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Realizable sample complexity for finite-VC classes

Statement

Let (X,Σ) be a nonempty measurable space, let H be a nonempty class of measurable binary functions with finite VC dimension d, put Z=X×{0,1}, and let P be a probability law on Z. Fix an integer m≥1 and a map A:Zm→H for which S↦RP(A(S)) and S↦R^S(A(S)) are measurable, where RP(h):=P{(x,y):h(x)≠y},R^S(h):=1m∑i=1m1{h(Xi)≠Yi}. Suppose P is realizable by H and A is consistent Pm-almost surely. For 0<ϵ<1, assume the events Em,ϵ:={S∈Zm:∃h∈H, RP(h)>ϵ, R^S(h)=0} and Bm,ϵ:={(S,S′)∈Zm×Zm:∃h∈H, RP(h)>ϵ, R^S(h)=0, R^S′(h)≥ϵ/2} are measurable in their respective product sigma-algebras; this holds when H is countable. For 0<ϵ,δ<1 and S∼Pm, if d=0 then RP(A(S))=0 almost surely for every such rule. If d≥1 and m≥64ϵ[dlog⁡ ⁣(64eϵ)+log⁡ ⁣(4δ)], where log⁡ is natural and e=exp⁡(1), then P{RP(A(S))≤ϵ}≥1−δ. This is a conditional upper bound for consistent risk-measurable learners; it asserts no existence or computational-efficiency result.

Statement references

Measurable PAC learning criteria for a possibly infinite VC class: defines a realizable distribution and the realizable PAC event for possibly infinite classes, with the required risk measurability.

Mohri et al., Chapter 3, Exercise 3.30: identifies the realizable VC upper-rate problem; the argument here supplies the complete finite-trace proof and explicit sample-size constants.

Bartlett and Mendelson, Theorem 5 and Appendix B: independent classification-risk generalization context; that theorem is not used as a realizable sample-size proof.

Facts & Assumptions

Given: A measurable binary class H with finite VC dimension d, a law P realized by some h∗∈H, a consistent risk-measurable rule A, m≥1, and 0<ϵ,δ<1.

[F1]

Realizable PAC learning means that every law with a zero-risk member satisfies the stated high-probability risk event; consistency means R^S(A(S))=0. The definition makes no learner-existence claim. Measurable PAC learning criteria for a possibly infinite VC class

[F2]

A finite indexed tuple is shattered when all binary label vectors occur, and d=VCdim⁡(H). Indexed shattering and VC dimension of a binary class

[F3]

ΠH(n) counts the largest number of traces on indexed n-tuples. Growth function of a binary hypothesis class

[F4]

If 1≤d≤n, then ΠH(n)≤(en/d)d. Sauer–Shelah growth bound for indexed samples

[F5]

Population risk is the zero-one error probability and empirical risk is the average error indicator; for each fixed h these are well-defined measurable quantities. Population and empirical risk, Population and empirical risks are measurable

[F6]

IID coordinates have the finite product joint law; finite product probability measures exist, and Tonelli permits integration of the nonnegative event indicators over the two samples. Identical distribution and IID families, For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, Independent random elements have product joint law, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

[F7]

Measurable functions of independent observations remain independent, and expectations of products of independent nonnegative variables factor. Measurable coordinatewise functions preserve independence, Expectations factor over finite products of independent random variables

[F8]

The expectation of a finite sum of integrable indicators is the sum of their expectations. Linearity, monotonicity, and the modulus bound for expectation

[F9]

For a nonnegative random variable W and a>0, P(W≥a)≤E[W]/a. Markov's inequality for random variables

[F10]

A finite union of events has probability at most the sum of their probabilities. Basic identities for a probability measure

[F12]

The nonnegative integral is monotone; on [1,2], 1/t≥1/2, and on [1,5/4], [5/4,3/2], and [3/2,2] it is respectively at most 1, 4/5, and 2/3. Consequently 1/2≤log⁡2≤47/60<4/5. Monotonicity and nonnegative homogeneity of the nonnegative integral

[A1]

No Axiom of Choice is used: the sample partitions, trace patterns, and event unions in the proof are finite, and existential witnesses are used pointwise rather than selected as a sample-dependent function.

Proof

Given: The hypotheses in the Statement.

1.1F1F2givenA1

If d=0, any two members of H agree at every x∈X, since otherwise the one-point tuple (x) would be shattered. Thus H has one function. Realizability makes that function have zero risk, and every H-valued rule outputs it, proving the d=0 clause.

