Egorov's theorem

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In measure theory, an area of mathematics, Egorov's theorem establishes a condition for the uniform convergence of a pointwise convergent sequence of measurable functions. The theorem is named after Dmitri Egorov, a Russian physicist and geometer, who published it in 1911.

Egorov's theorem can be used along with compactly supported continuous functions to prove Lusin's theorem for integrable functions.

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[edit] Statement of the theorem

Let (M,d) denote a separable metric space (such as the real numbers with the usual distance d(a,b) = |a − b| as metric). Given a sequence (fn) of M-valued measurable functions on some measure space (X,Σ,μ), and a measurable subset A of finite μ-measure such that (fn) converges μ-almost everywhere on A to a limit function f, the following result holds: for every ε > 0, there exists a measurable subset B of A such that μ(B) < ε, and (fn) converges to f uniformly on the relative complement A \ B.

Here, μ(B) denotes the μ-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 also called almost uniform convergence.

[edit] Discussion of assumptions

Note that the assumption μ(A) < ∞ is necessary. Under Lebesgue measure, consider the sequence of real-valued indicator functions

f_n(x) = 1_{[n,n+1]}(x),\qquad 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 R \ B for any set B of finite measure.

The separability of the metric space is needed to make sure that for M-valued, measuable functions f and g, the distance d(f(x),g(x)) is again a measurable real-valued function of x.

[edit] Proof

For natural numbers n and k, define the set En,k by the union

E_{n,k} = \bigcup_{m\ge n} \Bigl\{ x\in A \,\Big|\, d(f_m(x),f(x)) \ge \frac1k \Bigr\}.

These sets get smaller as n increases, meaning that En+1,k is always a subset of En,k, because union involves fewer sets. A point x, for which the sequence (fm(x)) converges to f(x), can for fixed k not be in every En,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\biggl(\bigcap_{n\in\mathbb{N}}E_{n,k}\biggr)=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}) < \frac\varepsilon{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

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 A \ B we therefore have uniform convergence.

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

\mu(B) \le\sum_{k\in\mathbb{N}}\mu(E_{n_k,k}) \le\sum_{k\in\mathbb{N}}\frac\varepsilon{2^k} =\varepsilon.

[edit] References

  1. Richard Beals (2004). Analysis: An Introduction. New York: Cambridge University Press. ISBN 0-521-60047-2.
  2. Dmitri Egoroff (1911). Sur les suites des fonctions measurables. C.R. Acad. Sci. Paris, 152:135–157.
  3. Eric W. Weisstein et al. (2005). Egorov's Theorem. Retrieved April 19, 2005.