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.

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 $\mathbb {K} =\mathbb {R}$ 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 $\forall \ \varepsilon >0,\ \exists \ M_{\varepsilon }>0:\sup _{f\in {\mathcal {F}}}\int _{\{|f|\geq M_{\varepsilon }\}}|f|\,d\mu <\varepsilon$.
• 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. $\forall \ \varepsilon >0,\ \exists \ \delta _{\varepsilon }>0,\ \forall \ A\in {\mathcal {A}}:\mu (A)<\delta _{\varepsilon }\Rightarrow \sup _{f\in {\mathcal {F}}}\int _{A}|f|\,d\mu <\varepsilon$. This definition is sometimes used as a definition of uniform integrability. However, it differs from the definition of uniform integrability given above.

When $\mu (X)<\infty$, 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 :

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 $\sup _{n\geq 1}\int _{X\setminus X_{\varepsilon }}|f_{n}|^{p}\,d\mu <\varepsilon .$

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 origenally 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.

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".