CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Convergence in $L^{p}$ implies convergence in measure

Statement

Let $1 \leq p \leq \infty$ , let $u_{n}, u \in L^{p} (μ)$ , and choose measurable representatives $f_{n}, f$ . If $∥ u_{n} - u ∥_{p} \to 0$ , then $f_{n} \to f$ in measure.

Facts & Assumptions

Given: Representatives $f_{n}, f$ of classes $u_{n}, u \in L^{p} (μ)$ with $∥ u_{n} - u ∥_{p} \to 0$ .

[L1]

Convergence in measure means the bad-set measures $μ ({∣ f_{n} - f ∣ > ε})$ tend to $0$ for every $ε > 0$ (Convergence in measure).

[L2]

The class norm agrees with the representative norm (The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

[L3]

Chebyshev-Markov gives $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ for nonnegative measurable $h$ (Chebyshev-Markov inequality for the integral).

[L4]

If $∥ g ∥_{\infty} < \infty$ , then $∣ g ∣ \leq ∥ g ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

Proof

technique · For $1 \le p < infinity$, apply Chebyshev-Markov to $|f_n-f|^p$. For $p = infinity$, the essential-supremum bound makes the bad set null for all large $n$

1.1L3L2givenalgebra

Assume $1 \leq p < \infty$ and fix $ε > 0$ . Apply [L3] to [L2, L3, given, algebra] $h : = ∣ f_{n} - f ∣^{p}$ and $t : = ε^{p}$ : $μ ({∣ f_{n} - f ∣ > ε}) = μ ({∣ f_{n} - f ∣^{p} > ε^{p}}) \leq ε^{- p} \int ∣ f_{n} - f ∣^{p} d μ = ε^{- p} ∥ u_{n} - u ∥_{p}^{p} .$ By [L2], the right-hand side tends to $0$ .

1.2L2L4given

Assume $p = \infty$ and fix $ε > 0$ . Because $∥ u_{n} - u ∥_{\infty} \to 0$ , [L2, L4, given] for all large $n$ one has $∥ u_{n} - u ∥_{\infty} < ε$ . Then [L4] gives $∣ f_{n} - f ∣ \leq ε$ almost everywhere, so $μ ({∣ f_{n} - f ∣ > ε}) = 0.$ Hence the bad-set measures are eventually $0$ .

2.1step 1.1step 1.2L1∎

Step 1.1 proves the finite- $p$ case and step 1.2 proves the [step 1.1, step 1.2, L1] $p = \infty$ case, so [L1] gives convergence in measure.

Depends on

Used by

Calderon-Zygmund operators map H¹ boundedly into L¹ Theorem

Dependency tree · two levels

18 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

John K. Hunter, Measure Theory, Chapter 13 and Chapter 15 (standard reference, not scraped)
Gerald B. Folland, Real Analysis, 2nd ed., Section 2.5 (standard reference, not scraped)

Convergence in LpLp implies convergence in measure