In the mathematical field of mathematical analysis, Lusin's theorem (or Luzin's theorem, named for Nikolai Luzin) or Lusin's criterion states that an almost-everywhere finite function is measurable if and only if it is a continuous function on nearly all its domain. In the informal formulation of J. E. Littlewood, "every measurable function is nearly continuous".

Classical statement

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For an interval [ab], let

 

be a measurable function. Then, for every ε > 0, there exists a compact E ⊆ [ab] such that f restricted to E is continuous and

 

Note that E inherits the subspace topology from [ab]; continuity of f restricted to E is defined using this topology.

Also for any function f, defined on the interval [a, b] and almost-everywhere finite, if for any ε > 0 there is a function ϕ, continuous on [a, b], such that the measure of the set

 

is less than ε, then f is measurable.[1]

General form

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Let   be a Radon measure space and Y be a second-countable topological space equipped with a Borel algebra, and let   be a measurable function. Given  , for every   of finite measure there is a closed set   with   such that   restricted to   is continuous. If   is locally compact and  , we can choose   to be compact and even find a continuous function   with compact support that coincides with   on   and such that

 .

Informally, measurable functions into spaces with countable base can be approximated by continuous functions on arbitrarily large portion of their domain.

On the proof

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The proof of Lusin's theorem can be found in many classical books. Intuitively, one expects it as a consequence of Egorov's theorem and density of smooth functions. Egorov's theorem states that pointwise convergence is nearly uniform, and uniform convergence preserves continuity.

Example

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The strength of Lusin's theorem might not be readily apparent, as can be demonstrated by example. Consider Dirichlet function, that is the indicator function   on the unit interval   taking the value of one on the rationals, and zero, otherwise. Clearly the measure of this function should be zero, but how can one find regions that are continuous, given that the rationals are dense in the reals? The requirements for Lusin's theorem can be satisfied with the following construction of a set  

Let   be any enumeration of  . Set

 

and

 .

Then the sequence of open sets   "knock out" all of the rationals, leaving behind a compact, closed set   which contains no rationals, and has a measure of more than  .

References

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Sources

  • N. Lusin. Sur les propriétés des fonctions mesurables, Comptes rendus de l'Académie des Sciences de Paris 154 (1912), 1688–1690.
  • G. Folland. Real Analysis: Modern Techniques and Their Applications, 2nd ed. Chapter 7
  • W. Zygmunt. Scorza-Dragoni property (in Polish), UMCS, Lublin, 1990
  • M. B. Feldman, "A Proof of Lusin's Theorem", American Math. Monthly, 88 (1981), 191-2
  • Lawrence C. Evans, Ronald F. Gariepy, "Measure Theory and fine properties of functions", CRC Press Taylor & Francis Group, Textbooks in mathematics, Theorem 1.14

Citations

  1. ^ "Luzin criterion - Encyclopedia of Mathematics".