Proof of The Integral over the Unit Cell of a Product of One-Variable Functions

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here.

Conventions used throughout. Transitivity of $\le$ on $\mathbb{R}$ is used freely: if $a\le b$ and $b\le c$ then $0\le b-a$ and $0\le c-b$ by claim 3 of Elementary Arithmetic in an Ordered Field, hence $0\le(b-a)+(c-b)=c-a$ by claim 2 of that lemma and the field axioms, hence $a\le c$ by claim 3 again. Each induction below is over the natural numbers, bounded by a natural number $r$, and is carried out by applying Principle of Induction for the Natural Numbers to the set $T$ of natural numbers $j$ such that either $r<j$, or $j\le r$ and the asserted property holds at $j$; by claim 3 of Properties of the Order on the Natural Numbers every natural number satisfies exactly one of $j\le r$ and $r<j$, so $T=\mathbb{N}$ gives the property at every $j\in[r]$. By Finite Product Notation in a Field a finite product $\prod_{k=1}^{j}a_{k}$ is determined by the map $k\mapsto a_{k}$ on the initial segment it is formed over, so two such products agree as soon as the two maps agree; this is used below without further comment. In the successor step one may assume $S(j)\le r$, and then $j\le r$: indeed $j<S(j)$ by claim 5 of Properties of the Order on the Natural Numbers, so either $S(j)=r$ and $j<r$, or $S(j)<r$ and $j<r$ by the transitivity of the strict order in claim 1 of that lemma; in both cases $j\le r$ by claim 1 of that lemma.

Notation. For a natural number $l$ with $1\le l$ we write $\mathcal{B}_{l}$ for the $\sigma$-algebra on $\mathbb{R}^{l}$ and $\lambda_{l}$ for the measure on it built in claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, together with the identification of $\mathbb{R}^{l}$ with $\mathbb{R}^{l-1}\times\mathbb{R}$ made there for $l\ge2$; in particular $\mathcal{B}_{1}=\mathcal{B}(\mathbb{R})$ and $\lambda_{1}=\lambda$ under the identification of $\mathbb{R}^{1}$ with $\mathbb{R}$. By Lebesgue Measure on $\mathbb{R}^n$, $\mathcal{B}_{n}=\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}$ is Lebesgue measure on $\mathbb{R}^{n}$; consequently $(Q,\mathcal{B}_{Q},\lambda_{Q})$ is, as recorded in The Flat Torus: Standing Notation (iii), the restriction of $(\mathbb{R}^{n},\mathcal{B}_{n},\lambda_{n})$ to $Q$ in the sense of claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions.

For $i\in[n]$ let $\hat{g}_{i}:\mathbb{R}\to\mathbb{R}$ be the map equal to $g_{i}$ on $J$ and to $0$ at every point of $\mathbb{R}$ outside $J$. For natural numbers $l,j$ with $1\le l\le n$ and $j\in[l]$ define

$$P_{l,j}:\mathbb{R}^{l}\to\mathbb{R},\qquad P_{l,j}(\theta)=\prod_{k=1}^{j}\hat{g}_{k}(\theta_{k}),$$

the product being that of Finite Product Notation in a Field formed for the map $k\mapsto\hat{g}_{k}(\theta_{k})$ on $[j]$, and put $H_{l}=P_{l,l}$. For $k\in[l]$ let $\pi_{k}:\mathbb{R}^{l}\to\mathbb{R}$ be the $k$th coordinate projection, $\pi_{k}(\theta)=\theta_{k}$.

