Reason: Proof of P8.4c (lem:copy-van-trees-data-from-n-agent-solution-2026a): trimming to the record support, van Trees regularity of the smoothed density, information bound in the injection direction, and pairing of cell coefficients with the profile response.
Claim 1.The record support. By claim 1 of the kernel instance, fˉ is Rs-measurable, R+∈Rs, g is finite and measurable with respect to B(Rd)⊗Rs, and g(θ,r)>0 if and only if r∈R+. Hence Ωtr=D−1(R+)∈F♯ and N=D−1(Rs∖R+)=(Θ,D)−1(Rd×(Rs∖R+)). Let F be the indicator of the rectangle Rd×(Rs∖R+), which is measurable (claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions). Then 1N=F(Θ,D), and F(θ,r)g(θ,r)=0 for every (θ,r), because g(θ,r)=0 when r∈/R+. By The Integral of an Indicator Function is the Measure of the Set and the density identity, μ♯(N)=∫Ω♯F(Θ,D)dμ♯=∫Fgd(λd⊗ρ)=0, the integral of the zero function being 0 (the only nonnegative simple minorant of 0 is 0, whose integral is 0). By additivity of the measureμ♯, μ♯(Ωtr)=μ♯(Ω♯)−μ♯(N)=1.
The trimmed copy. By claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions applied to (Ω♯,F♯,μ♯) and X0=Ωtr, the restriction (Ωtr,Ftr,μtr) with Ftr={A∈F♯:A⊆Ωtr} and μtr(A)=μ♯(A) is a measure space, and μtr(Ωtr)=1, so it is a probability space. Let Z:Ω♯→[0,∞] be F♯-measurable. The zero extension of Ztr is Z1Ωtr, which is measurable: {Z1Ωtr>a} equals {Z>a}∩Ωtr for a≥0 and Ω♯ for a<0. By the same claim, Ztr is Ftr-measurable and ∫ΩtrZtrdμtr=∫Ω♯Z1Ωtrdμ♯. Since Z=Z1Ωtr+Z1N pointwise, additivity of the integral (Linearity and Monotonicity of the Lebesgue Integral) gives ∫Zdμ♯=∫Z1Ωtrdμ♯+∫Z1Ndμ♯, and it remains to see that the last integral is 0. For n∈N put Zn=min(Z,n)1N, measurable because {Zn>a} is {Z>a}∩N for 0≤a<n, empty for a≥n, and Ω♯ for a<0. Then 0≤Zn≤n1N, so by monotonicity and positive homogeneity and The Integral of an Indicator Function is the Measure of the Set, ∫Zndμ♯≤nμ♯(N)=0. The sequence (Zn) is nondecreasing with pointwise limit Z1N (at a point of N with Z<∞ the terms equal Z from some index on; where Z=∞ they equal n; off N all vanish), so the monotone convergence theorem gives ∫Z1Ndμ♯=supn∫Zndμ♯=0. This proves the integral identity.
Laws and mean-square norms. Let (Z,G) be a measurable space and V:Ω♯→Z measurable. For A∈G, (Vtr)−1(A)=V−1(A)∩Ωtr belongs to F♯ and is contained in Ωtr, hence belongs to Ftr; so Vtr is measurable. Moreover, by The Integral of an Indicator Function is the Measure of the Set and the integral identity applied to Z=1V−1(A) (whose restriction is the indicator of (Vtr)−1(A) in Ωtr), μtr((Vtr)−1(A))=∫Ω♯1V−1(A)dμ♯=μ♯(V−1(A)); that is, the two image measures agree on G. If Z is a random variable on the copy, then Ztr is a random variable on the trimmed copy (the case Z=R), (Ztr)2=(Z2)tr, and the integral identity applied to Z2 gives Etr[(Ztr)2]=∫Ω♯Z2dμ♯; by Square-Integrable Random Variables and the Mean-Square Inner Product, Z is square-integrable if and only if Ztr is, and then the mean-square norms, the square roots of these two equal quantities, agree.
