Lusin's theorem

The set of Lusin® (by Nikolai Luzin ) is a mathematical theorem from the measure theory . It says that the domain of a measurable function can be restricted in such a way that the function is continuous on this restriction . Lusin provided the proof of this theorem in 1912, after the theorem was first suggested by Émile Borel in 1903 and formulated mathematically by Henri Lebesgue .

Motivation of the sentence

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

${\displaystyle D(x)={\begin{cases}1,&{\mbox{wenn }}x{\mbox{ rational}}\\0,&{\mbox{wenn }}x{\mbox{ irrational}}\end{cases}}}$

which maps all rational numbers to 1 and all irrational numbers to 0, you can see that there are measurable functions that are not continuous at any point. Lusin's theorem now shows that a measurable function is “almost continuous”. What is meant by “almost continuously” is clear from the sentence.

Lusin's theorem

In the following the Lebesgue measure . ${\displaystyle \lambda }$

Be a measurable amount with . If it is a measurable and bounded function, there is for each a compact set with such that the constraint is continuous. ${\displaystyle M\subset \mathbb {R} ^{n}}$ ${\displaystyle \lambda (M)<\infty }$ ${\displaystyle f\colon M\to {\mathbb {R} }}$ ${\displaystyle \varepsilon >0}$ ${\displaystyle K\subset M}$ ${\displaystyle \lambda (M\backslash K)<\varepsilon }$ ${\displaystyle f|_{K}}$

Evidence sketch: This theorem can be derived from Egorov's theorem. Since it belongs as a limited, measurable function and since the continuous functions are dense in this space , there is a sequence of continuous functions that converges to in the -norm . By going to a subsequence one can assume that outside a set of measure 0 there is point-wise convergence. According to Jegorow's theorem, there is uniform convergence outside a set of measure smaller than , and this set can be assumed to be open because of the regularity of the Lebesgue measure. The complement is then compact and the sequence converges uniformly on. Therefore the limit function is continuous. ${\displaystyle f}$ ${\displaystyle L^{1}(M)}$ ${\displaystyle (f_{n})_{n}}$ ${\displaystyle L^{1}}$ ${\displaystyle f}$ ${\displaystyle \varepsilon }$ ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle f|_{K}}$

It is possible to tighten the statement still further: Be measurable and measurable. Then for each there is a set with and a continuous function which corresponds to with . ${\displaystyle M\subset \mathbb {R} ^{n}}$ ${\displaystyle f\colon M\to \mathbb {R} }$ ${\displaystyle \varepsilon >0}$ ${\displaystyle K\subset M}$ ${\displaystyle \lambda (M\backslash K)<\varepsilon }$ ${\displaystyle {\tilde {f}}\colon M\to \mathbb {R} }$ ${\displaystyle K}$ ${\displaystyle f}$

example

There seems to be a contradiction to the above example when looking at and , because the function is not continuous at any point . Note, however, that Lusin's theorem does not claim that the function is continuous at every point . Rather, it says that another function, namely the restriction , is continuous at every point . In order for the above function to demonstrate is a counting of rational numbers in . To the given set . Then the union of these sets contains all rational points, it is relatively open with measure smaller than , and on the compact complement the function is constant 0, that is, it is the null function and therefore continuous. ${\displaystyle M=[0,1]}$ ${\displaystyle f=D|_{[0,1]}:[0,1]\to \mathbb {R} }$ ${\displaystyle f=D|_{[0,1]}}$ ${\displaystyle [0,1]}$ ${\displaystyle f}$ ${\displaystyle K}$ ${\displaystyle f|_{K}}$ ${\displaystyle K}$ ${\displaystyle f=D|_{[0,1]}}$ ${\displaystyle (r_{n})_{n}}$ ${\displaystyle [0,1]}$ ${\displaystyle \varepsilon >0}$ ${\displaystyle U_{n}=\left(r_{n}-{\tfrac {\varepsilon }{2^{n+2}}},r_{n}+{\tfrac {\varepsilon }{2^{n+2}}}\right)\cap [0,1]}$ ${\displaystyle \varepsilon }$ ${\displaystyle K}$ ${\displaystyle f|_{K}}$

generalization

Lusin's theorem does not only apply to functions on measurable sets in the . It can also be generalized to real-valued functions of locally compact spaces : ${\displaystyle {\mathbb {R} }^{n}}$

Be a measure space , and locally compact , a σ-algebra on which the Borel sets includes, and a regular level was. be a measurable function. ${\displaystyle (X,\Sigma ,\mu )}$ ${\displaystyle X}$ ${\displaystyle \Sigma }$ ${\displaystyle X}$ ${\displaystyle \mu }$ ${\displaystyle f\colon X\to {\mathbb {R} }}$ ${\displaystyle \Sigma }$

Then there is a compact set for each with and for each with , so that is continuous. ${\displaystyle A\in \Sigma }$ ${\displaystyle \mu (A)<\infty }$ ${\displaystyle \varepsilon >0}$ ${\displaystyle K\subset A}$ ${\displaystyle \mu (A\setminus K)<\varepsilon }$ ${\displaystyle f|_{K}}$

In the situation of this theorem one can even find a continuous function with compact support such that . ${\displaystyle g\colon X\to {\mathbb {R} }}$ ${\displaystyle f|_{K}=g|_{K}}$

literature

• Nikolai Lusin : Sur les propriétés des fonctions mesurables. In: Comptes Rendus de l'Académie des Sciences de Paris. Vol. 154, 1912, pp. 1688-1690, digitized .
• Donald L. Cohn: Measure Theory. Birkhäuser, Boston MA et al. 1980, ISBN 3-7643-3003-1 .