Proof of Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets

Each result cited is universally quantified over the data in its own statement. Throughout, "claim $n$ of the tuple lemma" refers to Euclidean Points as Tuples of Real Numbers, "of the Borel lemma" to The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, and "of the concatenation lemma" to Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space. For $k\in[q+p]$ the coordinate map $\pi_{k}:\mathbb{R}^{q+p}\to\mathbb{R}$, $\pi_{k}(z)=z_{k}$, is Borel by claim 1 of the Borel lemma. By claim 5 of Basic Properties of Initial Segments of the Natural Numbers, $[q]\subseteq[q+p]$ and $q+j\in[q+p]$ for $j\in[p]$.

Claim 1. For $z\in\mathbb{R}^{q+p}$ let $\mathrm{pr}^{q,p}_{1}(z)$ be the point of $\mathbb{R}^{q}$ whose $k$th component is $z_{k}$ for $k\in[q]$, and $\mathrm{pr}^{q,p}_{2}(z)$ the point of $\mathbb{R}^{p}$ whose $j$th component is $z_{q+j}$ for $j\in[p]$; both exist and are unique by claim 2 of the tuple lemma. For $x\in\mathbb{R}^{q}$ and $y\in\mathbb{R}^{p}$ the point $\iota(x,y)$ has $k$th component $x_{k}$ for $k\in[q]$ and $(q+j)$th component $y_{j}$ for $j\in[p]$, so $\mathrm{pr}^{q,p}_{1}(\iota(x,y))$ and $x$ have the same components and are equal by claim 1 of the tuple lemma; likewise $\mathrm{pr}^{q,p}_{2}(\iota(x,y))=y$. Given $z$, claim 1 of the concatenation lemma provides $x,y$ with $z=\iota(x,y)$, whence $\mathrm{pr}^{q,p}_{1}(z)=x$, $\mathrm{pr}^{q,p}_{2}(z)=y$ and $z=\iota(\mathrm{pr}^{q,p}_{1}(z),\mathrm{pr}^{q,p}_{2}(z))$; the same representation shows that any map $Q:\mathbb{R}^{q+p}\to\mathbb{R}^{q}$ with $Q(\iota(x,y))=x$ for all $x,y$ satisfies $Q(z)=x=\mathrm{pr}^{q,p}_{1}(z)$, so the projections are unique. The $k$th component of $\mathrm{pr}^{q,p}_{1}$ is the map $z\mapsto z_{k}=\pi_{k}(z)$, which is Borel, so $\mathrm{pr}^{q,p}_{1}$ is Borel by claim 2 of the Borel lemma (with $(X,\mathcal{F})=(\mathbb{R}^{q+p},\mathcal{B}(\mathbb{R}^{q+p}))$); the $j$th component of $\mathrm{pr}^{q,p}_{2}$ is $\pi_{q+j}$, so $\mathrm{pr}^{q,p}_{2}$ is Borel likewise. Finally, writing $z=\iota(x,y)$ as above, claim 3 of the concatenation lemma gives $\lVert z\rVert^{2}=\lVert x\rVert^{2}+\lVert y\rVert^{2}$; since $0\le\lVert y\rVert^{2}$ by claim 2 of Nonnegativity of Squares in an Ordered Field, $\lVert x\rVert^{2}\le\lVert z\rVert^{2}$ by claim 3 of Elementary Arithmetic in an Ordered Field, and $\lVert x\rVert\le\lVert z\rVert$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, both norms being nonnegative by The Euclidean Norm on $\mathbb{R}^n$; the bound for $\mathrm{pr}^{q,p}_{2}$ is the same with the roles of $x$ and $y$ exchanged.