Claim 1. (Zero extension of a real-valued function.) Let $(X,\mathcal{F},\mu)$ be a measure space, let $X_{0}\in\mathcal{F}$, and let $(X_{0},\mathcal{F}|_{X_{0}},\mu|_{X_{0}})$ be its restriction as in claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions. Let $f:X_{0}\to\mathbb{R}$ and let $\tilde{f}:X\to\mathbb{R}$ be the map equal to $f$ on $X_{0}$ and to $0$ off $X_{0}$. Then $f$ is measurable with respect to $\mathcal{F}|_{X_{0}}$ if and only if $\tilde{f}$ is measurable with respect to $\mathcal{F}$; and in that case $f$ is integrable with respect to $\mu|_{X_{0}}$ if and only if $\tilde{f}$ is integrable with respect to $\mu$, and then

$$\int_{X_{0}}f\,d\mu|_{X_{0}}=\int_{X}\tilde{f}\,d\mu .$$

Proof of Claim 1. Measurability of a real-valued function is that of Measurable Map, and the Set of Measurable Maps: the preimage of every member of $\mathcal{B}(\mathbb{R})$ lies in the $\sigma$-algebra. Let $B\in\mathcal{B}(\mathbb{R})$. If $0\notin B$ then $\tilde{f}^{-1}(B)=f^{-1}(B)$, and if $0\in B$ then $\tilde{f}^{-1}(B)=f^{-1}(B)\cup(X\setminus X_{0})$, since $\tilde{f}$ takes the value $0$ at every point off $X_{0}$ and agrees with $f$ on $X_{0}$. By claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions every member of $\mathcal{F}|_{X_{0}}$ lies in $\mathcal{F}$, and $X\setminus X_{0}\in\mathcal{F}$ because $\mathcal{F}$ is a $\sigma$-algebra; so measurability of $f$ gives measurability of $\tilde{f}$ in both cases. Conversely $f^{-1}(B)=\tilde{f}^{-1}(B)\cap X_{0}$, which lies in $\mathcal{F}$ and is contained in $X_{0}$, hence lies in $\mathcal{F}|_{X_{0}}$; so measurability of $\tilde{f}$ gives measurability of $f$.

Assume both are measurable. Let $f^{+}$ and $f^{-}$ be the positive and negative parts of $f$ as in Integrable Functions and Their Lebesgue Integral, and likewise $(\tilde{f})^{+}$ and $(\tilde{f})^{-}$ for $\tilde{f}$. Since $\max\{0,0\}=0$, the map $(\tilde{f})^{+}$ is the zero extension of $f^{+}$ and $(\tilde{f})^{-}$ is the zero extension of $f^{-}$. These four maps are nonnegative and real-valued, and they are measurable by Integrable Functions and Their Lebesgue Integral; for a nonnegative real-valued function, measurability in the sense of Measurable Map, and the Set of Measurable Maps and measurability as a $[0,\infty]$-valued function agree, as recorded in Nonnegative Measurable Maps and Their Lebesgue Integral. Claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, applied to $f^{+}$ and then to $f^{-}$, therefore gives

$$\int_{X_{0}}f^{+}\,d\mu|_{X_{0}}=\int_{X}(\tilde{f})^{+}\,d\mu,\qquad \int_{X_{0}}f^{-}\,d\mu|_{X_{0}}=\int_{X}(\tilde{f})^{-}\,d\mu .$$

By Integrable Functions and Their Lebesgue Integral, $f$ is integrable exactly when the two left-hand sides are finite, $\tilde{f}$ is integrable exactly when the two right-hand sides are finite, and in that case each of the two integrals in the claim is the difference of the corresponding pair of values. The two differences agree by the two displayed identities. This proves Claim 1.

Claim 2. (Adjoining a one-point factor.) Let $Y=\{z\}$ be a one-point set and let $(Y,\mathcal{G},\delta)$, with $\mathcal{G}=\{\emptyset,Y\}$, be the one-point measure space with unit mass at $z$ of claim 3 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions. Then $\delta$ is $\sigma$-finite. Let $l$ be a natural number with $1\le l$ and let $\varphi_{l}:\mathbb{R}^{l}\to\mathbb{R}^{l}\times Y$ be given by $\varphi_{l}(\theta)=(\theta,z)$. Then $\varphi_{l}$ is a bijection, the product $\sigma$-algebra satisfies

