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Dominated convergence for the measure integral

Statement

Let $(X, Σ, μ)$ be a measure space and let $f_{n} : X \to C$ be measurable with $f_{n} \to f$ almost everywhere. Suppose $g \geq 0$ is measurable and $\int g d μ < \infty$ , and $∣ f_{n} ∣ \leq g$ almost everywhere for every $n$ . Let $f_{0} (x)$ be the finite complex pointwise limit of $(f_{n} (x))$ where it exists, and $0$ elsewhere. Then $f_{0}$ is measurable, equals $f$ almost everywhere, is integrable, and $\int f_{n} d μ \to \int f_{0} d μ$ and $\int ∣ f_{n} - f_{0} ∣ d μ \to 0$ . If the specified $f$ is measurable, these same conclusions hold with $f$ in place of $f_{0}$ . In particular this is automatic when $μ$ is complete. On a noncomplete measure space an arbitrary almost-everywhere representative need not be measurable, so no integral of such a nonmeasurable $f$ is asserted.

Facts & Assumptions

Given: As stated in the item.

[F1]

Measure space and integral: integrability, signed/complex integrals, and the reference-measure convention are as defined there.

[F2]

Fatou's lemma: for nonnegative measurable functions, the integral of the pointwise liminf is at most the liminf of the integrals.

[F3]

Monotone convergence for the measure integral: the nonnegative integral is additive and positively homogeneous.

[F4]

Arithmetic and lattice operations preserve measurability whenever they are defined, Composition with a Borel measurable outer map preserves measurability, and Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable: component arithmetic, measurable compositions and pointwise real limits preserve measurability; The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts identifies complex Cauchy sequences with finite complex limits.

[A1]

The sequence and envelope are measurable, converge almost everywhere, and are dominated almost everywhere by the nonnegative integrable function $g$ .

Proof

Given: A measure space, almost-everywhere convergence, and an integrable dominating function $g \geq 0$ .

1.1F1F4A1algebra

The set on which $(f_{n})$ is Cauchy is measurable: write it as the countable intersection/union/intersection of sets ${∣ f_{n} - f_{k} ∣ < 1 / m}$ using [F4]. On this set the finite real and imaginary limits exist; their measurable limit functions, set to zero on its complement, define the measurable $f_{0}$ . The assumed almost-everywhere convergence gives $f_{0} = f$ outside a measurable null set. Enlarge that set by the countably many measurable domination exceptions to a measurable null set $N$ . Set $f_{n}$ and $f_{0}$ to zero on $N$ , which preserves their measurability. A nonnegative measurable function supported on $N$ has integral zero by the simple-minorant definition, so these changes preserve every defined integral. Convergence and domination now hold everywhere, and $∣ f_{0} ∣ \leq g$ proves integrability. This does not redefine the potentially nonmeasurable $f$ and then claim it was measurable.

2.1F1F2F3step 1.1

First take all functions real. The nonnegative functions $f_{n} + g$ converge pointwise to $f_{0} + g$ . Fatou gives $\int (f_{0} + g) \leq {lim inf}_{n} \int (f_{n} + g) = \int g + {lim inf}_{n} \int f_{n}$ , so $\int f_{0} \leq {lim inf}_{n} \int f_{n}$ . Applying Fatou to $g - f_{n} \geq 0$ gives $\int (g - f_{0}) \leq {lim inf}_{n} \int (g - f_{n}) = \int g - {lim sup}_{n} \int f_{n}$ , hence ${lim sup}_{n} \int f_{n} \leq \int f_{0}$ . Additivity of integrable signed integrals follows by writing each as a difference of its positive and negative parts and using [F3]. The inequalities imply convergence of integrals.

3.1F1F2F3step 2.1∎

For complex $f_{n}$ , apply step 2.1 separately to their real and imaginary parts, each dominated by $g$ ; their integral limits combine to $\int f_{0}$ . To prove $L^{1}$ convergence, apply Fatou to the nonnegative functions $2 g - ∣ f_{n} - f_{0} ∣$ , which converge to $2 g$ ; it gives $2 \int g \leq 2 \int g - {lim sup}_{n} \int ∣ f_{n} - f_{0} ∣$ , so the latter limsup is zero. Therefore $\int ∣ f_{n} - f_{0} ∣ \to 0$ as claimed.

Remarks

If $f$ is measurable and equals $f_{0}$ outside a measurable null set, their measurable difference has zero integral and zero $L^{1}$ norm; hence the same limits hold for $f$ . If the measure is complete, every subset of the exceptional null set is measurable, so every almost-everywhere representative is measurable. The completeness qualification is necessary: on a noncomplete space, a nonmeasurable subset $A$ of a measurable null set gives $f = 1_{A}$ and $f_{n} = g = 0$ . All stated almost-everywhere convergence and domination conditions hold, but $f$ has no measurable integral; the canonical representative is $f_{0} = 0$ .

Dominated convergence for the measure integral

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources