,\mathrm d^m v < \infty$$

before the order in its numerator is changed.

An explicit nonlinear Jacobian

Now write the old variable $q$ in terms of a new variable $\chi\in\mathbb R^N$ by

$$q_i=\Phi_i(\chi) = \chi_i+\lambda\chi_i^3, \qquad i=1,\ldots,N, \qquad \lambda>0.$$

If the field coordinates carry units, $\lambda$ has inverse-square field units so that $\lambda\chi_i^2$ is dimensionless. Each component is strictly increasing because

$$\frac{\partial\Phi_i}{\partial\chi_i} = 1+3\lambda\chi_i^2>0,$$

and it tends to $\pm\infty$ as $\chi_i\to\pm\infty$. Hence $\Phi$ is a global smooth diffeomorphism of $\mathbb R^N$. Its derivative is diagonal, so

$$J_\Phi(\chi) = \det D\Phi(\chi) = \prod_{i=1}^{N} \left(1+3\lambda\chi_i^2\right) >0.$$

The finite-dimensional change-of-variables theorem gives the exact identity

$$\begin{aligned} Z_N &= \int_{\mathbb R^N} e^{-S_E(q)}\,\mathrm d^N q \\ &= \int_{\mathbb R^N} e^{-S_E(\Phi(\chi))} \prod_{i=1}^{N} \left(1+3\lambda\chi_i^2\right) \,\mathrm d^N \chi. \end{aligned}$$

If the positive determinant is absorbed into the exponent, the transformed action is

$$\begin{aligned} S'_E(\chi) &= S_E(\Phi(\chi)) - \log J_\Phi(\chi) \\ &= S_E(\Phi(\chi)) - \sum_{i=1}^{N} \log\!\left(1+3\lambda\chi_i^2\right). \end{aligned}$$

The minus sign is forced by

$$e^{-S_E(\Phi)}J_\Phi = e^{-[S_E(\Phi)-\log J_\Phi]}.$$

The formal continuum change-of-field template, including an infinitesimal functional determinant, is exhibited in Zinn-Justin 2021, §7.5.3, pp. 136–137. That source motivates the continuum notation; the equality just derived is a finite-dimensional theorem and does not validate the formal continuum determinant.

Gaussian normalization checks the direction

Let $K$ be real symmetric positive-definite and take an invertible linear map

$$q=A\chi.$$

Then

$$q^{\mathsf T}Kq = \chi^{\mathsf T}A^{\mathsf T}KA\chi, \qquad \mathrm d^N q = |\det A|\,\mathrm d^N \chi.$$

The transformed Gaussian formula gives

$$\begin{aligned} |\det A|\, \frac{(2\pi)^{N/2}} {\sqrt{\det(A^{\mathsf T}KA)}} &= |\det A|\, \frac{(2\pi)^{N/2}} {|\det A|\sqrt{\det K}} \\ &= \frac{(2\pi)^{N/2}}{\sqrt{\det K}}, \end{aligned}$$

because

$$\det(A^{\mathsf T}KA) = (\det A)^2\det K.$$

This recovers the original normalization and independently checks the forward factor $|\det A|$. After an orthogonal diagonalization of $K$, Tonelli also justifies factorizing the nonnegative Gaussian into $N$ one-dimensional integrals.

Every conclusion in this example has a finite-regulator ceiling:

• $N$ is fixed and finite, and the base measure is ordinary Lebesgue measure on $\mathbb R^N$.
• If the original integral is restricted to a region $\Omega$, the new region is $\Phi^{-1}(\Omega)$. Keeping the old region silently changes the integral.
• Tonelli uses positivity of the Euclidean weight. A Lorentzian weight $e^{iS}$ is generally neither nonnegative nor absolutely integrable.
• The determinant is an ordinary $N\times N$ determinant. No infinite product, continuum field measure, regulator removal, anomaly, equivalence theorem, or renormalized statement follows.
• Bosonic Lebesgue substitution does not give the Berezinian rule for odd variables.

For the developed finite-regulator treatment, continue to Changes of Variables and Regulated Jacobians.

Common pitfalls

Separate measurability is not joint measurability. Measurability of every one-variable section does not by itself prove $\Sigma\otimes\mathcal T$-measurability of the original function.

Complete factors need not have a complete product. State whether the raw product sigma-algebra or its completion is in use. In the completion, section claims may have exceptional outer points.

Tonelli is for nonnegative integrands. It allows $+\infty$ and does not turn a signed $\infty-\infty$ expression into an integral.

Fubini needs absolute integrability. Two conditionally convergent iterated integrals can both exist and still disagree.

Almost every section does not mean every section. Altering an integrable function on a product-null set can make a particular exceptional section nonintegrable.

A Jacobian needs a map direction. The substitution $y=\Phi(x)$ uses $|\det D\Phi(x)|$; the density of the pushed-forward measure at $y$ uses $|\det D\Phi^{-1}(y)|$.

Ordinary measures use the absolute determinant. A signed determinant tracks orientation for differential forms, not positive Lebesgue volume.

Local invertibility is not global injectivity. A nonzero determinant does not prevent multiple preimages. A branch decomposition or multiplicity formula may be required.

The domain transforms too. A change of variables applies to the integrand, measure, and integration region or cycle together.

