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

lemmalem:euclidean-pairs-borel-toolkit-2026a
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· 13,238 chars · 37 deps · depth 16 Reason: Goal 3A: proof of the Euclidean pairs toolkit - projections and pairings coordinatewise, the product sigma-algebra by the generator criterion on Borel rectangles, the norm functions as sums of products of coordinate maps, and finite sets as finite unions of singletons.

Projections and pairings are handled coordinatewise through the componentwise criterion; the product sigma-algebra statement uses the generator criterion on Borel rectangles; the norm functions are sums of products of coordinate maps; finite sets are finite unions of singletons, each closed.

Proof

Each result cited is universally quantified over the data in its own statement. Throughout, "claim nn 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[q+p]k\in[q+p] the coordinate map πk:Rq+pR\pi_{k}:\mathbb{R}^{q+p}\to\mathbb{R}, πk(z)=zk\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][q+p][q]\subseteq[q+p] and q+j[q+p]q+j\in[q+p] for j[p]j\in[p].

Claim 1. For zRq+pz\in\mathbb{R}^{q+p} let pr1q,p(z)\mathrm{pr}^{q,p}_{1}(z) be the point of Rq\mathbb{R}^{q} whose kkth component is zkz_{k} for k[q]k\in[q], and pr2q,p(z)\mathrm{pr}^{q,p}_{2}(z) the point of Rp\mathbb{R}^{p} whose jjth component is zq+jz_{q+j} for j[p]j\in[p]; both exist and are unique by claim 2 of the tuple lemma. For xRqx\in\mathbb{R}^{q} and yRpy\in\mathbb{R}^{p} the point ι(x,y)\iota(x,y) has kkth component xkx_{k} for k[q]k\in[q] and (q+j)(q+j)th component yjy_{j} for j[p]j\in[p], so pr1q,p(ι(x,y))\mathrm{pr}^{q,p}_{1}(\iota(x,y)) and xx have the same components and are equal by claim 1 of the tuple lemma; likewise pr2q,p(ι(x,y))=y\mathrm{pr}^{q,p}_{2}(\iota(x,y))=y. Given zz, claim 1 of the concatenation lemma provides x,yx,y with z=ι(x,y)z=\iota(x,y), whence pr1q,p(z)=x\mathrm{pr}^{q,p}_{1}(z)=x, pr2q,p(z)=y\mathrm{pr}^{q,p}_{2}(z)=y and z=ι(pr1q,p(z),pr2q,p(z))z=\iota(\mathrm{pr}^{q,p}_{1}(z),\mathrm{pr}^{q,p}_{2}(z)); the same representation shows that any map Q:Rq+pRqQ:\mathbb{R}^{q+p}\to\mathbb{R}^{q} with Q(ι(x,y))=xQ(\iota(x,y))=x for all x,yx,y satisfies Q(z)=x=pr1q,p(z)Q(z)=x=\mathrm{pr}^{q,p}_{1}(z), so the projections are unique. The kkth component of pr1q,p\mathrm{pr}^{q,p}_{1} is the map zzk=πk(z)z\mapsto z_{k}=\pi_{k}(z), which is Borel, so pr1q,p\mathrm{pr}^{q,p}_{1} is Borel by claim 2 of the Borel lemma (with (X,F)=(Rq+p,B(Rq+p))(X,\mathcal{F})=(\mathbb{R}^{q+p},\mathcal{B}(\mathbb{R}^{q+p}))); the jjth component of pr2q,p\mathrm{pr}^{q,p}_{2} is πq+j\pi_{q+j}, so pr2q,p\mathrm{pr}^{q,p}_{2} is Borel likewise. Finally, writing z=ι(x,y)z=\iota(x,y) as above, claim 3 of the concatenation lemma gives z2=x2+y2\lVert z\rVert^{2}=\lVert x\rVert^{2}+\lVert y\rVert^{2}; since 0y20\le\lVert y\rVert^{2} by claim 2 of Nonnegativity of Squares in an Ordered Field, x2z2\lVert x\rVert^{2}\le\lVert z\rVert^{2} by claim 3 of Elementary Arithmetic in an Ordered Field, and xz\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 Euclidean Norm on Rn\mathbb{R}^n; the bound for pr2q,p\mathrm{pr}^{q,p}_{2} is the same with the roles of xx and yy exchanged.

