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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Borel sigma-algebra of continuous path space is generated by coordinates

Statement

Put C=C([0,∞),R), give C the uniform-on-compacts topology, and write πt(f)=f(t). Then B(C)=σ(πt:t≥0)=σ(πq:q∈Q∩[0,∞)). Here each expression on the right denotes the smallest sigma-algebra on C making every displayed coordinate map measurable. No choice principle is used.

Facts & Assumptions

Given: The path space C, its uoc metric d, and its coordinate maps as in the Statement. Write G for the sigma-algebra generated by all coordinates and GQ for that generated by the nonnegative rational coordinates.

[F1]

The uoc metric is d(f,g)=∑n≥12−n(1∧Mn(f,g)),Mn(f,g)=max⁡0≤t≤n∣f(t)−g(t)∣, and its topology is compact convergence. Uniform-on-compacts metric on continuous path space

[F3]

Countable suprema and pointwise limits of measurable real functions are measurable; finite sums, scalar multiples, positive parts, and absolute values preserve measurability. Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable Closure properties of measurable functions used by the integral

[F4]

The rationals are countable and dense in the reals. There is a fixed bijection between N2 and N, recursion constructs nested finite codes, finite products of countable sets are countable, and a subset of a countable set is countable. Q is countably infinite The rationals embed densely in the reals N×N≈N The recursion theorem A product of two at most countable sets is at most countable Every subset of an at most countable set is at most countable

Proof

technique · direct
1.1F1F2

Fix t≥0 and choose an integer n≥1 with t≤n. If ε>0, then d(f,g)<2−nmin⁡{1,ε} implies 2−n(1∧Mn(f,g))<2−nmin⁡{1,ε} and hence ∣πt(f)−πt(g)∣≤Mn(f,g)<ε. Thus every πt is continuous, so it is Borel measurable. Minimality in [F2] gives GQ⊆G⊆B(C).

1.2F2F3F6

Fix t≥0 and put qj=2−j⌊2jt⌋. Then qj∈Q∩[0,∞) and 0≤t−qj<2−j, so qj→t by [F6]. For every f∈C, continuity gives πqj(f)→πt(f). Every πqj is GQ-measurable by definition, so [F3] makes πt measurable. Hence G⊆GQ by minimality, and therefore G=GQ.

1.3F4construct

Let D be the family of paths which, for some integers N,m≥1, are affine on each interval [k/m,(k+1)/m] for 0≤k<mN, take rational values at all grid points k/m, and are constant after N. This family is countable without choice. Indeed, fix bijections e:N→Q and p:N→N2. Nested use of p−1 codes every finite sequence of naturals by one natural (and repeated application of p decodes it); composing entries with e codes every finite rational sequence. A further finite nesting codes (N,m) together with that sequence, giving a surjection from a subset of N onto D. Thus [F4] makes D at most countable.

1.4F1F4F5F6construct

The family D is uoc dense. Given f∈C and ε>0, use [F6] to choose N with ∑n>N2−n<ε/2. By [F5], f is uniformly continuous on [0,N]; choose m so that the oscillation of f over distances at most 1/m is less than ε/4. By rational density, choose rational ak with ∣ak−f(k/m)∣<ε/4 for the finitely many 0≤k≤mN (finite induction, not a choice principle), and let g∈D interpolate these values and remain constant after N. On a grid interval, convex interpolation and the triangle inequality give ∣g(t)−f(t)∣<ε/2. Consequently d(f,g)≤∑n=1N2−nMn(f,g)+∑n>N2−n<ε. This also covers t=0 and t=N.

1.5F1F3F4

Fix f∈C. For n≥1, continuity and rational density give Mn(g,f)=sup⁡q∈Q∩[0,n]∣πq(g)−f(q)∣. Enumerating that rational set, [F3] makes Mn( ⋅ ,f) GQ-measurable. The finite partial sums of the metric formula are measurable by [F3] and converge pointwise to d( ⋅ ,f), so this distance is measurable. Hence every metric ball, with an arbitrary center, belongs to GQ.

2.1F2F4F7step 1.1step 1.2step 1.4step 1.5∎

The balls B(h,r) with h∈D and positive rational r form a countable base: countability follows from [F4], while density from step 1.4 and the triangle inequality put such a ball around every point inside any prescribed open ball. For an open U, let VU be the subfamily of these basic balls which are contained in U. It is countable by [F4], every member belongs to GQ by step 1.5, and U=⋃VU. Thus every open set is in GQ; [F2] yields B(C)⊆GQ. Combining this with steps 1.1 and 1.2 proves all stated equalities. The zero path shows C and D are nonempty; singleton and zero-time coordinates cause no exception, and no countable family of nonempty sets was selected.

Source notes

Van der Vaart and Wellner use the standard fact that the Borel sigma-algebra on a separable continuous-function space is generated by evaluations. The proof above supplies the complete uoc and rational-coordinate argument, including an explicit choice-free countable basis.

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