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 $(X,\mathcal,\mu)$ be a measure space, i.e. \mu : \mathcal\to ,\infty/math> is a set function such that $\mu(\emptyset)=0$ and $\mu$ is countably-additive. All functions considered in the sequel will be functions $f:X\to \mathbb$, where $\mathbb=\R$ or $\mathbb$. We adopt the following definitions according to Bogachev's terminology.

* A set of functions $\mathcal \subset L^1(X,\mathcal,\mu)$ is called uniformly integrable if $\lim_ \sup_ \int_ , f, \, d\mu = 0$, i.e $\forall\ \varepsilon >0,\ \exists\ M_\varepsilon>0 : \sup_ \int_ , f, \, d\mu < \varepsilon$.
* A set of functions $\mathcal \subset L^1(X,\mathcal,\mu)$ is said to have uniformly absolutely continuous integrals if $\lim_\sup_ \int_A , f, \, d\mu = 0$, i.e. $\forall\ \varepsilon>0,\ \exists\ \delta_\varepsilon >0,\ \forall\ A\in\mathcal : \mu(A)<\delta_\varepsilon \Rightarrow \sup_ \int_A , f, \, d\mu < \varepsilon$. 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 $\mathcal \subset L^1(X,\mathcal,\mu)$ is uniformly integrable if and only if it is bounded in $L^1(X,\mathcal,\mu)$ 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 $(X,\mathcal,\mu)$ be a measure space with $\mu(X)<\infty$. Let $(f_n)\subset L^p(X,\mathcal,\mu)$ and $f$ be an $\mathcal$-measurable function. Then, the following are equivalent :

# $f\in L^p(X,\mathcal,\mu)$ and $(f_n)$ converges to $f$ in $L^p(X,\mathcal,\mu)$ ;

# The sequence of functions $(f_n)$ converges in $\mu$-measure to $f$ and $(, f_n, ^p)_$ is uniformly integrable ;

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

Infinite measure case

Let $(X,\mathcal,\mu)$ be a measure space and $1\leq p<\infty$. Let $(f_n)_ \subseteq L^p(X,\mathcal,\mu)$ and $f\in L^p(X,\mathcal,\mu)$. Then, $(f_n)$ converges to $f$ in $L^p(X,\mathcal,\mu)$ if and only if the following holds :
# The sequence of functions $(f_n)$ converges in $\mu$-measure to $f$ ;
#$(f_n)$ has uniformly absolutely continuous integrals;
# For every $\varepsilon>0$, there exists $X_\varepsilon\in \mathcal$ such that $\mu(X_\varepsilon)<\infty$ and $\sup_\int_ , f_n, ^p\, d\mu <\varepsilon.$
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 originally stated for measure spaces with finite measure. In this case, one can show that conditions 1 and 2 imply that the sequence $(, f_n, ^p)_$ is uniformly integrable.

Converse of the theorem

Let $(X,\mathcal,\mu)$ be measure space. Let $(f_n)_ \subseteq L^1(X,\mathcal,\mu)$ and assume that $\lim_\int_A f_n\,d\mu$ exists for every $A\in\mathcal$. Then, the sequence $(f_n)$ is bounded in $L^1(X,\mathcal,\mu)$ and has uniformly absolutely continuous integrals. In addition, there exists $f\in L^1(X,\mathcal,\mu)$ such that $\lim_\int_A f_n\,d\mu = \int_A f\, d\mu$ for every $A\in\mathcal$.

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

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

Citations

{{Measure theory

Theorems in measure theory