Claim 2. By claim 3 of the tuple lemma, the map $(u,v):s\mapsto\iota(u(s),v(s))$ has $k$th component $s\mapsto(u(s))_{k}$ for $k\in[q]$ and $(q+j)$th component $s\mapsto(v(s))_{j}$ for $j\in[p]$, by the description of $\iota$. Since $u$ is measurable with respect to $\mathcal{E}$ and $\mathcal{B}(\mathbb{R}^{q})=\mathcal{B}_{q}$, each of its components is measurable with respect to $\mathcal{E}$ and $\mathcal{B}(\mathbb{R})$ by claim 2 of the Borel lemma, and likewise for $v$; hence every component of $(u,v)$ is measurable, and $(u,v)$ is measurable with respect to $\mathcal{E}$ and $\mathcal{B}_{q+p}=\mathcal{B}(\mathbb{R}^{q+p})$ by claim 2 of the Borel lemma again. The case $(E,\mathcal{E})=(\mathbb{R}^{l},\mathcal{B}(\mathbb{R}^{l}))$ is the statement for Borel $u,v$. The swap $\sigma^{q,p}$ is the pairing of $\mathrm{pr}^{q,p}_{2}:\mathbb{R}^{q+p}\to\mathbb{R}^{p}$ and $\mathrm{pr}^{q,p}_{1}:\mathbb{R}^{q+p}\to\mathbb{R}^{q}$, formed with $\iota^{p,q}$; both are Borel by claim 1, so $\sigma^{q,p}$ is Borel by what was just shown, with $(E,\mathcal{E})=(\mathbb{R}^{q+p},\mathcal{B}(\mathbb{R}^{q+p}))$ and the roles of $q$ and $p$ exchanged.

Claim 3. By claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, $\mathcal{B}_{q+p}$ is generated by the Borel rectangles of $\mathbb{R}^{q+p}$, the sets $R=\{z\in\mathbb{R}^{q+p}:z_{i}\in A_{i}\text{ for every }i\in[q+p]\}$ with all $A_{i}\in\mathcal{B}(\mathbb{R})$; by claim 2 of Generator Criterion for Measurability it therefore suffices to show $\iota^{-1}(R)\in\mathcal{B}(\mathbb{R}^{q})\otimes\mathcal{B}(\mathbb{R}^{p})$ for every such $R$. By the description of $\iota$, a pair with entries $x$ and $y$ lies in $\iota^{-1}(R)$ exactly when $x_{k}\in A_{k}$ for all $k\in[q]$ and $y_{j}\in A_{q+j}$ for all $j\in[p]$; thus $\iota^{-1}(R)=R'\times R''$, where $R'=\{x\in\mathbb{R}^{q}:x_{k}\in A_{k}\text{ for }k\in[q]\}$ and $R''=\{y\in\mathbb{R}^{p}:y_{j}\in A_{q+j}\text{ for }j\in[p]\}$ are Borel rectangles of $\mathbb{R}^{q}$ and of $\mathbb{R}^{p}$. These belong to $\mathcal{B}_{q}=\mathcal{B}(\mathbb{R}^{q})$ and $\mathcal{B}_{p}=\mathcal{B}(\mathbb{R}^{p})$, since a generated $\sigma$-algebra contains its generators by claim 2 of Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra; so $R'\times R''$ is a measurable rectangle, which belongs to $\mathcal{B}(\mathbb{R}^{q})\otimes\mathcal{B}(\mathbb{R}^{p})$ by Measurable Rectangles and the Product σ-Algebra and the same claim. Hence $\iota$ is measurable.

A probability measure is $\sigma$-finite (take every $X_{m}$ in Measure, Measure Space, and Probability Measure to be the whole space, of measure $1<\infty$), so $\mu\otimes\nu$ exists by Existence and Uniqueness of the Product Measure, and it is a probability measure, as $(\mu\otimes\nu)(\mathbb{R}^{q}\times\mathbb{R}^{p})=\mu(\mathbb{R}^{q})\nu(\mathbb{R}^{p})=1$ by that theorem; hence $\mu\boxtimes\nu$ is a probability measure on $(\mathbb{R}^{q+p},\mathcal{B}(\mathbb{R}^{q+p}))$ by claim 1 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables. For $A\in\mathcal{B}(\mathbb{R}^{q})$ and $B\in\mathcal{B}(\mathbb{R}^{p})$, a point $z$ lies in $\iota(A\times B)$ exactly when $\iota^{-1}(z)$, which is the pair with entries $\mathrm{pr}^{q,p}_{1}(z)$ and $\mathrm{pr}^{q,p}_{2}(z)$ by claim 1 (as $z=\iota(\mathrm{pr}^{q,p}_{1}(z),\mathrm{pr}^{q,p}_{2}(z))$ and $\iota$ is injective), lies in $A\times B$; that is, $\iota(A\times B)=(\mathrm{pr}^{q,p}_{1})^{-1}(A)\cap(\mathrm{pr}^{q,p}_{2})^{-1}(B)$, an intersection of two members of $\mathcal{B}(\mathbb{R}^{q+p})$ by claim 1 and Measurable Map, and the Set of Measurable Maps, hence a member by σ-Algebra and Measurable Space. Since $\iota$ is a bijection, $\iota^{-1}(\iota(A\times B))=A\times B$, so by the definition of the image measure and Existence and Uniqueness of the Product Measure,

$$(\mu\boxtimes\nu)(\iota(A\times B))=(\mu\otimes\nu)(A\times B)=\mu(A)\,\nu(B).$$