$$\mathcal{B}_{l}\otimes\mathcal{G}=\{A\times Y:A\in\mathcal{B}_{l}\},$$

and the product measure $\lambda_{l}\otimes\delta$ is the image measure of $\lambda_{l}$ under $\varphi_{l}$. Moreover, for a map $F:\mathbb{R}^{l}\to\mathbb{R}$, the map $F\circ\varphi_{l}^{-1}$ is measurable with respect to $\mathcal{B}_{l}\otimes\mathcal{G}$ if and only if $F$ is measurable with respect to $\mathcal{B}_{l}$; in that case $F\circ\varphi_{l}^{-1}$ is integrable with respect to $\lambda_{l}\otimes\delta$ if and only if $F$ is integrable with respect to $\lambda_{l}$, and then

$$\int_{\mathbb{R}^{l}\times Y}F\circ\varphi_{l}^{-1}\,d(\lambda_{l}\otimes\delta)=\int_{\mathbb{R}^{l}}F\,d\lambda_{l}.$$

Proof of Claim 2. The constant sequence with every term $Y$ covers $Y$ by members of $\mathcal{G}$ of finite measure, since $\delta(Y)=1$; so $\delta$ is $\sigma$-finite in the sense of Measure, Measure Space, and Probability Measure. Because $Y$ has exactly one element, $\varphi_{l}$ is a bijection with inverse $(\theta,z)\mapsto\theta$, and $\varphi_{l}(A)=A\times Y$ for every $A\subseteq\mathbb{R}^{l}$.

By claim 2 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, applied to the measure space $(\mathbb{R}^{l},\mathcal{B}_{l},\lambda_{l})$ and to the bijection $\varphi_{l}$, the family $\varphi_{l}(\mathcal{B}_{l})=\{A\times Y:A\in\mathcal{B}_{l}\}$ is a $\sigma$-algebra on $\mathbb{R}^{l}\times Y$, the maps $\varphi_{l}$ and $\varphi_{l}^{-1}$ are measurable in the two directions between $(\mathbb{R}^{l},\mathcal{B}_{l})$ and $(\mathbb{R}^{l}\times Y,\varphi_{l}(\mathcal{B}_{l}))$, and the image measure $(\lambda_{l})_{\varphi_{l}}$ satisfies $(\lambda_{l})_{\varphi_{l}}(A\times Y)=\lambda_{l}(A)$ for every $A\in\mathcal{B}_{l}$.

The measurable rectangles of Measurable Rectangles and the Product σ-Algebra for $\mathcal{B}_{l}$ and $\mathcal{G}$ are the sets $A\times\emptyset=\emptyset=\emptyset\times Y$ and $A\times Y$ with $A\in\mathcal{B}_{l}$; all of them lie in $\varphi_{l}(\mathcal{B}_{l})$, because $\emptyset\in\mathcal{B}_{l}$. As $\mathcal{B}_{l}\otimes\mathcal{G}$ is the $\sigma$-algebra generated by those rectangles, and hence the smallest $\sigma$-algebra containing them, $\mathcal{B}_{l}\otimes\mathcal{G}\subseteq\varphi_{l}(\mathcal{B}_{l})$. Conversely each $A\times Y$ with $A\in\mathcal{B}_{l}$ is itself a measurable rectangle, so $\varphi_{l}(\mathcal{B}_{l})\subseteq\mathcal{B}_{l}\otimes\mathcal{G}$. The two families are therefore equal, which is the displayed description.

Both $\lambda_{l}$ and $\delta$ are $\sigma$-finite ($\lambda_{l}$ by claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$), so $\lambda_{l}\otimes\delta$ is defined by Existence and Uniqueness of the Product Measure and satisfies $(\lambda_{l}\otimes\delta)(A\times Y)=\lambda_{l}(A)\,\delta(Y)=\lambda_{l}(A)$ for $A\in\mathcal{B}_{l}$. Every member of $\mathcal{B}_{l}\otimes\mathcal{G}$ is of the form $A\times Y$ with $A\in\mathcal{B}_{l}$, by the description just proved, and on such a set $(\lambda_{l})_{\varphi_{l}}$ takes the same value $\lambda_{l}(A)$. The two measures therefore agree at every member of their common domain, so $\lambda_{l}\otimes\delta=(\lambda_{l})_{\varphi_{l}}$.

