Statement

Under the same hypothesis of the abstract Severini–Egorov theorem suppose that A is the union of a sequence of measurable sets of finite μ-measure, and (f_n) is a given sequence of M-valued measurable functions on some measure space (X,Σ,μ), such that (f_n) converges μ-almost everywhere on A to a limit function f, then A can be expressed as the union of a sequence of measurable sets H, A₁, A₂,... such that μ(H) = 0 and (f_n) converges to f uniformly on each set A_k.

Proof

It is sufficient to consider the case in which the set A is itself of finite μ-measure: using this hypothesis and the standard Severini–Egorov theorem, it is possible to define by mathematical induction a sequence of sets {A_k}_k=1,2,... such that

${\displaystyle \mu \left(A\setminus \bigcup _{k=1}^{N}A_{k}\right)\leq {\frac {1}{N}}}$

and such that (f_n) converges to f uniformly on each set A_k for each k. Choosing

${\displaystyle H=A\setminus \bigcup _{k=1}^{\infty }A_{k}}$

then obviously μ(H) = 0 and the theorem is proved.

Statement

Let (M,d) denote a separable metric space and (X,Σ) a measurable space: consider a measurable set A and a class ${\displaystyle {\mathfrak {A}}}$ containing A and its measurable subsets such that their countable in unions and intersections belong to the same class. Suppose there exists a non-negative measure μ such that μ(A) exists and

1. ${\displaystyle \mu (\cap A_{n})=\lim \mu (A_{n})}$ if ${\displaystyle A_{1}\supset A_{2}\supset \cdots }$ with ${\displaystyle A_{n}\in {\mathfrak {A}}}$ for all n
2. ${\displaystyle \mu (\cup A_{n})=\lim \mu (A_{n})}$ if ${\displaystyle A_{1}\subset A_{2}\subset \cdots }$ with ${\displaystyle \cup A_{n}\in {\mathfrak {A}}}$.

If (f_n) is a sequence of M-valued measurable functions converging μ-almost everywhere on ${\displaystyle A\in {\mathfrak {A}}}$ to a limit function f, then there exists a subset A′ of A such that 0 < μ(A) − μ(A′) < ε and where the convergence is also uniform.

Proof

Consider the indexed family of sets whose index set is the set of natural numbers ${\displaystyle m\in \mathbb {N} ,}$ defined as follows:

${\displaystyle A_{0,m}=\left\{x\in A|d(f_{n}(x),f(x))\leq 1\ \forall n\geq m\right\}}$

Obviously

${\displaystyle A_{0,1}\subseteq A_{0,2}\subseteq A_{0,3}\subseteq \dots }$

and

${\displaystyle A=\bigcup _{m\in \mathbb {N} }A_{0,m}}$

therefore there is a natural number m₀ such that putting A_0,m0=A₀ the following relation holds true:

${\displaystyle 0\leq \mu (A)-\mu (A_{0})\leq \varepsilon }$

Using A₀ it is possible to define the following indexed family

${\displaystyle A_{1,m}=\left\{x\in A_{0}\left|d(f_{m}(x),f(x))\leq {\frac {1}{2}}\ \forall n\geq m\right.\right\}}$

satisfying the following two relationships, analogous to the previously found ones, i.e.

${\displaystyle A_{1,1}\subseteq A_{1,2}\subseteq A_{1,3}\subseteq \dots }$

and

${\displaystyle A_{0}=\bigcup _{m\in \mathbb {N} }A_{1,m}}$

This fact enable us to define the set A_1,m1=A₁, where m₁ is a surely existing natural number such that

${\displaystyle 0\leq \mu (A)-\mu (A_{1})\leq \varepsilon }$

By iterating the shown construction, another indexed family of set {A_n} is defined such that it has the following properties:

• ${\displaystyle A_{0}\supseteq A_{1}\supseteq A_{2}\supseteq \cdots }$
• ${\displaystyle 0\leq \mu (A)-\mu (A_{m})\leq \varepsilon }$ for all ${\displaystyle m\in \mathbb {N} }$
• for each ${\displaystyle m\in \mathbb {N} }$ there exists k_m such that for all ${\displaystyle n\geq k_{m}}$ then ${\displaystyle d(f_{n}(x),f(x))\leq 2^{-m}}$ for all ${\displaystyle x\in A_{m}}$

and finally putting

${\displaystyle A'=\bigcup _{n\in \mathbb {N} }A_{n}}$

the thesis is easily proved.