Taking $B=\mathbb{R}^{p}$ gives $(\mu\boxtimes\nu)((\mathrm{pr}^{q,p}_{1})^{-1}(A))=\mu(A)\nu(\mathbb{R}^{p})=\mu(A)$, since $(\mathrm{pr}^{q,p}_{2})^{-1}(\mathbb{R}^{p})=\mathbb{R}^{q+p}$; so the image measure of $\mu\boxtimes\nu$ under $\mathrm{pr}^{q,p}_{1}$ is $\mu$, and taking $A=\mathbb{R}^{q}$ shows that the image under $\mathrm{pr}^{q,p}_{2}$ is $\nu$. Finally let $F:\mathbb{R}^{q+p}\to[0,\infty]$ be Borel. For real $c$, $\{w:F(\iota(w))>c\}=\iota^{-1}(\{z:F(z)>c\})$, the preimage under the measurable map $\iota$ of a member of $\mathcal{B}(\mathbb{R}^{q+p})$, so $F\circ\iota$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{q})\otimes\mathcal{B}(\mathbb{R}^{p})$ in the sense of Measurable Functions and the Lebesgue Integral: Standing Notation (iii); and the integral identity is claim 2 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables with $T=\iota$ and $g=F$.

Claim 4. By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, $\lVert x\rVert^{2}=\sum_{k=1}^{q}x_{k}^{2}$ for $x\in\mathbb{R}^{q}$; each $x\mapsto x_{k}^{2}$ is the product of the Borel map $\pi_{k}$ with itself, Borel by claim 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and the finite sum is Borel by claim 2 there. Thus $x\mapsto\lVert x\rVert^{2}$ is Borel. For real $c$, the set $\{x:\lVert x\rVert>c\}$ is $\mathbb{R}^{q}$ if $c<0$, and equals $\{x:\lVert x\rVert^{2}>c^{2}\}$ if $0\le c$, by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field applied to the nonnegative numbers $c$ and $\lVert x\rVert$; in both cases it belongs to $\mathcal{B}(\mathbb{R}^{q})$, so $x\mapsto\lVert x\rVert$ is Borel by Measurable Functions and the Lebesgue Integral: Standing Notation (iii). Now let $p=q$ and $z\in\mathbb{R}^{q+q}$. By Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, $\mathrm{pr}_{1}(z)\cdot\mathrm{pr}_{2}(z)=\sum_{k=1}^{q}z_{k}z_{q+k}$, and the difference $\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)$ has $k$th component $z_{k}-z_{q+k}$, so $\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert^{2}=\sum_{k=1}^{q}(z_{k}-z_{q+k})^{2}$ by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square; both are finite sums of products of finite linear combinations of the Borel maps $\pi_{k}$, $\pi_{q+k}$, hence Borel by claims 2 and 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and $z\mapsto\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert$ is Borel by the square-root argument just given. For the consequence, let $u,v:E\to\mathbb{R}^{q}$ be measurable. Each Euclidean space is a metric space under $d_{E}$ whose Borel $\sigma$-algebra in the sense of Borel Sigma-Algebra of a Metric Space is $\mathcal{B}(\mathbb{R}^{q})$ by The Borel $\sigma$-Algebras of Euclidean Space and of the Euclidean Metric Coincide, so claim 4 of Borel Measurability and Bounded Integration on a Metric Space applies: $s\mapsto\lVert u(s)\rVert^{2}$ is the composition of $u$ with the Borel map $x\mapsto\lVert x\rVert^{2}$, hence measurable. The pairing $(u,v)$ is measurable with respect to $\mathcal{E}$ and $\mathcal{B}(\mathbb{R}^{q+q})$ by claim 2, and $\mathrm{pr}_{1}((u,v)(s))=u(s)$, $\mathrm{pr}_{2}((u,v)(s))=v(s)$ by claim 1; so $s\mapsto u(s)\cdot v(s)$ and $s\mapsto\lVert u(s)-v(s)\rVert^{2}$ are the compositions of $(u,v)$ with the Borel maps $z\mapsto\mathrm{pr}_{1}(z)\cdot\mathrm{pr}_{2}(z)$ and $z\mapsto\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert^{2}$, measurable by the same claim.

