Vitali convergence theorem

In real analysis and measure theory, the Vitali convergence theorem, named after the Italian mathematician Giuseppe Vitali, is a generalization of the better-known dominated convergence theorem of Henri Lebesgue. It is a characterization of the convergence in L^p in terms of convergence in measure and a condition related to uniform integrability.

Preliminary definitions

Let [math]\displaystyle{ (X,\mathcal{A},\mu) }[/math] be a measure space, i.e. [math]\displaystyle{ \mu : \mathcal{A}\to [0,\infty] }[/math] is a set function such that [math]\displaystyle{ \mu(\emptyset)=0 }[/math] and [math]\displaystyle{ \mu }[/math] is countably-additive. All functions considered in the sequel will be functions [math]\displaystyle{ f:X\to \mathbb{K} }[/math], where [math]\displaystyle{ \mathbb{K}=\R }[/math] or [math]\displaystyle{ \mathbb{C} }[/math]. We adopt the following definitions according to Bogachev's terminology.

• A set of functions [math]\displaystyle{ \mathcal{F} \subset L^1(X,\mathcal{A},\mu) }[/math] is called uniformly integrable if [math]\displaystyle{ \lim_{M\to+\infty} \sup_{f\in\mathcal{F}} \int_{\{|f|\gt M\}} |f|\, d\mu = 0 }[/math], i.e [math]\displaystyle{ \forall\ \varepsilon \gt 0,\ \exists\ M_\varepsilon\gt 0 : \sup_{f\in\mathcal{F}} \int_{\{|f|\geq M_\varepsilon\}} |f|\, d\mu \lt \varepsilon }[/math].
• A set of functions [math]\displaystyle{ \mathcal{F} \subset L^1(X,\mathcal{A},\mu) }[/math] is said to have uniformly absolutely continuous integrals if [math]\displaystyle{ \lim_{\mu(A)\to 0}\sup_{f\in\mathcal{F}} \int_A |f|\, d\mu = 0 }[/math], i.e. [math]\displaystyle{ \forall\ \varepsilon\gt 0,\ \exists\ \delta_\varepsilon \gt 0,\ \forall\ A\in\mathcal{A} : \mu(A)\lt \delta_\varepsilon \Rightarrow \sup_{f\in \mathcal{F}} \int_A |f|\, d\mu \lt \varepsilon }[/math]. This definition is sometimes used as a definition of uniform integrability. However, it differs from the definition of uniform integrability given above.

When [math]\displaystyle{ \mu(X)\lt \infty }[/math], a set of functions [math]\displaystyle{ \mathcal{F} \subset L^1(X,\mathcal{A},\mu) }[/math] is uniformly integrable if and only if it is bounded in [math]\displaystyle{ L^1(X,\mathcal{A},\mu) }[/math] and has uniformly absolutely continuous integrals. If, in addition, [math]\displaystyle{ \mu }[/math] is atomless, then the uniform integrability is equivalent to the uniform absolute continuity of integrals.

Finite measure case

Let [math]\displaystyle{ (X,\mathcal{A},\mu) }[/math] be a measure space with [math]\displaystyle{ \mu(X)\lt \infty }[/math]. Let [math]\displaystyle{ (f_n)\subset L^p(X,\mathcal{A},\mu) }[/math] and [math]\displaystyle{ f }[/math] be an [math]\displaystyle{ \mathcal{A} }[/math]-measurable function. Then, the following are equivalent :

1. [math]\displaystyle{ f\in L^p(X,\mathcal{A},\mu) }[/math] and [math]\displaystyle{ (f_n) }[/math] converges to [math]\displaystyle{ f }[/math] in [math]\displaystyle{ L^p(X,\mathcal{A},\mu) }[/math] ;
2. The sequence of functions [math]\displaystyle{ (f_n) }[/math] converges in [math]\displaystyle{ \mu }[/math]-measure to [math]\displaystyle{ f }[/math] and [math]\displaystyle{ (|f_n|^p)_{n\geq 1} }[/math] is uniformly integrable ;

For a proof, see Bogachev's monograph "Measure Theory, Volume I".

Infinite measure case

Let [math]\displaystyle{ (X,\mathcal{A},\mu) }[/math] be a measure space and [math]\displaystyle{ 1\leq p\lt \infty }[/math]. Let [math]\displaystyle{ (f_n)_{n\geq 1} \subseteq L^p(X,\mathcal{A},\mu) }[/math] and [math]\displaystyle{ f\in L^p(X,\mathcal{A},\mu) }[/math]. Then, [math]\displaystyle{ (f_n) }[/math] converges to [math]\displaystyle{ f }[/math] in [math]\displaystyle{ L^p(X,\mathcal{A},\mu) }[/math] if and only if the following holds :

1. The sequence of functions [math]\displaystyle{ (f_n) }[/math] converges in [math]\displaystyle{ \mu }[/math]-measure to [math]\displaystyle{ f }[/math] ;
2. [math]\displaystyle{ (f_n) }[/math] has uniformly absolutely continuous integrals;
3. For every [math]\displaystyle{ \varepsilon\gt 0 }[/math], there exists [math]\displaystyle{ X_\varepsilon\in \mathcal{A} }[/math] such that [math]\displaystyle{ \mu(X_\varepsilon)\lt \infty }[/math] and [math]\displaystyle{ \sup_{n\geq 1}\int_{X\setminus X_\varepsilon} |f_n|^p\, d\mu \lt \varepsilon. }[/math]

When [math]\displaystyle{ \mu(X)\lt \infty }[/math], the third condition becomes superfluous (one can simply take [math]\displaystyle{ X_\varepsilon = X }[/math]) and the first two conditions give the usual form of Lebesgue-Vitali's convergence theorem originally stated for measure spaces with finite measure. In this case, one can show that conditions 1 and 2 imply that the sequence [math]\displaystyle{ (|f_n|^p)_{n\geq 1} }[/math] is uniformly integrable.

Converse of the theorem

Let [math]\displaystyle{ (X,\mathcal{A},\mu) }[/math] be measure space. Let [math]\displaystyle{ (f_n)_{n\geq 1} \subseteq L^1(X,\mathcal{A},\mu) }[/math] and assume that [math]\displaystyle{ \lim_{n\to\infty}\int_A f_n\,d\mu }[/math] exists for every [math]\displaystyle{ A\in\mathcal{A} }[/math]. Then, the sequence [math]\displaystyle{ (f_n) }[/math] is bounded in [math]\displaystyle{ L^1(X,\mathcal{A},\mu) }[/math] and has uniformly absolutely continuous integrals. In addition, there exists [math]\displaystyle{ f\in L^1(X,\mathcal{A},\mu) }[/math] such that [math]\displaystyle{ \lim_{n\to\infty}\int_A f_n\,d\mu = \int_A f\, d\mu }[/math] for every [math]\displaystyle{ A\in\mathcal{A} }[/math].

When [math]\displaystyle{ \mu(X)\lt \infty }[/math], this implies that [math]\displaystyle{ (f_n) }[/math] is uniformly integrable.

For a proof, see Bogachev's monograph "Measure Theory, Volume I".