CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31

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A Cauchy sequence in calligraphic $L^{p}$ can converge to two distinct functions

Statement refuted

In the representative space $L^{p} (μ)$ , the $p$ -distance always gives unique limits.

Facts & Assumptions

Given: A nonzero null-supported function from A nonzero function on a null set has zero $L^{p}$ seminorm.

[L1]

The previous counterexample supplies a measurable function $h \equiv̸ 0$ with $∥ h ∥_{p} = 0$ (A nonzero function on a null set has zero $L^{p}$ seminorm).

[L2]

$L^{p} (μ)$ is the representative function space before quotienting (The function space $L^{p} (μ)$ for $0 < p < \infty$ ).

Counterexample

technique · Take the constant sequence equal to a nonzero function supported on a null set. Its distance to the zero function is $0$, so it converges to both itself and $0$ in the pseudometric on calligraphic $L^p$

1.1L1L2

Let $f_{n} : = h$ for every $n$ , where $h$ is the function from [L1]. Then $(f_{n})$ is constant, hence Cauchy in the representative $p$ -distance.

2.1L1step 1.1

Also $∥ f_{n} - h ∥_{p} = 0, ∥ f_{n} - 0 ∥_{p} = ∥ h ∥_{p} = 0$ for every $n$ . So the same sequence converges both to $h$ and to $0$ , even though those two functions are distinct pointwise.

3.1step 2.1∎

Therefore the representative-space distance does not have unique limits before passing to almost-everywhere classes.

Depends on

Used by

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Sources

Sheldon Axler, Measure, Integration & Real Analysis, Section 7B (standard reference, not scraped)

A Cauchy sequence in calligraphic LpLp can converge to two distinct functions