Claim 2. By claim 3 of the tuple lemma, the map (u,v):sι(u(s),v(s))(u,v):s\mapsto\iota(u(s),v(s)) has kkth component s(u(s))ks\mapsto(u(s))_{k} for k[q]k\in[q] and (q+j)(q+j)th component s(v(s))js\mapsto(v(s))_{j} for j[p]j\in[p], by the description of ι\iota. Since uu is measurable with respect to E\mathcal{E} and B(Rq)=Bq\mathcal{B}(\mathbb{R}^{q})=\mathcal{B}_{q}, each of its components is measurable with respect to E\mathcal{E} and B(R)\mathcal{B}(\mathbb{R}) by claim 2 of the Borel lemma, and likewise for vv; hence every component of (u,v)(u,v) is measurable, and (u,v)(u,v) is measurable with respect to E\mathcal{E} and Bq+p=B(Rq+p)\mathcal{B}_{q+p}=\mathcal{B}(\mathbb{R}^{q+p}) by claim 2 of the Borel lemma again. The case (E,E)=(Rl,B(Rl))(E,\mathcal{E})=(\mathbb{R}^{l},\mathcal{B}(\mathbb{R}^{l})) is the statement for Borel u,vu,v. The swap σq,p\sigma^{q,p} is the pairing of pr2q,p:Rq+pRp\mathrm{pr}^{q,p}_{2}:\mathbb{R}^{q+p}\to\mathbb{R}^{p} and pr1q,p:Rq+pRq\mathrm{pr}^{q,p}_{1}:\mathbb{R}^{q+p}\to\mathbb{R}^{q}, formed with ιp,q\iota^{p,q}; both are Borel by claim 1, so σq,p\sigma^{q,p} is Borel by what was just shown, with (E,E)=(Rq+p,B(Rq+p))(E,\mathcal{E})=(\mathbb{R}^{q+p},\mathcal{B}(\mathbb{R}^{q+p})) and the roles of qq and pp exchanged.

Claim 3. By claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on Rl\mathbb{R}^l, Bq+p\mathcal{B}_{q+p} is generated by the Borel rectangles of Rq+p\mathbb{R}^{q+p}, the sets R={zRq+p:ziAi for every i[q+p]}R=\{z\in\mathbb{R}^{q+p}:z_{i}\in A_{i}\text{ for every }i\in[q+p]\} with all AiB(R)A_{i}\in\mathcal{B}(\mathbb{R}); by claim 2 of Generator Criterion for Measurability it therefore suffices to show ι1(R)B(Rq)B(Rp)\iota^{-1}(R)\in\mathcal{B}(\mathbb{R}^{q})\otimes\mathcal{B}(\mathbb{R}^{p}) for every such RR. By the description of ι\iota, a pair with entries xx and yy lies in ι1(R)\iota^{-1}(R) exactly when xkAkx_{k}\in A_{k} for all k[q]k\in[q] and yjAq+jy_{j}\in A_{q+j} for all j[p]j\in[p]; thus ι1(R)=R×R\iota^{-1}(R)=R'\times R'', where R={xRq:xkAk for k[q]}R'=\{x\in\mathbb{R}^{q}:x_{k}\in A_{k}\text{ for }k\in[q]\} and R={yRp:yjAq+j for j[p]}R''=\{y\in\mathbb{R}^{p}:y_{j}\in A_{q+j}\text{ for }j\in[p]\} are Borel rectangles of Rq\mathbb{R}^{q} and of Rp\mathbb{R}^{p}. These belong to Bq=B(Rq)\mathcal{B}_{q}=\mathcal{B}(\mathbb{R}^{q}) and Bp=B(Rp)\mathcal{B}_{p}=\mathcal{B}(\mathbb{R}^{p}), since a generated σ\sigma-algebra contains its generators by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra; so R×RR'\times R'' is a measurable rectangle, which belongs to B(Rq)B(Rp)\mathcal{B}(\mathbb{R}^{q})\otimes\mathcal{B}(\mathbb{R}^{p}) by Product Sigma-Algebra and the same claim. Hence ι\iota is measurable.

