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On a finite measure space, almost-everywhere convergence implies convergence in measure

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ , and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ $μ$ -almost everywhere, then $f_{n} \to f$ in measure.

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ and measurable functions $f_{n}, f : X \to R$ such that $f_{n} \to f$ almost everywhere.

[L1]

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

[L2]

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

[L3]

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

[L4]

If $A \subseteq B$ are measurable, then $μ (A) \leq μ (B)$ . (Measures are monotone)

Proof

technique · direct

1.1

Fix $ε > 0$ , and let $N$ be a measurable null set outside which $f_{n} (x) \to f (x)$ . For $r \geq 0$ put $E_{r} : = ⋃_{n \geq r} {∣ f_{n} - f ∣ > ε}$ . Then $(E_{r})$ is a decreasing sequence of measurable sets, each contained in $X$ , and $⋂_{r = 0}^{\infty} E_{r} \subseteq N$ because outside $N$ only finitely many indices can satisfy $∣ f_{n} (x) - f (x) ∣ > ε$ .

givenL1

2.1

Because $μ (X) < + \infty$ , [L3] applies to $(E_{r})$ . The intersection in step 1.1 is null, so $μ (E_{r}) ↓ μ ⁣ (⋂_{r = 0}^{\infty} E_{r}) = 0.$ For each $r$ one has ${∣ f_{r} - f ∣ > ε} \subseteq E_{r}$ , hence by [L4] $μ ({∣ f_{r} - f ∣ > ε}) \leq μ (E_{r}) \to 0$ . This is exactly [L2].

step 1.1L2L3L4

3.1

Since $ε > 0$ was arbitrary, the sequence converges in measure.

step 2.1L2∎

Depends on

Used by

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Sources

H. L. Royden and P. M. Fitzpatrick, Real Analysis, 4th ed., Proposition 3 (standard reference, not scraped)