Vitali convergence theorem explained

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

(X, l{A},\mu) be a measure space, i.e. \mu : l{A}\to [ 0 , inf ty] is a set function such that \mu(\emptyset) =0 and \mu is countably-additive. All functions considered in the sequel will be functions f:X\to K , where K = \R or C . We adopt the following definitions according to Bogachev's terminology.^[1]

• A set of functions

l{F} \subset L ^{1(X,l{A},\mu)} is called uniformly integrable if \lim _{M\to+infty} \sup _{f\inl{F} } \int_ |f|\, d\mu = 0, i.e \forall \varepsilon > 0 , \exists M _{\varepsilon>0 :} \sup _{f\inl{F} } \int_ |f|\, d\mu < \varepsilon.

• A set of functions

l{F} \subset L ^{1(X,l{A},\mu)} is said to have uniformly absolutely continuous integrals if \lim _{\mu(A)\to} \sup _{f\inl{F} } \int_A |f|\, d\mu = 0, i.e. \forall \varepsilon> 0 , \exists \delta _{\varepsilon} > 0 , \forall A\in l{A} : \mu(A)<\delta _{\varepsilon} ⇒ \sup _{f\in} } \int_A |f|\, d\mu < \varepsilon. This definition is sometimes used as a definition of uniform integrability. However, it differs from the definition of uniform integrability given above.

When

\mu(X)< inf ty , a set of functions l{F} \subset L ^{1(X,l{A},\mu)} is uniformly integrable if and only if it is bounded in L ^{1(X,l{A},\mu)} and has uniformly absolutely continuous integrals. If, in addition, \mu is atomless, then the uniform integrability is equivalent to the uniform absolute continuity of integrals.

Finite measure case

Let

(X, l{A},\mu) be a measure space with \mu(X)< inf ty . Let (f _{n)\subset} L ^{p(X,l{A},\mu)} and f be an l{A} -measurable function. Then, the following are equivalent : f\in L ^{p(X,l{A},\mu)} and (f _{n)} converges to f in L ^{p(X,l{A},\mu)} ;

1. The sequence of functions

(f _{n)} converges in \mu -measure to f and (|f ^{p)} _{n\geq 1} is uniformly integrable ;

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

Infinite measure case

Let

(X, l{A},\mu) be a measure space and 1 \leq p< inf ty . Let (f _{n)} _{n\geq} \subseteq L ^{p(X,l{A},\mu)} and f\in L ^{p(X,l{A},\mu)} . Then, (f _{n)} converges to f in L ^{p(X,l{A},\mu)} if and only if the following holds :

1. The sequence of functions

(f _{n)} converges in \mu -measure to f ;(f _{n)} has uniformly absolutely continuous integrals;

1. For every

\varepsilon> 0 , there exists X _{\varepsilon\in} l{A} such that \mu(X _{\varepsilon)<infty} and \sup _{n\geq} \int _{X\setminus X _{\varepsilon}} |f ^{p} _{n|} d\mu <\varepsilon. When \mu(X)< inf ty , the third condition becomes superfluous (one can simply take X _{\varepsilon} = X ) 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 (|f ^{p)} _{n\geq 1} is uniformly integrable.

Converse of the theorem

Let

(X, l{A},\mu) be measure space. Let (f _{n)} _{n\geq} \subseteq L ^{1(X,l{A},\mu)} and assume that \lim _{n\toinfty} \int _{A} f _{nd\mu} exists for every A\in l{A} . Then, the sequence (f _{n)} is bounded in L ^{1(X,l{A},\mu)} and has uniformly absolutely continuous integrals. In addition, there exists f\in L ^{1(X,l{A},\mu)} such that \lim _{n\toinfty} \int _{A} f _{nd\mu} = \int _{A} f d\mu for every A\in l{A} .

When

\mu(X)< inf ty , this implies that (f _{n)} is uniformly integrable.

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

Notes and References

1. Book: Bogachev, Vladimir I.. Measure Theory Volume I. Springer. 2007. 978-3-540-34513-8. New York. 267-271.