A probability measure is σ\sigma-finite (take every XmX_{m} in Measure, Measure Space, and Probability Measure to be the whole space, of measure 1<1<\infty), so μν\mu\otimes\nu exists by Existence and Uniqueness of the Product Measure, and it is a probability measure, as (μν)(Rq×Rp)=μ(Rq)ν(Rp)=1(\mu\otimes\nu)(\mathbb{R}^{q}\times\mathbb{R}^{p})=\mu(\mathbb{R}^{q})\nu(\mathbb{R}^{p})=1 by that theorem; hence μν\mu\boxtimes\nu is a probability measure on (Rq+p,B(Rq+p))(\mathbb{R}^{q+p},\mathcal{B}(\mathbb{R}^{q+p})) by claim 1 of Image Measures, Measures with Densities, and Change of Variables. For AB(Rq)A\in\mathcal{B}(\mathbb{R}^{q}) and BB(Rp)B\in\mathcal{B}(\mathbb{R}^{p}), a point zz lies in ι(A×B)\iota(A\times B) exactly when ι1(z)\iota^{-1}(z), which is the pair with entries pr1q,p(z)\mathrm{pr}^{q,p}_{1}(z) and pr2q,p(z)\mathrm{pr}^{q,p}_{2}(z) by claim 1 (as z=ι(pr1q,p(z),pr2q,p(z))z=\iota(\mathrm{pr}^{q,p}_{1}(z),\mathrm{pr}^{q,p}_{2}(z)) and ι\iota is injective), lies in A×BA\times B; that is, ι(A×B)=(pr1q,p)1(A)(pr2q,p)1(B)\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 B(Rq+p)\mathcal{B}(\mathbb{R}^{q+p}) by claim 1 and Measurable Function and Real-Valued Measurable Function, hence a member by Sigma-Algebra and Measurable Space. Since ι\iota is a bijection, ι1(ι(A×B))=A×B\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,

(μν)(ι(A×B))=(μν)(A×B)=μ(A)ν(B).(\mu\boxtimes\nu)(\iota(A\times B))=(\mu\otimes\nu)(A\times B)=\mu(A)\,\nu(B).

Taking B=RpB=\mathbb{R}^{p} gives (μν)((pr1q,p)1(A))=μ(A)ν(Rp)=μ(A)(\mu\boxtimes\nu)((\mathrm{pr}^{q,p}_{1})^{-1}(A))=\mu(A)\nu(\mathbb{R}^{p})=\mu(A), since (pr2q,p)1(Rp)=Rq+p(\mathrm{pr}^{q,p}_{2})^{-1}(\mathbb{R}^{p})=\mathbb{R}^{q+p}; so the image measure of μν\mu\boxtimes\nu under pr1q,p\mathrm{pr}^{q,p}_{1} is μ\mu, and taking A=RqA=\mathbb{R}^{q} shows that the image under pr2q,p\mathrm{pr}^{q,p}_{2} is ν\nu. Finally let F:Rq+p[0,]F:\mathbb{R}^{q+p}\to[0,\infty] be Borel. For real cc, {w:F(ι(w))>c}=ι1({z:F(z)>c})\{w:F(\iota(w))>c\}=\iota^{-1}(\{z:F(z)>c\}), the preimage under the measurable map ι\iota of a member of B(Rq+p)\mathcal{B}(\mathbb{R}^{q+p}), so FιF\circ\iota is measurable with respect to B(Rq)B(Rp)\mathcal{B}(\mathbb{R}^{q})\otimes\mathcal{B}(\mathbb{R}^{p}) in the sense of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable; and the integral identity is claim 2 of Image Measures, Measures with Densities, and Change of Variables with T=ιT=\iota and g=Fg=F.