The restricted product σ-algebra. Let (Y,Y) be a measurable space and put HY=Y×R+, a measurable rectangle, so HY∈Y⊗Rs. Write H1={E∈Y⊗Rs:E⊆HY} for the restriction of Y⊗Rs to HY, a σ-algebra on HY by claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, and H2=Y⊗R+, the σ-algebra on HY generated by the rectangles A×B with A∈Y and B∈R+. Every such rectangle belongs to Y⊗Rs (as B∈Rs) and is contained in HY, hence belongs to H1; since H2 is the smallest σ-algebra on HY containing these rectangles, H2⊆H1. Conversely let D be the family of all E∈Y⊗Rs with E∩HY∈H2. It contains Y×Rs (whose intersection with HY is the rectangle HY), is closed under complements (the intersection of the complement of E with HY is HY∖(E∩HY)) and under countable unions (intersection distributes over unions), and contains every rectangle A×B with A∈Y, B∈Rs, because (A×B)∩HY=A×(B∩R+) with B∩R+∈R+. Thus D is a σ-algebra containing the generators of Y⊗Rs, so D=Y⊗Rs; for E∈H1 this gives E=E∩HY∈H2. Hence H1=H2. From now on we use this with (Y,Y)=(Rd,B(Rd)), so that HY=H and H1=H2=B(Rd)⊗R+.
(the extension identity); and a function on H is H2-measurable if and only if its zero extension is B(Rd)⊗Rs-measurable. In particular, if ψ is a real-valued B(Rd)⊗Rs-measurable function on Rd×Rs, its restriction ψ∣H is H2-measurable: the preimage of a Borel set E′ under ψ∣H is ψ−1(E′)∩H∈H1=H2.
Random variables and record. Each Θq is square-integrable on the copy by claim 2 of Mean-Square Assembly of the Estimand Linearisation on the Synthetic Copy: Approximation of the Recentred Copy Endpoint by an Affine Function of the Parameter (available by its scope statement, as recorded in the adopted setting), so Θqtr is a square-integrable random variable on the trimmed copy by claim 1. For B∈R+, (Dtr)−1(B)=D−1(B) (as B⊆R+), which belongs to F♯ and is contained in Ωtr, hence to Ftr: Dtr is measurable with respect to Ftr and R+. By claim 1, Θtr is measurable with respect to Ftr and B(Rd), so the pair (Θtr,Dtr):Ωtr→H is measurable with respect to Ftr and H2; let νtr be its image measure under μtr, the joint law.
The density; hypothesis (i).gtr is H2-measurable (restriction of the measurable g), finite-valued, and strictly positive on H by claim 1 of the kernel instance. Let E∈H2. Then E∈B(Rd)⊗Rs and E⊆H, so (Θ,D)−1(E)⊆D−1(R+)=Ωtr and therefore (Θtr,Dtr)−1(E)=(Θ,D)−1(E). By claim 1 (image measures, with V=(Θ,D)), The Integral of an Indicator Function is the Measure of the Set, the density identity with F=1E, and (∗) applied to ϕ=1Egtr (whose zero extension is 1Eg, because E⊆H),
Thus νtr is the measure with densitygtr with respect to λd⊗ρ+, which is hypothesis (i) of The Multivariate van Trees Inequality for the data named in the claim (the σ-algebra Bd there being B(Rd) and its λd being Lebesgue measure, as noted above).
Hypothesis (ii). Let r∈R+. The function θ↦gtr(θ,r)=g(θ,r) on Rd is strictly positive and of class C1 by claim 2 of the kernel instance; its partial derivatives are ∂qgtr(θ,r)=∂qg(θ,r), so that ∂qgtr=(∂qg)∣H.
Hypothesis (iii). By claim 2 of the kernel instance, ∂qg is B(Rd)⊗Rs-measurable, so ∂qgtr is H2-measurable. The function ϕ(θ,r)=(1+∑k∣θk∣)∣∂qgtr(θ,r)∣ on H has zero extension 1H(1+∑k∣θk∣)∣∂qg∣≤(1+∑k∣θk∣)∣∂qg∣, so by (∗), monotonicity of the integral and claim 2 of the kernel instance, ∫Hϕd(λd⊗ρ+)≤∫(1+∑k∣θk∣)∣∂qg∣d(λd⊗ρ)<∞.
the third equality pointwise on H, the fourth by (∗) (the zero extension of (∂qgtr)2/gtr being the integrand of claim 2 of the kernel instance, read as 0 off H), and the inequality by that claim. Hence scq is square-integrable: hypothesis (iv′) of Score Identities and the Mixture-Weight Directional van Trees Inequality holds, and that lemma applies on the trimmed copy with l:=d, (Y,G,μ):=(R+,R+,ρ+), Θ:=Θtr, D:=Dtr, p:=gtr, and (in its claim 2) n=d and the shifts aq.