Let $F:\mathbb{R}^{l}\to\mathbb{R}$ and $B\in\mathcal{B}(\mathbb{R})$. Since $\varphi_{l}$ is a bijection with inverse $\varphi_{l}^{-1}$, we have $(F\circ\varphi_{l}^{-1})^{-1}(B)=\varphi_{l}\bigl(F^{-1}(B)\bigr)$, and, $\varphi_{l}$ being injective, $\varphi_{l}(F^{-1}(B))\in\varphi_{l}(\mathcal{B}_{l})$ holds if and only if $F^{-1}(B)\in\mathcal{B}_{l}$. This gives the asserted equivalence of measurability. Finally, claim 2 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables, applied to the measurable map $\varphi_{l}$ and to the function $F\circ\varphi_{l}^{-1}$ on $\mathbb{R}^{l}\times Y$, gives that $F\circ\varphi_{l}^{-1}$ is integrable with respect to $(\lambda_{l})_{\varphi_{l}}=\lambda_{l}\otimes\delta$ if and only if $(F\circ\varphi_{l}^{-1})\circ\varphi_{l}=F$ is integrable with respect to $\lambda_{l}$, with equality of the two integrals. This proves Claim 2.

Claim 3. (The extended factors.) For every $i\in[n]$ the map $\hat{g}_{i}$ is measurable with respect to $\mathcal{B}(\mathbb{R})$. If moreover a real number $M_{i}$ satisfies $|g_{i}(t)|\le M_{i}$ for every $t\in J$, then $0\le M_{i}$, $|\hat{g}_{i}(t)|\le M_{i}$ for every $t\in\mathbb{R}$, both $g_{i}$ and $\hat{g}_{i}$ are integrable with respect to $\lambda_{J}$ and $\lambda$ respectively, and

$$\int_{J}g_{i}\,d\lambda_{J}=\int_{\mathbb{R}}\hat{g}_{i}\,d\lambda .$$

Proof of Claim 3. Measurability of $\hat{g}_{i}$ follows from Claim 1 applied to the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, to $X_{0}=J$ and to $f=g_{i}$, whose zero extension is $\hat{g}_{i}$; the restriction of $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ to $J$ is $(J,\mathcal{B}_{J},\lambda_{J})$ by the statement.

Suppose $|g_{i}(t)|\le M_{i}$ for every $t\in J$. Since $0\le 0<1$ we have $0\in J$, and $0\le|g_{i}(0)|$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so $0\le M_{i}$ by transitivity. For $t\in J$ we have $|\hat{g}_{i}(t)|=|g_{i}(t)|\le M_{i}$, and for $t\in\mathbb{R}$ with $t\notin J$ we have $|\hat{g}_{i}(t)|=|0|=0\le M_{i}$ by claim 1 of Properties of the Absolute Value in an Ordered Field. Hence $|\hat{g}_{i}(t)|\le M_{i}\mathbf{1}_{J}(t)$ fails only where $\mathbf{1}_{J}(t)=0$, and there both sides are $0$; so in fact $|\hat{g}_{i}(t)|\le M_{i}\mathbf{1}_{J}(t)$ for every $t\in\mathbb{R}$, where $\mathbf{1}_{J}$ is the indicator of $J$.

The map $|\hat{g}_{i}|$ is measurable by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and $\mathbf{1}_{J}$ is measurable by claim 1 of that lemma. By claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (monotonicity, then homogeneity with the constant $M_{i}\in[0,\infty)$) and by The Integral of an Indicator Function is the Measure of the Set,

$$\int_{\mathbb{R}}|\hat{g}_{i}|\,d\lambda\le\int_{\mathbb{R}}M_{i}\mathbf{1}_{J}\,d\lambda=M_{i}\,\lambda(J)=M_{i}<\infty .$$

By Integrable Functions and Their Lebesgue Integral the map $\hat{g}_{i}$ is therefore integrable with respect to $\lambda$, and Claim 1 gives that $g_{i}$ is integrable with respect to $\lambda_{J}$ with the displayed equality of integrals. This proves Claim 3.

