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.

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]

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.

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]