Claim 4. By claim 1 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, x2=k=1qxk2\lVert x\rVert^{2}=\sum_{k=1}^{q}x_{k}^{2} for xRqx\in\mathbb{R}^{q}; each xxk2x\mapsto x_{k}^{2} is the product of the Borel map πk\pi_{k} with itself, Borel by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and the finite sum is Borel by claim 2 there. Thus xx2x\mapsto\lVert x\rVert^{2} is Borel. For real cc, the set {x:x>c}\{x:\lVert x\rVert>c\} is Rq\mathbb{R}^{q} if c<0c<0, and equals {x:x2>c2}\{x:\lVert x\rVert^{2}>c^{2}\} if 0c0\le c, by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field applied to the nonnegative numbers cc and x\lVert x\rVert; in both cases it belongs to B(Rq)\mathcal{B}(\mathbb{R}^{q}), so xxx\mapsto\lVert x\rVert is Borel by Measure Spaces and the Lebesgue Integral: Standing Notation §measurable. Now let p=qp=q and zRq+qz\in\mathbb{R}^{q+q}. By Difference, Dot Product, and Orthogonality in Rn\mathbb{R}^n, pr1(z)pr2(z)=k=1qzkzq+k\mathrm{pr}_{1}(z)\cdot\mathrm{pr}_{2}(z)=\sum_{k=1}^{q}z_{k}z_{q+k}, and the difference pr1(z)pr2(z)\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z) has kkth component zkzq+kz_{k}-z_{q+k}, so pr1(z)pr2(z)2=k=1q(zkzq+k)2\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 Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n; both are finite sums of products of finite linear combinations of the Borel maps πk\pi_{k}, πq+k\pi_{q+k}, hence Borel by claims 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and zpr1(z)pr2(z)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:ERqu,v:E\to\mathbb{R}^{q} be measurable. Each Euclidean space is a metric space under dEd_{E} whose Borel σ\sigma-algebra in the sense of Borel Sigma-Algebra of a Metric Space is B(Rq)\mathcal{B}(\mathbb{R}^{q}) by The Borel σ\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: su(s)2s\mapsto\lVert u(s)\rVert^{2} is the composition of uu with the Borel map xx2x\mapsto\lVert x\rVert^{2}, hence measurable. The pairing (u,v)(u,v) is measurable with respect to E\mathcal{E} and B(Rq+q)\mathcal{B}(\mathbb{R}^{q+q}) by claim 2, and pr1((u,v)(s))=u(s)\mathrm{pr}_{1}((u,v)(s))=u(s), pr2((u,v)(s))=v(s)\mathrm{pr}_{2}((u,v)(s))=v(s) by claim 1; so su(s)v(s)s\mapsto u(s)\cdot v(s) and su(s)v(s)2s\mapsto\lVert u(s)-v(s)\rVert^{2} are the compositions of (u,v)(u,v) with the Borel maps zpr1(z)pr2(z)z\mapsto\mathrm{pr}_{1}(z)\cdot\mathrm{pr}_{2}(z) and zpr1(z)pr2(z)2z\mapsto\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert^{2}, measurable by the same claim.

