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".
For an interval [a, b], let 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 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 is less than ε, then f is measurable.
be a Radon measure space and Y be a second-countable topological space equipped with a Borel algebra, and let
of finite measure there is a closed set
to be compact and even find a continuous function
with compact support that coincides with
and such that Informally, measurable functions into spaces with countable base can be approximated by continuous functions on arbitrarily large portion of their domain.
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.
The strength of Lusin's theorem might not be readily apparent, as can be demonstrated by example.
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
"knocks out" all of the rationals, leaving behind a compact, closed set
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