The mixture-weight information. The mixture weight of claim 2 of Score Identities and the Mixture-Weight Directional van Trees Inequality for the shifts aq is pˉ(θ,r)=21gtr(θ,r)+2d1∑qgtr(θ−aq,r) for (θ,r)∈H, which is gˉtr by definition and equals gˉ(θ,r), the mixture weight of the kernel instance, since gtr(θ−aq,r)=g(θ−aq,r)=gq(θ,r); by claim 1 of the kernel instance (applied to gˉ), gˉ>0 on H, so gˉtr is strictly positive. For u∈Rd, ∂ugtr=∑quq∂qgtr=(∂ug)∣H, so Iu as defined in the statement is the mixture-weight information ∫H(∂ugtr)2/pˉd(λd⊗ρ+) of that lemma (its integrand measurable by that claim); the zero extension of the integrand is 1H(∂ug)2/gˉ read as 0 off H, and (∗) gives Iu=∫1H(∂ug)2/gˉd(λd⊗ρ), finite by claim 2 of Score Identities and the Mixture-Weight Directional van Trees Inequality. Finally ∂tugtr=∑qtuq∂qgtr=t∂ugtr, so the integrand defining Itu is t2 times that defining Iu, and Itu=t2Iu by positive homogeneity of the integral (for t=0 both sides are 0).
Claim 3.Kernel bound. By claim 2, Iu1/2 is the left-hand side of the inequality of claim 3 of the kernel instance for the direction u=∑qwqaq (which is the u of that lemma for the weights w), and u=∑qwqmeq/N=(m/N)w by the scalar multiplication and addition of Rd. That claim gives Iu1/2≤(Jsym)1/2+2∑q∣wq∣(exp(κq)−1−κq)1/2 with κq=∥aq∥2/η; here ∥aq∥=m∥eq∥/N=m/N (claim 5 of Elementary Properties of the Euclidean Norm on Rn; ∥eq∥=1 by its claim 1), so κq=m2/(Nη)=κmv for every q, and ∑q∣wq∣=∥w∥1. This is the first display.
The representation of the coefficient functions. For τ∈[0,s] let χτ be the indicator of [τ,s]∩[0,s], a member of B[0,s]. We show
ac(τ)=c⋅vc+∫[0,s]χτ(u)c⋅Hucdu(τ∈[0,s]).(∗∗)
The map u↦Eu⋆vc is bounded measurable (its components are finite sums of products of the continuous, hence bounded and measurable, entries of E⋆ with constants; Continuous Real-Valued Functions on a Compact Interval are Bounded, claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions), so by claim 2 of Variation of Constants with Bounded Measurable Forcing and the Two-Parameter Fundamental Solution the map u↦Huc=ΦE(s,u)Eu⋆vc is bounded measurable, with ∣Huc∣≤Hˉ. Claim 1 of that lemma with t=s and v=vc gives ΦE(s,τ)vc−vc=∫[τ,s]Hucdu, the right-hand side being 0 when τ=s by the convention of that lemma. Fix a component γ and write h=Hc,γ. If τ<s: let h~τ be the function on R equal to h on [τ,s] and to 0 elsewhere. The restriction h∣[τ,s] is B[τ,s]-measurable (a preimage S∩[0,s] of a Borel set under h, with S Borel, has trace S∩[τ,s] on [τ,s]) and bounded, hence integrable with respect to λ[τ,s] (bounded by Hˉ, whose integral is Hˉ(s−τ) by The Integral of an Indicator Function is the Measure of the Set, positive homogeneity and claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval); by claim 2 of that toolkit on [τ,s], ∫[τ,s]h∣[τ,s]dλ[τ,s]=∫Rh~τdλ. The same h~τ is the zero extension of the bounded measurable function χτh on [0,s], so the same claim on [0,s] gives ∫[0,s]χτhdλ[0,s]=∫Rh~τdλ. Hence ∫[τ,s]h∣[τ,s]dλ[τ,s]=∫[0,s]χτhdλ[0,s]. If τ=s: χs is the indicator of {s}, and λ[0,s]({s})=λ({s})≤λ([s−ϵ,s])=λ[s−ϵ,s]([s−ϵ,s])=ϵ for every ϵ∈(0,s] (monotonicity of λ and claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval on [s−ϵ,s]), so λ[0,s]({s})=0 and ∣∫[0,s]χshdλ[0,s]∣≤∫[0,s]χsHˉdλ[0,s]=Hˉλ[0,s]({s})=0, in agreement with the convention. In both cases, summing over γ with the coefficients cγ and using linearity of the integral, ac(τ)−c⋅vc=c⋅(ΦE(s,τ)vc−vc)=∫[0,s]χτ(u)c⋅Hucdu, which is (∗∗). Consequently, for τ,τ′∈[0,s], writing Δ(τ,τ′) for the set of u∈[0,s] with χτ(u)=χτ′(u), which is [min(τ,τ′),max(τ,τ′)), one has ∣χτ−χτ′∣=1Δ(τ,τ′) and