For the inequalities, let $x,y\in\mathbb{R}^{q}$ and write $a=\lVert x\rVert$, $b=\lVert y\rVert$, both nonnegative. By claim 5 of Zero Products and Elementary Identities in a Field, $(a+b)^{2}=a^{2}+2ab+b^{2}$ and $(a-b)^{2}=a^{2}-2ab+b^{2}$, the latter from the former with $-b$ in place of $b$, using claim 2 of that lemma; as $0\le(a-b)^{2}=(a^{2}+b^{2})-2ab$ by claim 2 of Nonnegativity of Squares in an Ordered Field, claim 3 of Elementary Arithmetic in an Ordered Field gives $2ab\le a^{2}+b^{2}$, hence $(a+b)^{2}\le2a^{2}+2b^{2}$. Now $x-y=x+(-1)y$ by claims 2 and 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, and $\lVert(-1)y\rVert=|-1|\,b=b$ by claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so claim 6 there gives $\lVert x-y\rVert\le a+b$; as both sides are nonnegative, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\lVert x-y\rVert^{2}\le(a+b)^{2}\le2\lVert x\rVert^{2}+2\lVert y\rVert^{2}$. For the second inequality, $x=y+(x-y)$ by claim 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and the axioms of Vector Space over a Field, so $a\le b+\lVert x-y\rVert$ by claim 6 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, and squaring as before gives $\lVert x\rVert^{2}\le2\lVert y\rVert^{2}+2\lVert x-y\rVert^{2}$. Finally $|x\cdot y|\le ab$ is The Cauchy-Schwarz Inequality on $\mathbb{R}^n$, and $ab\le\tfrac12(a^{2}+b^{2})$ follows from $2ab\le a^{2}+b^{2}$ by multiplying with $2^{-1}$, which is positive by claims 8 and 7 of Elementary Order Arithmetic in an Ordered Field, using claim 5 of Elementary Arithmetic in an Ordered Field and $2^{-1}(2ab)=ab$.

Claim 5. Let $a\in\mathbb{R}^{q}$. If $x\in\mathbb{R}^{q}\setminus\{a\}$, then $x-a\ne0_{\mathbb{R}^{q}}$ by claim 2 of Elementary Identities in a Vector Space, so $r=\lVert x-a\rVert=d_{E}(x,a)$ is positive by claims 1, 2 and 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, and the open ball $B(x,r)$ does not contain $a$, since $d_{E}(x,a)=r$ is not less than $r$; thus $B(x,r)\subseteq\mathbb{R}^{q}\setminus\{a\}$, and $\mathbb{R}^{q}\setminus\{a\}$ is open. Hence $\{a\}$ is closed. A finite set $F$ is either empty, and then closed by claim 1 of Complements, Unions and Intersections of Closed Sets in a Topological Space, or has $n$ elements for some $n\in\mathbb{N}$ by Finite Set, that is, is the image of a bijection $[n]\to F$ by Number of Elements of a Set, and then $F=\bigcup_{i=1}^{n}\{a_{i}\}$ with $a_{i}$ the image of $i$, a finite union of closed sets, closed by claim 2 of that lemma. Closed sets belong to $\mathcal{B}(\mathbb{R}^{q})$ by Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background (iv), and so do their complements by σ-Algebra and Measurable Space. For the last assertion, if $F=\varnothing$ the map is the constant $0$, Borel by claim 1 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps; otherwise, with $a_{1},\dots,a_{n}$ as above, the map equals $\sum_{i=1}^{n}w(a_{i})\mathbf{1}_{\{a_{i}\}}$ pointwise: at $x=a_{j}$ every summand with $i\ne j$ vanishes by injectivity of $i\mapsto a_{i}$, so the sum is $w(a_{j})$ by claim 7 of Properties of Finite Sums, and at $x\notin F$ every summand vanishes, so the sum is $0$ by the same claim. Each $\mathbf{1}_{\{a_{i}\}}$ is Borel by claim 1 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, as $\{a_{i}\}\in\mathcal{B}(\mathbb{R}^{q})$, and the finite linear combination is Borel by claim 2 there.