Claim 4. (Partial products.) Let $l$ be a natural number with $1\le l\le n$. Then $P_{l,j}$ is measurable with respect to $\mathcal{B}_{l}$ for every $j\in[l]$. If moreover real numbers $M_{i}$ with $|g_{i}(t)|\le M_{i}$ for every $t\in J$ are given for every $i\in[n]$, then

$$|P_{l,j}(\theta)|\le\prod_{k=1}^{j}M_{k}\qquad\text{for every }j\in[l]\text{ and every }\theta\in\mathbb{R}^{l}.$$

Proof of Claim 4. For $k\in[l]$ the projection $\pi_{k}$ is measurable with respect to $\mathcal{B}_{l}$ and $\mathcal{B}(\mathbb{R})$ by claim 1 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets. Hence the map $\theta\mapsto\hat{g}_{k}(\theta_{k})$, which is $\hat{g}_{k}\circ\pi_{k}$, is measurable with respect to $\mathcal{B}_{l}$: for $B\in\mathcal{B}(\mathbb{R})$ one has $(\hat{g}_{k}\circ\pi_{k})^{-1}(B)=\pi_{k}^{-1}\bigl(\hat{g}_{k}^{-1}(B)\bigr)$, and $\hat{g}_{k}^{-1}(B)\in\mathcal{B}(\mathbb{R})$ by Claim 3.

We induct on $j$, bounded by $l$. By claim 1 of Properties of Finite Products, applied for each fixed $\theta$ to the map $k\mapsto\hat{g}_{k}(\theta_{k})$ on $[l]$, we have $P_{l,1}=\hat{g}_{1}\circ\pi_{1}$ and, whenever $S(m)\in[l]$,

$$P_{l,S(m)}(\theta)=P_{l,m}(\theta)\,\hat{g}_{S(m)}(\theta_{S(m)})\qquad(\theta\in\mathbb{R}^{l}),$$

that is, $P_{l,S(m)}$ is the pointwise product of $P_{l,m}$ and $\hat{g}_{S(m)}\circ\pi_{S(m)}$. The case $j=1$ is the measurability just proved, and the successor step is claim 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps applied to $P_{l,m}$ and $\hat{g}_{S(m)}\circ\pi_{S(m)}$. This proves the measurability assertion.

Assume now the bounds $M_{i}$ are given, so that $0\le M_{i}$ and $|\hat{g}_{i}(t)|\le M_{i}$ for every $t\in\mathbb{R}$ by Claim 3. We induct on $j$ again, bounded by $l$, fixing $\theta\in\mathbb{R}^{l}$. For $j=1$ we have $|P_{l,1}(\theta)|=|\hat{g}_{1}(\theta_{1})|\le M_{1}=\prod_{k=1}^{1}M_{k}$, the last equality by claim 1 of Properties of Finite Products. Suppose the bound holds at $m$ and $S(m)\in[l]$. Writing $A=\prod_{k=1}^{m}M_{k}$, claim 5 of Properties of Finite Products gives $0\le A$, and claim 1 of Properties of the Absolute Value in an Ordered Field gives $0\le|\hat{g}_{S(m)}(\theta_{S(m)})|$. By claim 4 of Properties of the Absolute Value in an Ordered Field and by claim 5 of Elementary Arithmetic in an Ordered Field, applied first with the nonnegative factor $|\hat{g}_{S(m)}(\theta_{S(m)})|$ and then with the nonnegative factor $A$ (in the first application the factor multiplies on the right, which is the same as multiplying on the left by the commutativity of multiplication in $\mathbb{R}$),

$$|P_{l,S(m)}(\theta)|=|P_{l,m}(\theta)|\,|\hat{g}_{S(m)}(\theta_{S(m)})|\le A\,|\hat{g}_{S(m)}(\theta_{S(m)})|\le A\,M_{S(m)}=\prod_{k=1}^{S(m)}M_{k},$$

the last equality by claim 1 of Properties of Finite Products and the two inequalities combined by transitivity. This proves Claim 4.

