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 Lp in terms of convergence in measure and a condition related to uniform integrability.

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


 * A set of functions $$\mathcal{F} \subset L^1(X,\mathcal{A},\mu)$$ is called uniformly integrable if $$\lim_{M\to+\infty} \sup_{f\in\mathcal{F}} \int_{\{|f|>M\}} |f|\, d\mu = 0$$, i.e $$\forall\ \varepsilon >0,\ \exists\ M_\varepsilon>0
 * \sup_{f\in\mathcal{F}} \int_{\{|f|\geq M_\varepsilon\}} |f|\, d\mu < \varepsilon$$.

\mu(A)<\delta_\varepsilon \Rightarrow \sup_{f\in \mathcal{F}} \int_A |f|\, d\mu < \varepsilon$$. This definition is sometimes used as a definition of uniform integrability. However, it differs from the definition of uniform integrability given above.
 * A set of functions $$\mathcal{F} \subset L^1(X,\mathcal{A},\mu)$$ is said to have uniformly absolutely continuous integrals if $$\lim_{\mu(A)\to 0}\sup_{f\in\mathcal{F}} \int_A |f|\, d\mu = 0$$, i.e. $$\forall\ \varepsilon>0,\ \exists\ \delta_\varepsilon >0,\ \forall\ A\in\mathcal{A} :

When $$\mu(X)<\infty$$, a set of functions $$\mathcal{F} \subset L^1(X,\mathcal{A},\mu)$$ is uniformly integrable if and only if it is bounded in $$L^1(X,\mathcal{A},\mu)$$ 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 $$(X,\mathcal{A},\mu)$$ be a measure space with $$\mu(X)<\infty$$. Let $$(f_n)\subset L^p(X,\mathcal{A},\mu)$$ and $$f$$ be an $$\mathcal{A}$$-measurable function. Then, the following are equivalent :


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

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

Infinite measure case
Let $$(X,\mathcal{A},\mu)$$ be a measure space and $$1\leq p<\infty$$. Let $$(f_n)_{n\geq 1} \subseteq L^p(X,\mathcal{A},\mu)$$ and $$f\in L^p(X,\mathcal{A},\mu)$$. Then, $$(f_n)$$ converges to $$f$$ in $$L^p(X,\mathcal{A},\mu)$$ if and only if the following holds : When $$\mu(X)<\infty$$, 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 $$(|f_n|^p)_{n\geq 1}$$ is uniformly integrable.
 * 1) The sequence of functions $$(f_n)$$ converges in $$\mu$$-measure to $$f$$ ;
 * 2) $$(f_n)$$ has uniformly absolutely continuous integrals;
 * 3) For every $$\varepsilon>0$$, there exists $$X_\varepsilon\in \mathcal{A}$$ such that $$\mu(X_\varepsilon)<\infty$$ and  $$\sup_{n\geq 1}\int_{X\setminus X_\varepsilon} |f_n|^p\, d\mu <\varepsilon.$$

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

When $$\mu(X)<\infty$$, this implies that $$(f_n)$$ is uniformly integrable.

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