Littlewood's First Principle

Littlewood stated the principles in his 1944 Lectures on the Theory of Functions as:

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 main result is the following: Let $E$ be a Borel measurable set. For every $\varepsilon >0$ there exists a compact set $K$ with $K\subset E$ such that $m(E)<m(K)+\varepsilon$. Note that this implies $0<m(K)-m(E)<\varepsilon$ by monotonicity.

Proof: By construction of the Borel measure, there exists a sequence of intervals $\{(a_{k},b_{k})\}_{k=1}^{\infty }$ such that $E\subset \bigcup _{k=1}^{\infty }(a_{k},b_{k})$ and

$$m(E)>\sum _{k=1}^{\infty }(b_{k}-a_{k})-\varepsilon$$

Taking $U=\bigcup _{k=1}^{\infty }(a_{k},b_{k})$ gives the result.

Approximation by Compact Sets

The main result is the following: Let </math> E </math> be a Borel measurable set. For every </math>\varepsilon > 0</math> there exists an open set </math> U </math> with </math> E\subset U </math> such that </math> m(E)>m(U)-\varepsilon </math>. Note that this implies </math> 0< m(U)-m(E)<\varepsilon </math> by monotonicity.

Proof: First suppose </math> E </math> is bounded. If </math> E </math> is also closed, then we're done; </math> E </math> is compact. Otherwise, the boundary </math> \partial{E}=\overline{E}\backslash E </math> is nonempty. Now, </math> \partial{E} </math> is a Borel measurable set, so we can approximate it from above by the previous section. That is, there exists an open set </math> U\supset \partial{E} </math> such that </math> m(U) < m(\overline{E}\backslash E) +\varepsilon = \varepsilon </math>. Now define </math> K= \overline{E} \backslash U </math>. By construction, this set is contained in </math> U </math>. Moreover, it is closed, being the intersection of two closed sets. Finally, </math> K </math> is bounded because </math> E </math> is bounded. By the Heine-Borel Theorem, </math> K </math> is compact, and </math> K </math> satisfies the constraints of the theorem because

$$m(K)=m({\overline {E}}\backslash U)=m(E)-M(E\cap U)=m(E)-m(U)+m(U/E)\geq m(E)-m(U)+m({\overline {E}}\backslash E)\geq m(E)-\varepsilon$$

Now assume </math> E </math> is unbounded. Let </math> E_j = E \cap [j, j+1) </math>. By the preceding argument, for each </math> \varepsilon > 0 </math> there exist compact </math> K_j </math> such that </math> K_j \subset E_j </math> and </math> m(K_j) > m(E_j)-\varepsilon / 2^k </math>. Let </math> H_n = \bigcup_{j=-n}^{n} K_j </math>. Then </math> H_n </math> is compact and </math> m(H_n) \geq m(\bigcup_{j=-n}^n E_j) + \varepsilon </math>, and since </math> \lim_{n\to\infty} m(E)=m(\bigcup_{j=-n}^n E_j) </math>, the result follows.

Littlewood's First Principle

The previous two results imply the following formulation of Littlewood's First Principle: 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)-\epsilon < m(K) </math> and </math> m(E)+\epsilon > 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>.