Claim 5. (Integrability of the full products.) Assume the bounds $M_{i}$ of Claim 4 are given, and let $l$ be a natural number with $1\le l\le n$. Put $Q_{l}=\{\theta\in\mathbb{R}^{l}:\theta_{k}\in J\ \text{for every}\ k\in[l]\}$. Then $Q_{l}\in\mathcal{B}_{l}$, $\lambda_{l}(Q_{l})=1$, the map $H_{l}$ vanishes at every point of $\mathbb{R}^{l}$ outside $Q_{l}$, and $H_{l}$ is integrable with respect to $\lambda_{l}$.

Proof of Claim 5. The set $Q_{l}$ is the Borel rectangle with every factor equal to $J$, so $Q_{l}\in\mathcal{B}_{l}$ and $\lambda_{l}(Q_{l})$ is the product in $[0,\infty]$ of $l$ factors each equal to $\lambda(J)=1$, both by claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$; that product is $1$, since all its factors are finite, so that it is an ordinary product of real numbers, and $1\cdot1=1$. If $\theta\in\mathbb{R}^{l}$ and $\theta\notin Q_{l}$, then $\theta_{k}\notin J$ for some $k\in[l]$, so $\hat{g}_{k}(\theta_{k})=0$ and $H_{l}(\theta)=0$ by claim 4 of Properties of Finite Products.

Write $C=\prod_{k=1}^{l}M_{k}$, so $0\le C$ by claim 5 of Properties of Finite Products. For $\theta\in Q_{l}$ we have $|H_{l}(\theta)|\le C=C\,\mathbf{1}_{Q_{l}}(\theta)$ by Claim 4, and for $\theta\notin Q_{l}$ we have $|H_{l}(\theta)|=0=C\,\mathbf{1}_{Q_{l}}(\theta)$; so $|H_{l}|\le C\,\mathbf{1}_{Q_{l}}$ pointwise on $\mathbb{R}^{l}$. The map $H_{l}$ is measurable by Claim 4, hence so is $|H_{l}|$ by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and $\mathbf{1}_{Q_{l}}$ is measurable by claim 1 of that lemma. By claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion and by The Integral of an Indicator Function is the Measure of the Set,

$$\int_{\mathbb{R}^{l}}|H_{l}|\,d\lambda_{l}\le\int_{\mathbb{R}^{l}}C\,\mathbf{1}_{Q_{l}}\,d\lambda_{l}=C\,\lambda_{l}(Q_{l})=C<\infty ,$$

so $H_{l}$ is integrable with respect to $\lambda_{l}$ by Integrable Functions and Their Lebesgue Integral. This proves Claim 5.

Claim 6. (The product formula on $\mathbb{R}^{l}$.) Assume the bounds $M_{i}$ of Claim 4 are given and write $c_{k}=\int_{\mathbb{R}}\hat{g}_{k}\,d\lambda$ for $k\in[n]$, which is defined by Claim 3. Then for every natural number $l$ with $1\le l\le n$,

$$\int_{\mathbb{R}^{l}}H_{l}\,d\lambda_{l}=\prod_{k=1}^{l}c_{k}.$$

Proof of Claim 6. We induct on $l$, bounded by $n$. For $l=1$ we have $\mathcal{B}_{1}=\mathcal{B}(\mathbb{R})$, $\lambda_{1}=\lambda$ and, by claim 1 of Properties of Finite Products, $H_{1}(\theta)=\hat{g}_{1}(\theta_{1})$ for every $\theta\in\mathbb{R}^{1}$; under the identification of $\mathbb{R}^{1}$ with $\mathbb{R}$ this reads $H_{1}=\hat{g}_{1}$, so both sides equal $c_{1}$, the right-hand side by claim 1 of Properties of Finite Products.

