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Dominated convergence is a Vitali corollary

Statement

Let $(X, A, μ)$ be a sigma-finite measure space. Let $f_{n} : X \to R$ be measurable, let $g \in L^{1} (μ)$ be nonnegative, and suppose $∣ f_{n} ∣ \leq g$ almost everywhere for every $n$ and $f_{n} \to f$ almost everywhere. Then $f \in L^{1} (μ)$ and $f_{n} \to f$ in $L^{1} (μ)$ .

Facts & Assumptions

Given: A sigma-finite measure space $(X, A, μ)$ , measurable functions $f_{n}, f : X \to R$ , and a nonnegative integrable function $g$ with $∣ f_{n} ∣ \leq g$ almost everywhere and $f_{n} \to f$ almost everywhere.

[L1]

A dominated family is uniformly integrable. (Dominated families are uniformly integrable)

[L2]

On a sigma-finite measure space, convergence in measure together with uniform integrability and tightness implies convergence in $L^{1}$ . (Vitali convergence theorem on finite and sigma-finite measure spaces)

[L3]

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

[L4]

If $h : X \to [0, + \infty]$ is measurable and $t > 0$ , then $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ . (Chebyshev-Markov inequality for the integral)

[L5]

For an increasing sequence of nonnegative measurable functions, the integrals increase to the integral of the limit. (Monotone convergence for the integral)

[L6]

If two integrable functions are equal almost everywhere, then their integrals over every measurable set agree. (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree)

[L7]

Sigma-finiteness means that $X = ⋃_{m} X_{m}$ for some measurable sets $X_{m}$ of finite measure. (Finite, sigma-finite, and semifinite measures)

[L8]

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)

[L9]

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

[L10]

Countable unions of null sets are null. (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1

Let $N_{*}$ be a null set outside which $f_{n} (x) \to f (x)$ , and for each $n \in N$ let $N_{n}$ be a null set outside which $∣ f_{n} ∣ \leq g$ . Put $N : = N_{*} \cup ⋃_{n = 0}^{\infty} N_{n} .$ By [L10], the set $N$ is null. On $X ∖ N$ one has $f_{n} (x) \to f (x)$ for every $x$ and $∣ f_{n} (x) ∣ \leq g (x)$ for every $n$ , so also $∣ f (x) ∣ \leq g (x)$ there.

L10construct

1.2

By [L1], the family ${f_{n} : n \in N}$ is uniformly integrable.

2.1

By [L7], choose measurable sets $Y_{m}$ of finite measure with $X = ⋃_{m} Y_{m}$ , and put $X_{m} : = ⋃_{j = 0}^{m} Y_{j}$ . Then $X_{m}$ is measurable of finite measure, $X_{m} ↑ X$ , and hence $g χ_{X_{m}} ↑ g$ . Fact [L5] gives $\int_{X_{m}} g d μ ↑ \int g d μ$ , so $\int_{X ∖ X_{m}} g d μ \to 0$ . Fix $ε > 0$ and choose $m$ with $\int_{X ∖ X_{m}} g d μ < ε$ . For each $n$ , put $u_{n, m} : = ∣ f_{n} ∣ χ_{(X ∖ X_{m}) ∖ N}$ . Then $u_{n, m} = ∣ f_{n} ∣ χ_{X ∖ X_{m}}$ almost everywhere and $u_{n, m} \leq g χ_{X ∖ X_{m}}$ pointwise, so [L6], [L8], and [L9] give $\int_{X ∖ X_{m}} ∣ f_{n} ∣ d μ = \int_{X ∖ X_{m}} u_{n, m} d μ \leq \int_{X ∖ X_{m}} g d μ < ε .$ So the family ${f_{n}}$ is tight.

step 1.1L5L6L7L8L9algebra

2.2

Let $ε, η > 0$ . Choose $m$ so that $\int_{X ∖ X_{m}} g d μ < ε η / 8$ . On the finite measure space $X_{m}$ , step 1.1 and [L3] make $f_{n} \to f$ in measure. Define $h_{n} : = ∣ f_{n} - f ∣ χ_{(X ∖ X_{m}) ∖ N}$ . Then ${h_{n} > ε} = {∣ f_{n} - f ∣ > ε} \cap ((X ∖ X_{m}) ∖ N),$ and $h_{n} \leq 2 g χ_{X ∖ X_{m}}$ pointwise. Since $N$ is null, $μ ({∣ f_{n} - f ∣ > ε} \cap (X ∖ X_{m})) = μ ({h_{n} > ε}) .$ Therefore [L4], [L8], and [L9] give $μ ({∣ f_{n} - f ∣ > ε} \cap (X ∖ X_{m})) \leq ε^{- 1} \int_{X ∖ X_{m}} h_{n} d μ \leq \frac{2}{ε} \int_{X ∖ X_{m}} g d μ < η / 4.$ Combining this with convergence in measure on $X_{m}$ shows $f_{n} \to f$ in measure on all of $X$ .

step 1.1L3L4L8L9choosealgebra

3.1

Step 1.2 gives uniform integrability, step 2.1 gives tightness, and step 2.2 gives convergence in measure. Applying [L2] yields $f \in L^{1} (μ)$ and $f_{n} \to f$ in $L^{1} (μ)$ .

step 1.2step 2.1step 2.2L2∎

Depends on

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Sources

Terence Tao, 245A Notes 4: Modes of convergence, Theorem 29 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Exercise 7.23 (standard reference, not scraped)