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VC-subgraph covering entropy with its finite-r logarithmic bound

Statement

Let (X,A,Q) be a probability space, let 1≤r<∞, and let F be a nonempty VC-subgraph class of measurable real functions with finite VC index V. Suppose a measurable envelope F satisfies ∣f(x)∣≤F(x) for every f∈F and x∈X, and

0<∥F∥Q,r:=(∫Fr dQ)1/r<∞.

For every 0<ϵ<1, put K0:=3e2/(e−1) and K:=4K0. Then

N ⁣(ϵ∥F∥Q,r,F,Lr(Q))≤[K(2/ϵ)rlog⁡ ⁣(3e(2/ϵ)r)]V−1≤(Aϵ)ArV,A:=12eK+2.

The exponent in the first bound is the finite-r logarithmic form. No Axiom of Choice or Countable Choice is used in this proof.

Facts & Assumptions

Given: A probability measure Q, a nonempty VC-subgraph class F with index V, an envelope F with 0<∥F∥Q,r<∞, 1≤r<∞, and 0<ϵ<1.

[F1]

The subgraph of f is Cf={(x,t):t<f(x)}, and the class has first-unshattered index V (VC-subgraph classes, envelopes, and the first-unshattered index).

[F2]

Lr(Q) distance is ∥f−g∥Q,r; covering centers belong to the covered class (Covering, packing, bracketing numbers, and metric entropy).

[F3]

A nonempty measurable set class of finite VC index W satisfies N(δ,C,L1(μ))≤[K0δ−1log⁡(3e/δ)]W−1 for 0<δ<1 and any probability measure μ (Logarithmic L-r covering bound for VC classes of sets).

[F4]

If h≥0 is measurable, then h dQ is a measure with total mass ∫h dQ (The measure with density f relative to μ, The indefinite integral of a nonnegative measurable function is a measure).

[F6]

For the finite grid UJ with trapezoidal weights from A finite uniform grid approximates interval length, every interval [a,b)⊆[−1,1] has grid mass within 1/J of (b−a)/2.

[F7]

Coordinate lifts on a product space are measurable, and arithmetic and order operations on measurable real functions preserve measurability; strict threshold sets of measurable real functions are measurable (The product sigma-algebra and its finite iterates, Arithmetic and lattice operations preserve measurability whenever they are defined, Threshold characterisations of real-valued and extended-real-valued measurability).

[F8]

For r>0, t↦tr is increasing on [0,∞) and sends nonnegative measurable functions to measurable functions (Positive real powers preserve measurability).

[F9]

For each δ>0, there is an integer J≥1 with 1/J≤δ (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[F12]

e=exp⁡(1)>1, since the exponential is strictly increasing and exp⁡(0)=1 (The real exponential function and the number e by a power series, The exponential function is strictly increasing).

Proof

Given: The objects and hypotheses of the statement.

1.1F4F7F8givenalgebra

Put M:=∫Fr dQ=∥F∥Q,rr. By [F8] Fr is measurable; the hypotheses give 0<M<∞. By [F7], the density Fr/M is measurable. Define R by dR=(Fr/M) dQ; its total mass is one by [F4].

2.1F5F6F9givenalgebrastep 1.1

Set δ:=ϵr/2r+1. Then 0<δ<1, so [F9] supplies an integer J≥1 with 1/J≤δ. Use the grid UJ and probability νJ of [F6], and form the product probability Q~:=R⊗νJ on X×UJ using [F5].

3.1F1F7givenstep 2.1

Let X+:={x:F(x)>0} and define Af:={(x,u):x∈X+, uF(x)<f(x)} for f∈F. The coordinate projections are measurable because their preimages of measurable sets are rectangles; hence [F7] makes X+ and each Af product-measurable. On X+×UJ, the map ψ(x,u):=(x,F(x)u) is injective and Af=ψ−1(Cf). If a finite set of pairs is shattered by {Af:f∈F}, its image under ψ is a same-size set shattered by {Cf:f∈F}. Thus the nonempty set class CJ:={Af:f∈F} has finite VC index VJ≤V.

4.1F6F8step 1.1step 3.1given

Fix f,g∈F. For x∈X+, put a(x):=min⁡(f(x),g(x))/F(x) and b(x):=max⁡(f(x),g(x))/F(x). The envelope gives −1≤a(x)≤b(x)≤1, and the section (Af△Ag)x is exactly {u∈UJ:a(x)≤u<b(x)}. By [F6], its νJ-mass differs from ∣f(x)−g(x)∣/(2F(x)) by at most 1/J. The set X∖X+ has R-measure zero, since the density Fr/M vanishes there.

4.2F3F12givenchoosestep 2.1step 3.1

Apply [F3] to CJ under Q~ at radius δ. Since e>1, K0=3e2/(e−1)>1, and δ<1 makes log⁡(3e/δ)>1; thus the bracketed base is greater than one, so VJ≤V yields N(δ,CJ,L1(Q~))≤[K0δ−1log⁡(3e/δ)]V−1<∞. Choose a cover with this many centers Ag1,…,Agk, each represented by some gj∈F. The cover uses finitely many centers, so this uses only finite choice.

5.1F2F5F8F10step 1.1step 4.1algebra

Integrating the section estimate under R and using [F5] gives ∣Q~(Af△Ag)−12∫X+∣f−g∣F dR∣≤1J. For x∈X+, put a:=∣f(x)−g(x)∣≤2F(x). If a=0 the inequality ar≤2r−1F(x)r−1a is immediate; if a>0 and r>1, [F8] raises a≤2F(x) to the positive power r−1 and multiplication by a gives it; for r=1 it is equality. Hence [F2] gives ∥f−g∥Q,rr≤2r−1M∫X+∣f−g∣F dR≤2rM(Q~(Af△Ag)+1J). The last inequality uses the discrepancy bound and the definition of R.

6.1step 2.1step 5.1step 4.2algebra

For every f∈F, some center satisfies Q~(Af△Agj)≤δ. Steps 5.1 and 2.1 then give ∥f−gj∥Q,rr≤2rM(δ+1/J)≤2r+1Mδ=ϵrM. Thus the function class is covered by the same k centers at radius ϵ∥F∥Q,r, and its covering number is no larger than the set covering number in step 4.2.

7.1F11F12step 4.2step 6.1algebra

Put L:=log⁡(3e(2/ϵ)r). From δ−1=2(2/ϵ)r and log⁡(3e/δ)=L+log⁡2≤2L (since 3e(2/ϵ)r>2, so L>log⁡2), the base in step 4.2 is at most 4K0(2/ϵ)rL=K(2/ϵ)rL. Since this base exceeds one, step 6.1 proves the first asserted entropy bound.

8.1F11F12step 7.1algebra

Put η:=(ϵ/2)r∈(0,1). For u≥1, the function h(u):=u−log⁡u has h(1)=1 and h′(u)=1−1/u≥0; the mean value theorem [F11] gives log⁡u≤u. Hence Kη−1log⁡(3e/η)≤3eKη−2=3eK(2/ϵ)2r. Let A:=12eK+2. From [F12], e−1<e, so K0=3e2/(e−1)>3e>3 and K=4K0>4. Thus x:=3eK>1 and A/2=6eK+1≥x, whence (A/2)2r≥x2r≥x because r≥1. This gives 3eK(2/ϵ)2r≤(A/ϵ)2r.

9.1step 7.1step 8.1algebra∎

The first bound is the step 7.1 base raised to V−1. Since A/ϵ>1, A≥2, and V≥1, we have 2r(V−1)≤ArV. Thus the first bound is at most (A/ϵ)ArV, proving the second inequality.

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