Suppose the identity holds at $l$ and that $S(l)\le n$. Since $1\le l$ we have $2\le S(l)$, so the identification of $\mathbb{R}^{S(l)}$ with $\mathbb{R}^{l}\times\mathbb{R}$ is in force; we write a point of $\mathbb{R}^{S(l)}$ as $(\theta',t)$ with $\theta'\in\mathbb{R}^{l}$ and $t\in\mathbb{R}$. By claim 1 of Properties of Finite Products,

$$H_{S(l)}(\theta',t)=H_{l}(\theta')\,\hat{g}_{S(l)}(t)\qquad(\theta'\in\mathbb{R}^{l},\ t\in\mathbb{R}).$$

Let $(Y,\mathcal{G},\delta)$ and $\varphi_{S(l)}$ be as in Claim 2, and put $f=H_{S(l)}\circ\varphi_{S(l)}^{-1}$. By Claims 2 and 5 the map $f$ is measurable with respect to $\mathcal{B}_{S(l)}\otimes\mathcal{G}$ and integrable with respect to $\lambda_{S(l)}\otimes\delta$, with

$$\int_{\mathbb{R}^{S(l)}\times Y}f\,d(\lambda_{S(l)}\otimes\delta)=\int_{\mathbb{R}^{S(l)}}H_{S(l)}\,d\lambda_{S(l)} .$$

Apply claim 4 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ with the dimension $S(l)$, with the $\sigma$-finite measure space $(Y,\mathcal{G},\delta)$ of Claim 2, with the index $i=S(l)$, and to the map $f$. The insertion map there is $\Psi_{S(l)}\bigl(t,(\theta',y)\bigr)=\bigl((\theta',t),y\bigr)$, so that

$$f\bigl(\Psi_{S(l)}(t,(\theta',y))\bigr)=H_{S(l)}(\theta',t)=H_{l}(\theta')\,\hat{g}_{S(l)}(t).$$

The claim provides $N\in\mathcal{B}_{l}\otimes\mathcal{G}$ with $(\lambda_{l}\otimes\delta)(N)=0$ such that for every $(\theta',y)\notin N$ the map $t\mapsto H_{l}(\theta')\hat{g}_{S(l)}(t)$ is integrable with respect to $\lambda$, and such that the map $\Phi$ on $\mathbb{R}^{l}\times Y$ equal to $\int_{\mathbb{R}}H_{l}(\theta')\hat{g}_{S(l)}(t)\,d\lambda(t)$ off $N$ and to $0$ on $N$ is integrable with respect to $\lambda_{l}\otimes\delta$ with

$$\int_{\mathbb{R}^{l}\times Y}\Phi\,d(\lambda_{l}\otimes\delta)=\int_{\mathbb{R}^{S(l)}\times Y}f\,d(\lambda_{S(l)}\otimes\delta).$$

Let $\Theta=c_{S(l)}\,\bigl(H_{l}\circ\varphi_{l}^{-1}\bigr)$ on $\mathbb{R}^{l}\times Y$. By Claims 2 and 5 the map $H_{l}\circ\varphi_{l}^{-1}$ is integrable with respect to $\lambda_{l}\otimes\delta$, so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, applied with the constants $c_{S(l)}$ and $0$ and with the companion function $H_{l}\circ\varphi_{l}^{-1}$ in both slots, $\Theta$ is integrable with respect to $\lambda_{l}\otimes\delta$ and

$$\int_{\mathbb{R}^{l}\times Y}\Theta\,d(\lambda_{l}\otimes\delta)=c_{S(l)}\int_{\mathbb{R}^{l}\times Y}H_{l}\circ\varphi_{l}^{-1}\,d(\lambda_{l}\otimes\delta)=c_{S(l)}\int_{\mathbb{R}^{l}}H_{l}\,d\lambda_{l},$$

the second equality by Claim 2.

