Inreal analysis andmeasure theory, theVitali convergence theorem, named after theItalianmathematicianGiuseppe Vitali, is a generalization of the better-knowndominated convergence theorem ofHenri Lebesgue. It is a characterization of the convergence inL^p in terms of convergence in measure and a condition related touniform integrability.

Let${\displaystyle (X,\mathcal{A},\mu )}$ be ameasure space, i.e.${\displaystyle \mu :\mathcal{A}\to [0,\mathrm{\infty }]}$ is a set function such that${\displaystyle \mu (\mathrm{\varnothing })=0}$ and${\displaystyle \mu }$ is countably-additive. All functions considered in the sequel will be functions${\displaystyle f:X\to \mathbb{K}}$, where${\displaystyle \mathbb{K}=\mathbb{R}}$ or${\displaystyle \mathbb{C}}$. We adopt the following definitions according to Bogachev's terminology.^[1]

• A set of functions${\displaystyle \mathcal{F}\subset L^1(X,\mathcal{A},\mu )}$ is calleduniformly integrable if${\displaystyle \underset{M\to +\mathrm{\infty }}{lim}\underset{f\in \mathcal{F}}{sup}{\int }_{\{|f|>M\}}|f|d\mu =0}$, i.e.
• A set of functions${\displaystyle \mathcal{F}\subset L^1(X,\mathcal{A},\mu )}$ is said to haveuniformly absolutely continuous integrals if${\displaystyle \underset{\mu (A)\to 0}{lim}\underset{f\in \mathcal{F}}{sup}{\int }_A|f|d\mu =0}$, i.e.. This definition is sometimes used as a definition of uniform integrability. However, it differs from the definition of uniform integrability given above.

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

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

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

For a proof, see Bogachev's monograph "Measure Theory, Volume I".^[1]

Let${\displaystyle (X,\mathcal{A},\mu )}$ be a measure space and. Let${\displaystyle (f_n{)}_{n\ge 1}\subseteq L^p(X,\mathcal{A},\mu )}$ and${\displaystyle f\in L^p(X,\mathcal{A},\mu )}$. Then,${\displaystyle (f_n)}$ converges to${\displaystyle f}$ in${\displaystyle L^p(X,\mathcal{A},\mu )}$ if and only if the following holds :

1. The sequence of functions${\displaystyle (f_n)}$ converges in${\displaystyle \mu }$-measure to${\displaystyle f}$ ;
2. ${\displaystyle (f_n)}$ has uniformly absolutely continuous integrals;
3. For every${\displaystyle \epsilon >0}$, there exists${\displaystyle X_{\epsilon }\in \mathcal{A}}$ such that and

When, the third condition becomes superfluous (one can simply take${\displaystyle X_{\epsilon }=X}$) 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${\displaystyle (|f_n{|}^p{)}_{n\ge 1}}$ is uniformly integrable.

Let${\displaystyle (X,\mathcal{A},\mu )}$ be measure space. Let${\displaystyle (f_n{)}_{n\ge 1}\subseteq L^1(X,\mathcal{A},\mu )}$ and assume that${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_A{f}_{n}d\mu }$ exists for every${\displaystyle A\in \mathcal{A}}$. Then, the sequence${\displaystyle (f_n)}$ is bounded in${\displaystyle L^1(X,\mathcal{A},\mu )}$ and has uniformly absolutely continuous integrals. In addition, there exists${\displaystyle f\in L^1(X,\mathcal{A},\mu )}$ such that${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_A{f}_{n}d\mu ={\int }_{A}fd\mu }$ for every${\displaystyle A\in \mathcal{A}}$.

When, this implies that${\displaystyle (f_n)}$ is uniformly integrable.

For a proof, see Bogachev's monograph "Measure Theory, Volume I".^[1]

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