Littlewood's First Principle: Difference between revisions
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The previous two results imply the following formulation of Littlewood's First Principle: | The previous two results imply the following formulation of Littlewood's First Principle: | ||
'''Theorem.''' If <math> E </math> is a Borel measurable set and <math> m(E)<+\infty </math>, then for every <math> \varepsilon > 0</math> there is a set <math> A </math> that is a finite union of open intervals such that <math> m(A\ | '''Theorem.''' If <math> E </math> is a Borel measurable set and <math> m(E)<+\infty </math>, then for every <math> \varepsilon > 0</math> there is a set <math> A </math> that is a finite union of open intervals such that <math> m(A\Delta E) < \varepsilon </math>, where <math> \Delta </math> denotes symmetric difference. | ||
''Proof:'' Using the previous two sections find <math> K </math> and <math> U </math> such that <math> K\subset E \subset U </math> and <math> m(E)-\varepsilon < m(K) </math> and <math> m(E)+\varepsilon > m(U) </math>. As noted in the proof of the first section, <math> U </math> is really a countable union of open intervals. These intervals form an open cover for <math> K </math> and since <math> K </math> is compact, only finitely many cover <math> K </math>. Defining <math> A </math> to be the union of these finitely many intervals, the result follows, since <math> m(A/E) \leq m(U/E) </math> and <math> m(E/A) \leq m(E/K) </math>. | ''Proof:'' Using the previous two sections find <math> K </math> and <math> U </math> such that <math> K\subset E \subset U </math> and <math> m(E)-\varepsilon < m(K) </math> and <math> m(E)+\varepsilon > m(U) </math>. As noted in the proof of the first section, <math> U </math> is really a countable union of open intervals. These intervals form an open cover for <math> K </math> and since <math> K </math> is compact, only finitely many cover <math> K </math>. Defining <math> A </math> to be the union of these finitely many intervals, the result follows, since <math> m(A/E) \leq m(U/E) </math> and <math> m(E/A) \leq m(E/K) </math>. | ||
==References== | ==References== |
Latest revision as of 23:39, 19 December 2020
Littlewood three principles of real analysis in his 1944 Lectures on the Theory of Functions [1]:
There are three principles, roughly expressible in the following terms: Every (measurable) set is nearly a finite sum of intervals; every function (of class Lp) is nearly continuous; every convergent sequence of functions is nearly uniformly convergent.
This article examines the first principle and attempts to divine how restrictive the constraint of (Borel) measurability truly is for a set. This will be done in the form of two important results; namely, showing how well such a set can be approximated from above by open sets and how well such a set can be approximated from below by compact sets.
These two results will demonstrate how convenient Borel measurable sets are to work with and how one may more easily prove results about them by using this approximation and passing to the limit by continuity argument.
Approximation by Open Sets
The following theorem shows how well Borel measurable sets can be approximated from above by open sets.
Theorem. Let be a Borel measurable set. For every there exists an open set with such that .
Proof: By construction of the Borel measure, there exists a sequence of intervals such that and
Approximation by Compact Sets
The next theorem demonstrates how well Borel measurable sets can be approximated from below by compact sets.
Theorem. Let be a Borel measurable set. For every there exists an open set with such that .
Proof: First suppose is bounded. If is also closed, then we're done; is compact. Otherwise, the boundary is nonempty. Now, is a Borel measurable set, so we can approximate it from above by the previous section. That is, there exists an open set such that . Now define . By construction, this set is contained in . Moreover, it is closed, being the intersection of two closed sets. Finally, is bounded because is bounded. By the Heine-Borel Theorem, is compact, and satisfies the constraints of the theorem because
Littlewood's First Principle
The previous two results imply the following formulation of Littlewood's First Principle:
Theorem. If is a Borel measurable set and , then for every there is a set that is a finite union of open intervals such that , where denotes symmetric difference.
Proof: Using the previous two sections find and such that and and . As noted in the proof of the first section, is really a countable union of open intervals. These intervals form an open cover for and since is compact, only finitely many cover . Defining to be the union of these finitely many intervals, the result follows, since and .