CounterexampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Taking out an unbounded factor needs integrability

Statement refuted

Under the AC conditional-class convention, omitting product integrability from the signed $L^{1}$ taking-out rule can leave its left side undefined even when both factors are integrable and the product of the factor with the conditional mean is zero.

Facts & Assumptions

Given: The proposed signed taking-out rule with product integrability omitted; a countable atomic witness will be constructed.

[F1]

The Dirac set function is the indicator that the specified point belongs to an event. (The Dirac set function at a point)

[F2]

Each Dirac set function is a probability measure. (A Dirac set function is a probability measure)

[F3]

Countable nonnegative weighted sums are defined eventwise. (Nonnegative scalar multiples and countable weighted sums of measures)

[F4]

Nonnegative weighted sums of measures are measures. (Nonnegative scalar multiples and countable weighted sums of measures are measures)

[F5]

Integer powers at the positive base two are defined. (Integer powers $a^{m}$ )

[F6]

For |r|<1 the geometric series from n=0 sums to 1/(1-r). (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges)

[F7]

A measure of total mass one is a probability measure. (Probability measures and probability spaces)

[F8]

The signed $L^{1}$ conditional expectation requires an integrable real input. (Conditional expectation as an ae class)

[F9]

The taking-out rule requires integrability of the input product. (Taking out what is known)

[F10]

Integrals on the countable atomic space are sums, by increasing partial sums for nonnegative functions. (Monotone convergence for the integral)

Counterexample

technique · direct

1.1

Let $Ω = N_{\geq 1} \times {- 1, 1}$ with its power-set sigma-algebra. By [F5]–[F6], $S = \sum_{n \geq 1} 2^{- 3 n} = (1 / 8) / (1 - 1 / 8) = 1 / 7$ . Set $w_{n} = 2^{- 3 n - 1} / S$ . These weights are positive finite numbers.

F5F6

2.1

Define $P = \sum_{n \geq 1} w_{n} δ_{(n, - 1)} + \sum_{n \geq 1} w_{n} δ_{(n, 1)}$ . The Dirac probabilities [F1]–[F2] and weighted-sum construction [F3]–[F4] make this a measure on all subsets. Its mass is $2 \sum_{n} w_{n} = S / S = 1$ , so [F7] makes it a probability measure. In particular each atom (n,s) has mass w_n.

step 1.1F1F2F3F4F7

3.1

Let $G = σ ((n, s) \mapsto n)$ , $X (n, s) = s 2^{n}$ and $Z (n, s) = 2^{2 n}$ . Z is finite G-measurable, X is real measurable, and [F10] evaluates their absolute moments as sums. Using [F6], $E ∣ X ∣ = S^{- 1} \sum_{n \geq 1} 2^{- 2 n} = 7 / 3$ and $E ∣ Z ∣ = S^{- 1} \sum_{n \geq 1} 2^{- n} = 7$ . Thus each is integrable.

step 1.1step 2.1F5F6F10

4.1

Each G-event is a union of two-point fibres. On the nth fibre the X integral is $w_{n} (- 2^{n}) + w_{n} 2^{n} = 0$ ; summing is legitimate by the finite absolute moment in step 3.1. Consequently the zero function has every defining event integral and is a version of $E [X ∣ G]$ by [F8]. Hence $Z E [X ∣ G] = 0$ is integrable.

step 3.1F8

5.1

But $Z X (n, s) = s 2^{3 n}$ . Its positive-part integral is $\sum_{n \geq 1} w_{n} 2^{3 n} = \sum_{n \geq 1} (2 S)^{- 1} = \infty$ , and the negative-part integral has exactly the same value. These sums are nonnegative integrals by [F10]. Thus the signed expectation would require infinity minus infinity; ZX is not an $L^{1}$ input to [F8]. The left side $E [Z X ∣ G]$ of [F9] is undefined in that sense although the proposed right side is zero. This proves the failure when the product-integrability hypothesis is omitted.

step 2.1step 4.1F8F9F10∎

Source notes

Durrett Theorem 4.1.14, printed pp.212–213, states the integrable-product hypothesis. The constructed atomic counterexample, its normalization and all moments are independently calculated here.

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Sources

Durrett, Probability: Theory and Examples, 5th ed. (standard reference, not scraped)