# Lusin's theorem

Lusin's theorem (Redirected from Luzin's theorem) Jump to navigation Jump to search This article is about the theorem of real analysis. For the separation theorem in descriptive set theory, see Lusin's separation theorem.

In the mathematical field of real 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".

Contents 1 Classical statement 2 General form 3 On the proof 4 References Classical statement For an interval [a, b], let {displaystyle f:[a,b]rightarrow mathbb {C} } be a measurable function. Then, for every ε > 0, there exists a compact E ⊆ [a, b] such that f restricted to E is continuous and {displaystyle mu (E)>b-a-varepsilon .} Note that E inherits the subspace topology from [a, b]; 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 {displaystyle {xin [a,b]:f(x)neq phi (x)}} is less than ε, then f is measurable.[1] General form Let {displaystyle (X,Sigma ,mu )} be a Radon measure space and Y be a second-countable topological space equipped with a Borel algebra, and let {displaystyle f:Xrightarrow Y} be a measurable function. Given {displaystyle varepsilon >0} , for every {displaystyle Ain Sigma } of finite measure there is a closed set {displaystyle E} with {displaystyle mu (Asetminus E)

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