CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Convergence in probability need not imply $L^{p}$ convergence

Statement refuted

Convergence in probability need not imply $L^{p}$ convergence, for $1 \leq p < \infty$ .

Facts & Assumptions

Given: $1 \leq p < \infty$ , events $E_{n}$ with $P (E_{n}) = 1 / (n + 1)$ , and $X_{n} = (n + 1)^{1 / p} 1_{E_{n}}$ for $n \in N$ .

[L1]

Convergence in probability is fixed-threshold tail convergence (Convergence in probability).

[L2]

$L^{p}$ convergence requires the $p$ th moment of the difference to tend to zero ( $L^{p}$ convergence for random variables).

Counterexample

technique · direct

1.1

For fixed $ε > 0$ , $X_{n}$ is nonzero only on $E_{n}$ , hence [L1] $P (∣ X_{n} ∣ > ε) \leq 1 / (n + 1) \to 0$ . Thus $X_{n} \to 0$ in probability by [L1].

2.1

Nevertheless, [step 1.1, L2] $E ∣ X_{n} ∣^{p} = (n + 1) P (E_{n}) = 1$ for all $n$ . By [L2], $X_{n}$ does not converge to zero in $L^{p}$ .

step 1.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Roch, Lecture 3: Modes of convergence, Example 3.11 (standard reference, not scraped)

Convergence in probability need not imply LpLp convergence