Time cells. Fix c∈L. By claim 2 of Injection Weights from a Bounded Profile Along Mean-Field Clocks: Label-Rate Form of the Fluctuation Covariance, Weight Bounds, the Prior Quadratic Form, and the Step-Function Injection (horizon s): Cˉc is nondecreasing; the time cells Jˉc,1,…,Jˉc,Jc belong to B[0,s], are pairwise disjoint with union [0,s], so that ∑j1Jˉc,j=1 on [0,s] and ∑jλ[0,s](Jˉc,j)=λ[0,s]([0,s])=s (additivity of λ[0,s] and claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval); and N∫[0,s]1Jˉc,jϕcdu≤μc,j≤μmax, where by the convention of that lemma ∫J⋅du denotes the integral over [0,s] of the integrand multiplied by 1J. By the definition of the injection weights, Nmwc,j=∫[0,s]1Jˉc,jfcdu. Call jfull if Cˉsc≥bjc and partial otherwise; by (CP) and claim 1 of Mean-Square Assembly of the Estimand Linearisation on the Synthetic Copy: Approximation of the Recentred Copy Endpoint by an Affine Function of the Parameter, for full j the set {u∈[0,s]:Cˉuc≥bjc} has the least element τˉc,j and αc,j=ac(τˉc,j), while αc,j=0 for partial j. Two observations. (T1) Let j be full and u∈Jˉc,j; then Δ(τˉc,j,u)⊆Jˉc,j. Indeed, write τˉ=τˉc,j; by minimality and monotonicity, Cˉu′c≥bjc holds exactly when u′≥τˉ; and u∈Jˉc,j means Cˉuc≤bjc and, when j≥2, Cˉuc>bj−1c. If u<τˉ and u′∈[u,τˉ), then Cˉu′c<bjc and Cˉu′c≥Cˉuc, so u′∈Jˉc,j. If u≥τˉ and u′∈[τˉ,u), then bjc≤Cˉu′c≤Cˉuc≤bjc, so Cˉu′c=bjc∈Ic,j and u′∈Jˉc,j (for j=1, Jˉc,1={Cˉc≤b1c} contains u′ as well). (T2) At most one partial j has Jˉc,j=∅: if j<j′ are partial and u∈Jˉc,j′, then j′≥2 and Cˉsc≥Cˉuc>bj′−1c≥bjc, contradicting the partiality of j.
The pairing bound. By the definition of z and the dot product, α⋅z=∑qαqzq=Nm∑qwqαq=∑c∑jαc,j∫[0,s]1Jˉc,jfcdu, while by the display for c⋅ψˉs and ∑j1Jˉc,j=1, c⋅ψˉs=∑c∑j∫[0,s]1Jˉc,jacfcdu. Hence, by linearity and monotonicity of the integral (all integrands being bounded measurable),
Fix c. For full j and u∈Jˉc,j, (T1) and the bound following (∗∗) give ∣αc,j−ac(u)∣=∣ac(τˉc,j)−ac(u)∣≤∣c∣Hˉλ[0,s](Jˉc,j) (monotonicity of λ[0,s]), so the j-th term is at most ∣c∣Hˉλ[0,s](Jˉc,j)⋅2Λ∫[0,s]1Jˉc,jϕcdu≤∣c∣Hˉ2ΛNμmaxλ[0,s](Jˉc,j); summing over the full j and using ∑jλ[0,s](Jˉc,j)=s bounds their total by 2∣c∣HˉΛsμmax/N=2∣c∣Φˉ2ΛEΛsμmax/N. For partial j the term is ∫[0,s]1Jˉc,j∣ac∣∣fc∣du≤2∣c∣Φˉ2⋅2Λμmax/N, and it vanishes when Jˉc,j=∅; by (T2) the partial terms total at most 2∣c∣Φˉ2Λμmax/N. Summing over the l(l−1) labels gives ∣α⋅z−c⋅ψˉs∣≤einj.
The remaining bounds. For q∈L, α⋅aq=(m/N)α⋅eq=(m/N)αq, and ∣αq∣≤2∣c∣Φˉ2 (it is ac(τˉq) or 0), which gives the bound on maxq∣α⋅aq∣. Finally ∥α∥2=∑qαq2≤d⋅2∣c∣2Φˉ4 (claim 1 of Elementary Properties of the Euclidean Norm on Rn), and taking nonnegative square roots, ∥α∥≤2d∣c∣Φˉ2. This completes the proof.