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.

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

• A set of functions ${\displaystyle {\mathcal {F}}\subset L^{1}(X,{\mathcal {A}},\mu )}${\displaystyle {\mathcal {F}}\subset L^{1}(X,{\mathcal {A}},\mu )} is called uniformly integrable if ${\displaystyle \lim _{M\to +\infty }\sup _{f\in {\mathcal {F}}}\int _{\{|f|>M\}}|f|,円d\mu =0}${\displaystyle \lim _{M\to +\infty }\sup _{f\in {\mathcal {F}}}\int _{\{|f|>M\}}|f|,円d\mu =0}, i.e ${\displaystyle \forall \ \varepsilon >0,\ \exists \ M_{\varepsilon }>0:\sup _{f\in {\mathcal {F}}}\int _{\{|f|\geq M_{\varepsilon }\}}|f|,円d\mu <\varepsilon }${\displaystyle \forall \ \varepsilon >0,\ \exists \ M_{\varepsilon }>0:\sup _{f\in {\mathcal {F}}}\int _{\{|f|\geq M_{\varepsilon }\}}|f|,円d\mu <\varepsilon }.
• A set of functions ${\displaystyle {\mathcal {F}}\subset L^{1}(X,{\mathcal {A}},\mu )}${\displaystyle {\mathcal {F}}\subset L^{1}(X,{\mathcal {A}},\mu )} is said to have uniformly absolutely continuous integrals if ${\displaystyle \lim _{\mu (A)\to 0}\sup _{f\in {\mathcal {F}}}\int _{A}|f|,円d\mu =0}${\displaystyle \lim _{\mu (A)\to 0}\sup _{f\in {\mathcal {F}}}\int _{A}|f|,円d\mu =0}, i.e. ${\displaystyle \forall \ \varepsilon >0,\ \exists \ \delta _{\varepsilon }>0,\ \forall \ A\in {\mathcal {A}}:\mu (A)<\delta _{\varepsilon }\Rightarrow \sup _{f\in {\mathcal {F}}}\int _{A}|f|,円d\mu <\varepsilon }${\displaystyle \forall \ \varepsilon >0,\ \exists \ \delta _{\varepsilon }>0,\ \forall \ A\in {\mathcal {A}}:\mu (A)<\delta _{\varepsilon }\Rightarrow \sup _{f\in {\mathcal {F}}}\int _{A}|f|,円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 ${\displaystyle \mu (X)<\infty }${\displaystyle \mu (X)<\infty }, a set of functions ${\displaystyle {\mathcal {F}}\subset L^{1}(X,{\mathcal {A}},\mu )}${\displaystyle {\mathcal {F}}\subset L^{1}(X,{\mathcal {A}},\mu )} is uniformly integrable if and only if it is bounded in ${\displaystyle L^{1}(X,{\mathcal {A}},\mu )}${\displaystyle L^{1}(X,{\mathcal {A}},\mu )} and has uniformly absolutely continuous integrals. If, in addition, ${\displaystyle \mu }${\displaystyle \mu } is atomless, then the uniform integrability is equivalent to the uniform absolute continuity of integrals.

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

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

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

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

1. The sequence of functions ${\displaystyle (f_{n})}${\displaystyle (f_{n})} converges in ${\displaystyle \mu }${\displaystyle \mu }-measure to ${\displaystyle f}${\displaystyle f} ;
2. ${\displaystyle (f_{n})}${\displaystyle (f_{n})} has uniformly absolutely continuous integrals;
3. For every ${\displaystyle \varepsilon >0}${\displaystyle \varepsilon >0}, there exists ${\displaystyle X_{\varepsilon }\in {\mathcal {A}}}${\displaystyle X_{\varepsilon }\in {\mathcal {A}}} such that ${\displaystyle \mu (X_{\varepsilon })<\infty }${\displaystyle \mu (X_{\varepsilon })<\infty } and ${\displaystyle \sup _{n\geq 1}\int _{X\setminus X_{\varepsilon }}|f_{n}|^{p},円d\mu <\varepsilon .}${\displaystyle \sup _{n\geq 1}\int _{X\setminus X_{\varepsilon }}|f_{n}|^{p},円d\mu <\varepsilon .}

When ${\displaystyle \mu (X)<\infty }${\displaystyle \mu (X)<\infty }, the third condition becomes superfluous (one can simply take ${\displaystyle X_{\varepsilon }=X}${\displaystyle 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 ${\displaystyle (|f_{n}|^{p})_{n\geq 1}}${\displaystyle (|f_{n}|^{p})_{n\geq 1}} is uniformly integrable.

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

When ${\displaystyle \mu (X)<\infty }${\displaystyle \mu (X)<\infty }, this implies that ${\displaystyle (f_{n})}${\displaystyle (f_{n})} is uniformly integrable.

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

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