Finite and functional determinants are different objects. A formal $\det(\delta\phi/\delta\chi)$ requires a regulator and definition; anomalous measure variation is not a consequence of the finite theorem.

Where the next treatments begin

• $L^p$ Spaces, Inequalities, and Weak Convergence develops norm, duality, and weak-convergence control beyond the $L^1$ criterion used here.
• Changes of Variables and Regulated Jacobians develops finite-regulator field redefinitions and determinant bookkeeping.
• Anatomy of a Loop Integral treats loop measures, regulators, and the physical meaning of loop integration.
• Regulated Jacobians and Measure Variation treats anomaly-facing functional Jacobians and Ward identities.
• Graded Algebra, Grassmann Variables, and Berezin Integration develops odd-variable substitutions and inverse determinant behavior.
• Differential Forms, Integration, and Stokes’ Theorem develops oriented integration on manifolds.
• Foundations of Quantum Field Theory supplies the broader finite-regulator physical setting.

Exercises

Tonelli on a triangle. Let

$$E=\{(x,y)\in[0,1]^2:0\leq y\leq x\leq1\}.$$

Compute the integral of $\mathbf1_E$ in both orders.

Solution

For fixed $x$, the allowed $y$-interval is $[0,x]$, so its length is $x$. For fixed $y$, the allowed $x$-interval is $[y,1]$, so its length is $1-y$. Tonelli gives

$$\int_0^1\int_0^1 \mathbf1_E(x,y)\,\mathrm dy\,\mathrm dx = \int_0^1x\,\mathrm dx = \frac12$$

and

$$\int_0^1\int_0^1 \mathbf1_E(x,y)\,\mathrm dx\,\mathrm dy = \int_0^1(1-y)\,\mathrm dy = \frac12.$$

Diagnose a double series. For

$$a_{mn} = \mathbf1_{\{n=m\}} - \mathbf1_{\{n=m+1\}},$$

compute the two iterated sums and identify the missing Fubini hypothesis.

Solution

Every row sums to zero, so summing rows first gives zero. The first column sums to one and every later column sums to zero, so summing columns first gives one. Absolute integrability fails because

$$\sum_{m,n}|a_{mn}|=2\sum_{m=1}^{\infty}1=\infty.$$

Thus Fubini does not apply, and the signed product integral would require the undefined subtraction $\infty-\infty$.

Transform a one-dimensional density. Let $Y=aX+b$ with $a\neq0$, and suppose $X$ has density $\rho_X$. Find the density of $Y$ and verify its normalization.

Solution

The inverse map is

$$x=\frac{y-b}{a}, \qquad \left|\frac{\mathrm dx}{\mathrm dy}\right| = \frac1{|a|}.$$

Therefore

$$\rho_Y(y) = \frac1{|a|} \rho_X\!\left(\frac{y-b}{a}\right).$$

One more change of variables checks

$$\int_{\mathbb R}\rho_Y(y)\,\mathrm dy = \int_{\mathbb R}\rho_X(x)\,\mathrm dx.$$

Thus a probability density remains normalized.

Finite Gaussian variable change. Let $q=A\chi$ with $A\in GL(N,\mathbb R)$ and $K>0$. Transform the Gaussian quadratic form and measure, then verify that its normalization is unchanged.

Solution

The transformed data are

$$q^{\mathsf T}Kq = \chi^{\mathsf T}A^{\mathsf T}KA\chi, \qquad \mathrm d^N q = |\det A|\,\mathrm d^N \chi.$$

Using

$$\det(A^{\mathsf T}KA) = (\det A)^2\det K,$$

the transformed normalization is

$$|\det A| \frac{(2\pi)^{N/2}} {\sqrt{\det(A^{\mathsf T}KA)}} = \frac{(2\pi)^{N/2}}{\sqrt{\det K}}.$$

The equality is finite-dimensional. It does not define an infinite-dimensional functional determinant.

References

• Sheldon Axler, Measure, Integration & Real Analysis, Graduate Texts in Mathematics 282, Springer (2020), Axler 2020, Chapter 5, pp. 117–140, author-hosted open-access PDF. Product sigma-algebras and measures, measurable sections, Tonelli, Fubini, and the non-sigma-finite diagonal counterexample.
• Semyon Dyatlov, Lecture Notes for 18.155: Distributions, Elliptic Regularity, and Applications to PDEs, MIT (2022), Dyatlov 2022, §10.1.3, Theorem 10.5, p. 108, PDF. Exact $C^1$-diffeomorphism and $L^1$ change-of-variables statement.
• Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Wiley (1999), Folland 1999, §§2.5–2.6, pp. 64–76, publisher record. Structural reference for product measures, Fubini–Tonelli, Euclidean integration, and change of variables.
• Terence Tao, An Introduction to Measure Theory, Graduate Studies in Mathematics 126, American Mathematical Society (2011), Tao 2011, Remark 0.0.3, pp. xiv–xv, and §1.7, pp. 179–206, author-hosted preliminary PDF. Product measures, Tonelli, Fubini, double-series rearrangement, and non-sigma-finite qualifications.
• Jean Zinn-Justin, Quantum Field Theory and Critical Phenomena, 5th ed., Oxford University Press (2021), Zinn-Justin 2021, Chapter 7, §7.5.3, pp. 136–137, OUP. Formal continuum change-of-field template used only to motivate the finite-dimensional comparison.