TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Riesz's subsequence theorem for convergence in measure

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ in measure, then there is a strictly increasing sequence $(n_{k})_{k \geq 0}$ of natural numbers such that $f_{n_{k}} \to f$ $μ$ -almost everywhere.

The construction below uses the least admissible index at each stage, so no choice principle is spent.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , measurable functions $f_{n}, f : X \to R$ , and convergence in measure of $(f_{n})$ to $f$ .

[L1]

Convergence in measure means that for every real $ε > 0$ , $μ ({∣ f_{n} - f ∣ > ε}) \to 0$ . (Convergence in measure)

[L2]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L3]

For measurable $(E_{k})$ one has $μ (⋃_{k} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k})$ . (Finite and countable subadditivity of measures)

[L4]

If $(E_{r})$ is a decreasing sequence of measurable sets and one $E_{r_{0}}$ has finite measure, then $μ (⋂_{r} E_{r}) = \inf_{r} μ (E_{r})$ . (Continuity from above when one set has finite measure)

Proof

technique · direct

1.1

For each $k \geq 0$ , [L1] applied with $ε : = 2^{- (k + 1)}$ yields an index after which $μ ({∣ f_{n} - f ∣ > 2^{- (k + 1)}}) < 2^{- (k + 1)}$ . Define $n_{0}$ to be the least admissible index for $k = 0$ , and recursively define $n_{k + 1}$ to be the least admissible index larger than $n_{k}$ . Then $(n_{k})_{k \geq 0}$ is strictly increasing and $μ ({∣ f_{n_{k}} - f ∣ > 2^{- (k + 1)}}) < 2^{- (k + 1)} (k \geq 0) .$

L1choose

2.1

Put $A_{k} : = {∣ f_{n_{k}} - f ∣ > 2^{- (k + 1)}}$ and $E_{r} : = ⋃_{k \geq r} A_{k}$ . Then $(E_{r})$ is a decreasing sequence of measurable sets, and step 1.1 together with [L3] gives $μ (E_{r}) \leq \sum_{k = r}^{\infty} μ (A_{k}) \leq \sum_{k = r}^{\infty} 2^{- (k + 1)} = 2^{- r} .$ In particular $μ (E_{0}) \leq 1 < + \infty$ .

step 1.1L3algebra

3.1

Let $N : = ⋂_{r = 0}^{\infty} E_{r}$ . By [L4] and step 2.1, $μ (N) = \lim_{r \to \infty} μ (E_{r}) = 0.$ If $x \in X ∖ N$ , then $x \notin E_{r}$ for some $r$ , hence $x \notin A_{k}$ for every $k \geq r$ . Therefore $∣ f_{n_{k}} (x) - f (x) ∣ \leq 2^{- (k + 1)}$ for all $k \geq r$ , so $f_{n_{k}} (x) \to f (x)$ . By [L2], $f_{n_{k}} \to f$ almost everywhere.

step 2.1L2L4

4.1

The subsequence constructed in step 1.1 has the required almost-everywhere limit.

step 3.1∎

Depends on

Used by

Dependency tree · two levels

13 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

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 2.30 (standard reference, not scraped)
H. L. Royden and P. M. Fitzpatrick, Real Analysis, 4th ed., Theorem 4 (standard reference, not scraped)