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 ${\displaystyle (X,{\mathcal {A)),\mu )}$ be a measure space, i.e. ${\displaystyle \mu :{\mathcal {A))\to [0,\infty ]}$ is a set function such that ${\displaystyle \mu (\emptyset )=0}$ 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 called uniformly integrable if ${\displaystyle \lim _{M\to +\infty }\sup _{f\in {\mathcal {F))}\int _{\{|f|>M\))|f|\,d\mu =0}$, i.e ${\displaystyle \forall \ \varepsilon >0,\ \exists \ M_{\varepsilon }>0:\sup _{f\in {\mathcal {F))}\int _{\{|f|\geq M_{\varepsilon }\))|f|\,d\mu <\varepsilon }$.
• A set of functions ${\displaystyle {\mathcal {F))\subset L^{1}(X,{\mathcal {A)),\mu )}$ is said to have uniformly absolutely continuous integrals if ${\displaystyle \lim _{\mu (A)\to 0}\sup _{f\in {\mathcal {F))}\int _{A}|f|\,d\mu =0}$, i.e. ${\displaystyle \forall \ \varepsilon >0,\ \exists \ \delta _{\varepsilon }>0,\ \forall \ A\in {\mathcal {A)):\mu (A)<\delta _{\varepsilon }\Rightarrow \sup _{f\in {\mathcal {F))}\int _{A}|f|\,d\mu <\varepsilon }$. This definition is sometimes used as a definition of uniform integrability. However, it differs from the definition of uniform integrability given above.

When ${\displaystyle \mu (X)<\infty }$, 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.

Finite measure case

Let ${\displaystyle (X,{\mathcal {A)),\mu )}$ be a measure space with ${\displaystyle \mu (X)<\infty }$. 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 ${\displaystyle (|f_{n}|^{p})_{n\geq 1))$ is uniformly integrable ;

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

Infinite measure case

Let ${\displaystyle (X,{\mathcal {A)),\mu )}$ be a measure space and ${\displaystyle 1\leq p<\infty }$. Let ${\displaystyle (f_{n})_{n\geq 1}\subseteq L^{p}(X,{\mathcal {A)),\mu )}$ 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 ${\displaystyle \varepsilon >0}$, there exists ${\displaystyle X_{\varepsilon }\in {\mathcal {A))}$ such that ${\displaystyle \mu (X_{\varepsilon })<\infty }$ and ${\displaystyle \sup _{n\geq 1}\int _{X\setminus X_{\varepsilon ))|f_{n}|^{p}\,d\mu <\varepsilon .}$

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

Converse of the theorem

Let ${\displaystyle (X,{\mathcal {A)),\mu )}$ be measure space. Let ${\displaystyle (f_{n})_{n\geq 1}\subseteq L^{1}(X,{\mathcal {A)),\mu )}$ and assume that ${\displaystyle \lim _{n\to \infty }\int _{A}f_{n}\,d\mu }$ 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 ${\displaystyle \lim _{n\to \infty }\int _{A}f_{n}\,d\mu =\int _{A}f\,d\mu }$ for every ${\displaystyle A\in {\mathcal {A))}$.

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

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

Citations

1. ^ a b c Bogachev, Vladimir I. (2007). Measure Theory Volume I. New York: Springer. pp. 267–271. ISBN 978-3-540-34513-8.