Egorov's theorem

Formal statement and proof

Statement

Let ${\displaystyle (f_n)}$ be a sequence of ${\displaystyle M}$-valued measurable functions, where ${\displaystyle M}$ is a separable metric space, on some measure space ${\displaystyle (X,\sigma ,\mu )}$, and suppose there is a measurable subset ${\displaystyle A\subseteq X}$, with finite ${\displaystyle \mu }$-measure, such that ${\displaystyle (f_n)}$ converges ${\displaystyle \mu }$-almost everywhere on ${\displaystyle A}$ to a limit function ${\displaystyle f}$. The following result holds: for every ${\displaystyle \epsilon >0}$, there exists a measurable subset ${\displaystyle B}$ of ${\displaystyle A}$ such that ${\displaystyle \mu (B)<\epsilon }$, and ${\displaystyle (f_n)}$ converges to ${\displaystyle f}$ uniformly on ${\displaystyle A\setminus B}$.

Here, ${\displaystyle \mu (B)}$ denotes the ${\displaystyle \mu }$-measure of ${\displaystyle B}$. In words, the theorem says that pointwise convergence almost everywhere on ${\displaystyle A}$ implies the apparently much stronger uniform convergence everywhere except on some subset ${\displaystyle B}$ of arbitrarily small measure. This type of convergence is called almost uniform convergence.

Discussion of assumptions and a counterexample

• The hypothesis ${\displaystyle \mu (A)<\mathrm{\infty }}$ is necessary. To see this, it is simple to construct a counterexample when μ is the Lebesgue measure: consider the sequence of real-valued indicator functions $${\displaystyle f_n(x)=1_{[n,n+1]}(x),n\in \mathbb{N},x\in \mathbb{R},}$$ defined on the real line. This sequence converges pointwise to the zero function everywhere but does not converge uniformly on ${\displaystyle \mathbb{R}\setminus B}$ for any set ${\displaystyle B}$ of finite measure: a counterexample in the general ${\displaystyle n}$-dimensional real vector space ${\displaystyle {\mathbb{R}}^n}$ can be constructed as shown by Cafiero (1959, p. 302).
• The separability of the metric space is needed to make sure that for ${\displaystyle M}$-valued, measurable functions ${\displaystyle f}$ and ${\displaystyle g}$, the distance ${\displaystyle d(f(x),g(x))}$ is again a measurable real-valued function of ${\displaystyle x}$.

Proof

Fix ${\displaystyle \epsilon >0}$. For natural numbers n and k, define the set E_n,k by the union

$${\displaystyle E_{n,k}=\bigcup _{m\ge n}\left\{x\in A||f_m(x)-f(x)|\ge \frac{1}{k}\right\}.}$$

These sets get smaller as n increases, meaning that E_n+1,k is always a subset of E_n,k, because the first union involves fewer sets. A point x, for which the sequence (f_m(x)) converges to f(x), cannot be in every E_n,k for a fixed k, because f_m(x) has to stay closer to f(x) than 1/k eventually. Hence by the assumption of μ-almost everywhere pointwise convergence on A,

$${\displaystyle \mu \left(\bigcap _{n\in \mathbb{N}}{E}_{n,k}\right)=0}$$

for every k. Since A is of finite measure, we have continuity from above; hence there exists, for each k, some natural number n_k such that

$${\displaystyle \mu (E_{n_k,k})<\frac{\epsilon }{2^k}.}$$

For x in this set we consider the speed of approach into the 1/k-neighbourhood of f(x) as too slow. Define

$${\displaystyle B=\bigcup _{k\in \mathbb{N}}{E}_{n_k,k}}$$

as the set of all those points x in A, for which the speed of approach into at least one of these 1/k-neighbourhoods of f(x) is too slow. On the set difference ${\displaystyle A\setminus B}$ we therefore have uniform convergence. Explicitly, for any ${\displaystyle ϵ}$, let ${\displaystyle \frac{1}{k}<ϵ}$, then for any ${\displaystyle n>n_k}$, we have ${\displaystyle |f_n-f|<ϵ}$ on all of ${\displaystyle A\setminus B}$.

Appealing to the sigma additivity of μ and using the geometric series, we get

$${\displaystyle \mu (B)\le \sum _{k\in \mathbb{N}}\mu (E_{n_k,k})<\sum _{k\in \mathbb{N}}\frac{\epsilon }{2^k}=\epsilon .}$$

Generalizations

Luzin's version

Nikolai Luzin's generalization of the Severini–Egorov theorem is presented here according to Saks (1937, p. 19).

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

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

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

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

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

Korovkin's version

The proof of the Korovkin version follows closely the version on Kharazishvili (2000, pp. 183–184), which however generalizes it to some extent by considering admissible functionals instead of non-negative measures and inequalities ${\displaystyle \le }$ and ${\displaystyle \ge }$ respectively in conditions 1 and 2.

Statement

Let $(M,d)$ denote a separable metric space and $(X,\mathrm{\Sigma })$ 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 $\mu$ such that $\mu (A)$ 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 $\mu$-almost everywhere on $A\in \mathfrak{A}$ to a limit function $f$, then there exists a subset $A^{\mathrm{\prime }}$ of $A$ such that $0<\mu (A)-\mu (A^{\mathrm{\prime }})<\epsilon$0 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:

$${\displaystyle A_{0,m}=\left\{x\in A|d(f_n(x),f(x))\le 1\mathrm{\forall }n\ge m\right\}}$$

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:

$${\displaystyle 0\le \mu (A)-\mu (A_0)\le \epsilon }$$

Using A₀ it is possible to define the following indexed family

$${\displaystyle A_{1,m}=\left\{x\in A_0|d(f_m(x),f(x))\le \frac{1}{2}\mathrm{\forall }n\ge m\right\}}$$

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

$${\displaystyle 0\le \mu (A)-\mu (A_1)\le \epsilon }$$

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

• ${\displaystyle A_0\supseteq A_1\supseteq A_2\supseteq \cdots }$
• ${\displaystyle 0\le \mu (A)-\mu (A_m)\le \epsilon }$ for all ${\displaystyle m\in \mathbb{N}}$
• for each ${\displaystyle m\in \mathbb{N}}$ there exists k_m such that for all ${\displaystyle n\ge k_m}$ then ${\displaystyle d(f_n(x),f(x))\le 2^{-m}}$ for all ${\displaystyle x\in A_m}$

and finally putting

$${\displaystyle A^{\prime }=\bigcup _{n\in \mathbb{N}}{A}_n}$$

the thesis is easily proved.