Egorov's theorem

The first proof of the theorem was given by Carlo Severini in 1910: he used the result as a tool in his research on series of orthogonal functions. His work remained apparently unnoticed outside Italy, probably due to the fact that it is written in Italian, appeared in a scientific journal with limited diffusion and was considered only as a means to obtain other theorems. A year later Dmitri Egorov published his independently proved results, and the theorem became widely known under his name: however, it is not uncommon to find references to this theorem as the Severini–Egoroff theorem. The first mathematicians to prove independently the theorem in the nowadays common abstract measure space setting were , and in : an earlier generalization is due to Nikolai Luzin, who succeeded in slightly relaxing the requirement of finiteness of measure of the domain of convergence of the pointwise converging functions in the long (242 pages) paper . Further generalizations were given much later by Pavel Korovkin, in the paper , and by Gabriel Mokobodzki in the paper : in particular Korovkin extended the result to a class of non–negative set functions more general than measures. ==Formal statement and proof==

Statement Let (f_n) be a sequence of M-valued measurable functions, where M is a separable metric space, on some measure space (X,\sigma,\mu), and suppose there is a measurable subset A \subseteq X, with finite \mu-measure, such that (f_n) converges \mu-almost everywhere on A to a limit function f. The following result holds: for every \varepsilon > 0, there exists a measurable subset B of A such that \mu (B), and (f_n) converges to f uniformly on A\setminus B. Here, \mu (B) denotes the \mu-measure of B. In words, the theorem says that pointwise convergence almost everywhere on A implies the apparently much stronger uniform convergence everywhere except on some subset B of arbitrarily small measure. This type of convergence is called almost uniform convergence. Discussion of assumptions and a counterexample • The hypothesis \mu (A) 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 f_n(x) = 1_{[n,n+1]}(x), \qquad n\in\N,\ x\in\R, defined on the real line. This sequence converges pointwise to the zero function everywhere but does not converge uniformly on \R\setminus B for any set B of finite measure: a counterexample in the general n-dimensional real vector space \R^n can be constructed as shown by . • The separability of the metric space is needed to make sure that for M-valued, measurable functions f and g, the distance d(f(x), g(x)) is again a measurable real-valued function of x. Proof Fix \varepsilon > 0. For natural numbers n and k, define the set En,k by the union : E_{n,k} = \bigcup_{m\ge n} \left\{ x\in A \,\Big|\, |f_m(x) - f(x)| \ge \frac1k \right\}. These sets get smaller as n increases, meaning that En+1,k is always a subset of En,k, because the first union involves fewer sets. A point x, for which the sequence (fm(x)) converges to f(x), cannot be in every En,k for a fixed k, because fm(x) has to stay closer to f(x) than 1/k eventually. Hence by the assumption of μ-almost everywhere pointwise convergence on A, :\mu\left(\bigcap_{n\in\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 nk such that :\mu(E_{n_k,k}) For x in this set we consider the speed of approach into the 1/k-neighbourhood of f(x) as too slow. Define :B = \bigcup_{k\in\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 A\setminus B we therefore have uniform convergence. Explicitly, for any \epsilon, let \frac 1 k , then for any n > n_k, we have |f_n - f| on all of A\setminus B. Appealing to the sigma additivity of μ and using the geometric series, we get :\mu(B) \le \sum_{k\in\N} \mu(E_{n_k,k}) ==Generalizations==

Luzin's version Nikolai Luzin's generalization of the Severini–Egorov theorem is presented here according to . 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 (fn) is a given sequence of M-valued measurable functions on some measure space (X,Σ,μ), such that (fn) 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 (fn) converges to f uniformly on each set Ak. 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 {Ak}k=1,2,... such that :\mu\left (A \setminus \bigcup_{k=1}^{N} A_k \right)\leq\frac{1}{N} and such that (fn) converges to f uniformly on each set Ak for each k. Choosing :H=A\setminus\bigcup_{k=1}^{\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 , which however generalizes it to some extent by considering admissible functionals instead of non-negative measures and inequalities \leq and \geq respectively in conditions 1 and 2. Statement Let (M,d) denote a separable metric space and (X,\Sigma ) a measurable space: consider a measurable set A and a class \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 • \mu(\cap A_n) = \lim \mu(A_n) if A_1 \supset A_2 \supset \cdots with A_n\in\mathfrak{A} for all n • \mu(\cup A_n) = \lim \mu(A_n) if A_1 \subset A_2 \subset \cdots with \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^\prime of A such that 00 and where the convergence is also uniform. Proof Consider the indexed family of sets whose index set is the set of natural numbers m\in\N, defined as follows: :A_{0,m}=\left\{x\in A|d(f_n(x),f(x)) \le 1\ \forall n\geq m\right\} Obviously :A_{0,1}\subseteq A_{0,2}\subseteq A_{0,3}\subseteq\dots and :A=\bigcup_{m\in\N}A_{0,m} therefore there is a natural number m0 such that putting A0,m0=A0 the following relation holds true: :0\leq\mu(A)-\mu(A_0)\leq\varepsilon Using A0 it is possible to define the following indexed family :A_{1,m}=\left\{x\in A_0\left| d(f_m(x),f(x)) \le \frac12 \ \forall n\geq m\right.\right\} satisfying the following two relationships, analogous to the previously found ones, i.e. :A_{1,1}\subseteq A_{1,2}\subseteq A_{1,3}\subseteq\dots and :A_0=\bigcup_{m\in\N}A_{1,m} This fact enable us to define the set A1,m1=A1, where m1 is a surely existing natural number such that :0\leq\mu(A)-\mu(A_1)\leq\varepsilon By iterating the shown construction, another indexed family of set {An} is defined such that it has the following properties: • A_0\supseteq A_1\supseteq A_2\supseteq\cdots • 0\leq\mu(A)-\mu(A_m)\leq\varepsilon for all m\in\N • for each m\in\N there exists km such that for all n \geq k_m then d(f_n(x),f(x)) \le 2^{-m} for all x \in A_m and finally putting :A'=\bigcup_{n\in\N}A_n the thesis is easily proved. ==Notes==