For $(\theta',y)\notin N$, the map $t\mapsto H_{l}(\theta')\hat{g}_{S(l)}(t)$ is the constant $H_{l}(\theta')$ times the map $\hat{g}_{S(l)}$, which is integrable with respect to $\lambda$ by Claim 3; so claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, applied with the constants $H_{l}(\theta')$ and $0$ and with the companion function $\hat{g}_{S(l)}$ in both slots, gives

$$\Phi(\theta',y)=\int_{\mathbb{R}}H_{l}(\theta')\hat{g}_{S(l)}(t)\,d\lambda(t)=H_{l}(\theta')\,c_{S(l)}=\Theta(\theta',y),$$

using $\varphi_{l}^{-1}(\theta',y)=\theta'$ and the commutativity of multiplication in $\mathbb{R}$. Hence the set of points of $\mathbb{R}^{l}\times Y$ at which $\Phi$ and $\Theta$ differ is contained in $N$, a set of $(\lambda_{l}\otimes\delta)$-measure zero, so $\Phi=\Theta$ almost everywhere. Both maps are integrable with respect to $\lambda_{l}\otimes\delta$, so the almost-everywhere comparison The Lebesgue Integral and Null Sets (iii) gives that their integrals agree. Combining the four displayed identities of this step with the induction hypothesis and with claim 1 of Properties of Finite Products,

$$\int_{\mathbb{R}^{S(l)}}H_{S(l)}\,d\lambda_{S(l)}=c_{S(l)}\int_{\mathbb{R}^{l}}H_{l}\,d\lambda_{l}=c_{S(l)}\prod_{k=1}^{l}c_{k}=\prod_{k=1}^{S(l)}c_{k},$$

the last equality using the commutativity of multiplication in $\mathbb{R}$. This proves Claim 6.

Proof of claim 1 of the statement. For $x\in Q$ we have $x_{i}\in J$ for every $i\in[n]$ by the statement, so $g_{i}(x_{i})$ is defined and $G(x)=\prod_{i=1}^{n}g_{i}(x_{i})$ is a real number; thus $G:Q\to\mathbb{R}$ is well defined. Let $\hat{G}:\mathbb{R}^{n}\to\mathbb{R}$ be the map equal to $G$ on $Q$ and to $0$ off $Q$. We show $\hat{G}=H_{n}$. If $x\in Q$ then $x_{i}\in J$ for every $i\in[n]$, so the maps $i\mapsto\hat{g}_{i}(x_{i})$ and $i\mapsto g_{i}(x_{i})$ on $[n]$ coincide, and therefore their finite products coincide: $H_{n}(x)=G(x)=\hat{G}(x)$. If $x\in\mathbb{R}^{n}$ and $x\notin Q$ then $x_{i}\notin J$ for some $i\in[n]$, so $\hat{g}_{i}(x_{i})=0$ and $H_{n}(x)=0=\hat{G}(x)$ by claim 4 of Properties of Finite Products.

By Claim 4, taken with $l=n$ and $j=n$, the map $H_{n}=\hat{G}$ is measurable with respect to $\mathcal{B}_{n}=\mathcal{B}(\mathbb{R}^{n})$. Claim 1, applied to the measure space $(\mathbb{R}^{n},\mathcal{B}_{n},\lambda_{n})$, to $X_{0}=Q$ and to $f=G$, whose zero extension is $\hat{G}$, therefore gives that $G$ is measurable with respect to $\mathcal{B}_{Q}$.

Proof of claim 2 of the statement. Assume the bounds $M_{i}$. By Claim 3 each $g_{i}$ is integrable with respect to $\lambda_{J}$ with $\int_{J}g_{i}\,d\lambda_{J}=c_{i}$. By Claim 5 with $l=n$, the map $\hat{G}=H_{n}$ is integrable with respect to $\lambda_{n}$, so by Claim 1 the map $G$ is integrable with respect to $\lambda_{Q}$ and

$$\int_{\mathbb{T}^{n}}G\,dx=\int_{Q}G\,d\lambda_{Q}=\int_{\mathbb{R}^{n}}H_{n}\,d\lambda_{n},$$

the first equality being the notation fixed in The Flat Torus: Standing Notation (iii). By Claim 6 with $l=n$ the right-hand side equals $\prod_{k=1}^{n}c_{k}$. Since the maps $k\mapsto c_{k}$ and $k\mapsto\int_{J}g_{k}\,d\lambda_{J}$ on $[n]$ coincide, their finite products coincide, which is the asserted identity.