CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Almost-sure martingale convergence need not preserve expectation

Statement

Assume AC. On $([0, 1], B, λ)$ with $F_{0} = {\emptyset, [0, 1]}$ and $F_{n} = σ ((0, 2^{- 1}], \dots, (0, 2^{- n}])$ , take $M_{0} = 1$ and $M_{n} = 2^{n} 1_{(0, 2^{- n}]}$ for $n \geq 1$ . This integrable martingale has $E M_{n} = 1$ for every $n$ but almost-sure limit $M_{\infty} = 0$ with expectation zero. Thus almost-sure convergence need not preserve expectations.

Facts & Assumptions

Given: The hypotheses, objects, and conventions in the Statement.

[F1]

An L1-bounded martingale need not converge in L1 verifies this process is a nonnegative martingale, computes its means, and proves its pointwise limit.

[F2]

Closed martingale characterization explains the missing uniform-integrability hypothesis.

[F3]

The Axiom of Choice is inherited from the martingale construction.

Counterexample

1.1F1

By F1, $M_{n} \to 0$ almost surely while $E M_{n} = 1$ for every $n$ . Therefore $\lim_{n} E M_{n} = 1 \neq 0 = E [\lim_{n} M_{n}] .$ This is the required explicit failure.

2.1F2F3step 1.1∎

If $(M_{n})$ were uniformly integrable, F2 would force $L^{1}$ convergence to its almost-sure limit. That would imply $E ∣ M_{n} ∣ \to 0$ , contradicting the computed value one. Thus the failure is exactly outside the closed/UI regime. AC has only the inherited role in F3.

Depends on

Used by

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Dependency tree · two levels

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Sources

van der Vaart, Martingales, Diffusions and Financial Mathematics, martingale-convergence warnings in §§2.5 and 2.8 (standard reference, not scraped)