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<sup>p</sup> in terms of convergence in measure and a condition related to uniform integrability.

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 $\mathbb{K}=\R$ 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 <math>\forall\ \varepsilon >0,\ \exists\ M_\varepsilon>0

\sup_{f\in\mathcal{F}} \int_{\{|f|\geq M_\varepsilon\}} |f| d\mu < \varepsilon</math>.

* 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. <math>\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</math>. 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.

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 : 

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

# 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".

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 :

# 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 $X_\varepsilon\in \mathcal{A}$ 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 originally stated for measure spaces with finite measure. In this case, one can show that conditions 1 and 2 imply that the sequence $(|f_n|^p)_{n\geq 1}$ is uniformly integrable.

Let $(X,\mathcal{A},\mu)$ be measure space. Let $(f_n)_{n\geq 1} \subseteq L^1(X,\mathcal{A},\mu)$ and assume that $\lim_{n\to\infty}\int_A f_nd\mu$ 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 $\lim_{n\to\infty}\int_A f_nd\mu = \int_A f d\mu$ 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".