Lusin's theorem

The set of Lusin (after Nikolai Nikolaevich Luzin ) is a mathematical theorem from measure theory. He states that the domain of a measurable function can be restricted so that the functionality of this restriction is continuous. Lusin provided the proof of this theorem in 1912, when the record was in 1903 indicated by Émile Borel first and mathematically formulated by Henri Lebesgue.

Motivation of the sentence

From the definition of the Lebesgue measure follows immediately that every continuous function is measurable. The example of the Dirichlet function

Which maps all rational numbers to 1 and all irrational numbers to 0, we see that there are measurable functions which are continuous at any point. The set of Lusin now shows that a measurable function is " almost continuous". What is " almost constantly " to be understood, it is clear from the record.

Set of Lusin

The following denotes the Lebesgue measure.

Let be a measurable set with. Let be a measurable and bounded function, then there exists for every a compact set with such that the restriction is continuous.

Sketch of proof: This theorem can be derived from the set of Yegorov. Because as limited, measurable function belongs to and since the continuous functions are dense in this space, there is a sequence of continuous functions that converges in the norm against. While passing to a subsequence, we can assume that outside a set of measure 0 exists pointwise convergence. By the theorem of Egorov is then uniform convergence outside a set of measure smaller than before, and this amount can be assumed to be open because of the regularity of the Lebesgue measure. The complement is then compact, and the sequence converges uniformly. Therefore, the limit function is continuous.

It is possible that the statement still to tighten: Be measurable and measurable. Then, for each of a set and a continuous extension of that matches up with.

Example

It seems a contradiction to the above example to exist when you and regards, because the function is not in any point ever. Note, however, that the set of Lusin not claim that the function at each point of is continuous. He says, rather, that a different function, namely the restriction is continuous at every point. To demonstrate this for the function above, was an enumeration of the rational numbers in. For a given set. Then the union of these sets contains all rational points, it is relatively open with degree less than, and on the compact complement the function is constant 0, that is, the zero function and therefore continuous.

Generalization

The set of Lusin not only applies to functions on measurable quantities. He can be generalized real-valued functions on locally compact spaces:

In the situation of this theorem one can even find a continuous function with compact support, so that.