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On a finite measure space, the truncated L^1 metric metrises convergence in measure

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ . For measurable $f, g : X \to R$ , define $d (f, g) : = \int \min {∣ f - g ∣, 1} d μ .$

Then:

$d (f, g) = 0$ exactly when $f = g$ $μ$ -almost everywhere, so $d$ descends to a metric on almost-everywhere equivalence classes of measurable functions.
For a sequence $(f_{n})$ of measurable functions and a measurable $f$ , $d (f_{n}, f) \to 0$ if and only if $f_{n} \to f$ in measure.

Facts & Assumptions

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

[L1]

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

[L2]

For a measurable set $E$ and a nonnegative measurable function $u$ , $\int_{E} u d μ = \int u χ_{E} d μ$ . (Integral over a measurable subset)

[L3]

The nonnegative integral is monotone and homogeneous. (Monotonicity and nonnegative homogeneity of the nonnegative integral)

[L4]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

Proof

technique · direct

1.1

One has $d (f, f) = 0$ and $d (f, g) = d (g, f)$ by the pointwise identities $\min {∣ f - f ∣, 1} = 0$ and $\min {∣ f - g ∣, 1} = \min {∣ g - f ∣, 1}$ . Also $∣ f - h ∣ \leq ∣ f - g ∣ + ∣ g - h ∣$ implies $\min {∣ f - h ∣, 1} \leq \min {∣ f - g ∣, 1} + \min {∣ g - h ∣, 1},$ so [L3] gives $d (f, h) \leq d (f, g) + d (g, h)$ .

L3algebra

1.2

Suppose $d (f_{n}, f) \to 0$ . Fix $ε > 0$ and put $A_{n} : = {∣ f_{n} - f ∣ > ε}$ . On $A_{n}$ one has $\min {∣ f_{n} - f ∣, 1} \geq \min {ε, 1}$ , so $\min {ε, 1} μ (A_{n}) \leq d (f_{n}, f) .$ Hence $μ (A_{n}) \to 0$ , which is exactly [L1].

L1L2L3algebra

1.3

Conversely, assume $f_{n} \to f$ in measure. Let $ε \in (0, 1)$ and put $A_{n} : = {∣ f_{n} - f ∣ > ε}$ . Then $d (f_{n}, f) \leq \int_{X ∖ A_{n}} ε d μ + \int_{A_{n}} 1 d μ \leq ε μ (X) + μ (A_{n}) .$ By [L1], $μ (A_{n}) \to 0$ , so ${lim sup}_{n \to \infty} d (f_{n}, f) \leq ε μ (X)$ . Since $ε \in (0, 1)$ was arbitrary and $μ (X) < + \infty$ , it follows that $d (f_{n}, f) \to 0$ .

L1L2L3algebra

1.4

If $d (f, g) = 0$ , then [L4] applied to $\min {∣ f - g ∣, 1}$ shows $\min {∣ f - g ∣, 1} = 0$ almost everywhere, which is equivalent to $f = g$ almost everywhere. Conversely, if $f = g$ almost everywhere, then [L4] gives $d (f, g) = 0$ .

2.1

Let $[f]$ denote the almost-everywhere class of $f$ . If $f \sim f^{'}$ and $g \sim g^{'}$ , then step 1.4 gives $d (f, f^{'}) = d (g, g^{'}) = 0$ , so step 1.1 yields $d (f, g) \leq d (f, f^{'}) + d (f^{'}, g^{'}) + d (g^{'}, g) = d (f^{'}, g^{'}) .$ The same argument with the pairs reversed gives $d (f^{'}, g^{'}) \leq d (f, g)$ , so $d (f, g) = d (f^{'}, g^{'})$ . Thus the value $d ([f], [g])$ is well defined and step 1.1 makes it a metric on the quotient.

step 1.1step 1.4algebra

3.1

Step 2.1 identifies the quotient metric, and steps 1.2 and 1.3 prove that its convergence is exactly convergence in measure. So the truncated distance metrises convergence in measure on finite measure spaces.

step 1.2step 1.3step 2.1∎

Depends on

Used by

Implication table for the main modes of convergence on a finite measure space Remark

Dependency tree · two levels

9 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., Exercise 32 (standard reference, not scraped)