1.2F5F6F7F8F9F11F12F13givenA1

Suppose d≥1 and put L=log⁡(64e/ϵ)>1 and D=log⁡(4/δ)>0; the sample-size premise gives mϵ≥64(dL+D)>8. Let ES:={∃h∈H:RP(h)>ϵ, R^S(h)=0} and let BS,S′ be the event that some such h also has R^S′(h)≥ϵ/2, where S′ is an independent IID sample. For each fixed S∈ES, fix one witness h for this S only, with no sample-dependent selection function, and let Kh=∑i=1m1{h(Xi′)≠Yi′}. By [F5]–[F8], Kh is a sum of independent Bernoulli indicators with mean mp, where p=RP(h)>ϵ. Factorization [F7] gives E[2−Kh]=(1−p/2)m≤e−mp/2 by [F13]. On Kh<mp/2, 2−Kh>2−mp/2, so [F9] gives P(Kh<mp/2)≤exp⁡[−mp(1−log⁡2)/2]. From [F11]–[F12], 1/2≤log⁡2<4/5, so this probability is at most e−mp/10≤e−mϵ/10≤e−4/5≤e−log⁡2=1/2. Since mp/2>mϵ/2, this proves PS′(BS,S′∣S)≥1/2 for each S∈ES. Integrating the measurable sections by [F6] yields P(ES)≤2P(BS,S′).

2.1F2F3F6F10step 1.2A1

View 2m IID observations as a combined sample T. By [F6], permuting its coordinates preserves its product law, so the joint law of the first and second halves equals the average over all uniform choices of an m-index training subset I⊆[2m] and its complement. For fixed T, each h determines an error-index set Jh={i:h(Xi)≠Yi}, and the number of distinct such sets is at most ΠH(2m) because the labels Yi only flip coordinates of the trace. If ∣Jh∣=k>m, no training subset can be disjoint from all errors. If k≤m, the fraction of training subsets disjoint from Jh is ∏j=0k−1(m−j)/(2m−j)≤2−k: after j specified errors have been assigned to the ghost half, the next specified index lies there with probability (m−j)/(2m−j)≤1/2. Therefore only error patterns with k≥mϵ/2 can witness B, and the finite union bound [F10] gives P(BS,S′)≤ΠH(2m)2−mϵ/2.

3.1F1F4step 1.2step 2.1givenalgebra

Combining steps 1.2 and 2.1, P(ES)≤2ΠH(2m)2−mϵ/2. The sample-size hypothesis implies m≥d, so [F4] at n=2m gives P(ES)≤2(2em/d)d2−mϵ/2. Put NA:={S:R^S(A(S))>0}. This event is measurable because S↦R^S(A(S)) is measurable, and it is Pm-null by the almost-sure consistency assumption. For S∉NA with RP(A(S))>ϵ, A(S) witnesses ES, so the failure event {S:RP(A(S))>ϵ} is contained in ES∪NA.

4.1F11F12F13step 3.1givenalgebra

Put L=log⁡(64e/ϵ), D=log⁡(4/δ), and M=64(dL+D)/ϵ. Then L>1, D>0, and the hypothesis gives m≥M. For real t≥M, define q(t)=log⁡2+dlog⁡(2et/d)−(log⁡2)ϵt/2. Its derivative is d/t−(log⁡2)ϵ/2≤ϵ/(64L)−ϵ/4<0, using log⁡2≥1/2 from [F11]–[F12]; hence q(m)≤q(M). The identity log⁡(2eM/d)=L+log⁡2+log⁡(L+D/d), together with log⁡(L+D/d)≤log⁡L+D/(dL) (from [F11], [F13]), log⁡L≤L, log⁡2≤L, and log⁡2≥1/2, gives q(M)≤L−13dL+D/L−16D. Consequently q(M)−log⁡δ≤L−13dL+D/L−15D−log⁡4<0, because d≥1, L>1, and D>0. Exponentiating shows 2(2em/d)d2−mϵ/2≤δ.

5.1step 3.1step 4.1givenA1∎

By step 3.1 the failure event is contained in ES∪NA, with P(NA)=0, and step 4.1 bounds P(ES) by δ. Hence PS∼Pm{RP(A(S))≤ϵ}≥1−δ. The argument used no arbitrary-index choice axiom.

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