For the inequalities, let x,yRqx,y\in\mathbb{R}^{q} and write a=xa=\lVert x\rVert, b=yb=\lVert y\rVert, both nonnegative. By claim 5 of Zero Products and Elementary Identities in a Field, (a+b)2=a2+2ab+b2(a+b)^{2}=a^{2}+2ab+b^{2} and (ab)2=a22ab+b2(a-b)^{2}=a^{2}-2ab+b^{2}, the latter from the former with b-b in place of bb, using claim 2 of that lemma; as 0(ab)2=(a2+b2)2ab0\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 2aba2+b22ab\le a^{2}+b^{2}, hence (a+b)22a2+2b2(a+b)^{2}\le2a^{2}+2b^{2}. Now xy=x+(1)yx-y=x+(-1)y by claims 2 and 3 of Euclidean Space Rn\mathbb{R}^n is a Real Vector Space, and (1)y=1b=b\lVert(-1)y\rVert=|-1|\,b=b by claim 5 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, so claim 6 there gives xya+b\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 xy2(a+b)22x2+2y2\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+(xy)x=y+(x-y) by claim 3 of Euclidean Space Rn\mathbb{R}^n is a Real Vector Space and the axioms of Vector Space over a Field, so ab+xya\le b+\lVert x-y\rVert by claim 6 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, and squaring as before gives x22y2+2xy2\lVert x\rVert^{2}\le2\lVert y\rVert^{2}+2\lVert x-y\rVert^{2}. Finally xyab|x\cdot y|\le ab is Cauchy-Schwarz Inequality for the Euclidean Dot Product, and ab12(a2+b2)ab\le\tfrac12(a^{2}+b^{2}) follows from 2aba2+b22ab\le a^{2}+b^{2} by multiplying with 212^{-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 21(2ab)=ab2^{-1}(2ab)=ab.

Claim 5. Let aRqa\in\mathbb{R}^{q}. If xRq{a}x\in\mathbb{R}^{q}\setminus\{a\}, then xa0Rqx-a\ne0_{\mathbb{R}^{q}} by claim 2 of Elementary Identities in a Vector Space, so r=xa=dE(x,a)r=\lVert x-a\rVert=d_{E}(x,a) is positive by claims 1, 2 and 3 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, and the open ball B(x,r)B(x,r) does not contain aa, since dE(x,a)=rd_{E}(x,a)=r is not less than rr; thus B(x,r)Rq{a}B(x,r)\subseteq\mathbb{R}^{q}\setminus\{a\}, and Rq{a}\mathbb{R}^{q}\setminus\{a\} is open. Hence {a}\{a\} is closed. A finite set FF is either empty, and then closed by claim 1 of Complements, Unions and Intersections of Closed Sets in a Topological Space, or has nn elements for some nNn\in\mathbb{N} by Finite Set, that is, is the image of a bijection [n]F[n]\to F by Number of Elements of a Set, and then F=i=1n{ai}F=\bigcup_{i=1}^{n}\{a_{i}\} with aia_{i} the image of ii, a finite union of closed sets, closed by claim 2 of that lemma. Closed sets belong to B(Rq)\mathcal{B}(\mathbb{R}^{q}) by Euclidean Space and Lebesgue Measure: Standing Notation §borel, and so do their complements by Sigma-Algebra and Measurable Space. For the last assertion, if F=F=\varnothing the map is the constant 00, Borel by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; otherwise, with a1,,ana_{1},\dots,a_{n} as above, the map equals i=1nw(ai)1{ai}\sum_{i=1}^{n}w(a_{i})\mathbf{1}_{\{a_{i}\}} pointwise: at x=ajx=a_{j} every summand with iji\ne j vanishes by injectivity of iaii\mapsto a_{i}, so the sum is w(aj)w(a_{j}) by claim 7 of Properties of Finite Sums, and at xFx\notin F every summand vanishes, so the sum is 00 by the same claim. Each 1{ai}\mathbf{1}_{\{a_{i}\}} is Borel by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, as {ai}B(Rq)\{a_{i}\}\in\mathcal{B}(\mathbb{R}^{q}), and the finite linear combination is Borel by claim 2 there.

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