Vitali convergence theorem

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

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

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

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

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

Infinite measure case

Let $(X,𝒜,\mu )$ be a measure space and $1\le p<\mathrm{\infty }$. Let $(f_n{)}_{n\ge 1}\subseteq L^p(X,𝒜,\mu )$ and $f\in L^p(X,𝒜,\mu )$. Then, $(f_n)$ converges to $f$ in $L^p(X,𝒜,\mu )$ 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 $\epsilon >0$, there exists $X_{\epsilon }\in 𝒜$ such that $\mu (X_{\epsilon })<\mathrm{\infty }$ and $\underset{n\ge 1}{\sup }\underset{X\setminus X_{\epsilon }}{\int }|f_n{|}^{p}d\mu <\epsilon .$

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

Converse of the theorem

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

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

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