Problem 7 - A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry

Question

If \(\mu^{*}(N)=0\) show that for any set \(E, \mu^{*}(E \cup N)=\mu^{*}(E-N)=\mu^{*}(E)\). Hence show that \(E \cup N\) and \(E-N\) are Lebesgue measurable if and only if \(E\) is measurable.

Step-by-Step Solution

Answer

Given a null set \(N\), we have the following: \(\mu^{*}(E \cup N) = \mu^{*}(E)\) and \(\mu^{*}(E-N) = \mu^{*}(E)\). Furthermore, if \(E\) is measurable, the measurability is preserved under union and difference with a null set. Hence, both \(E \cup N\) and \(E-N\) are also Lebesgue measurable.

Step 1: Show that \(\mu^{*}(E \cup N) = \mu^{*}(E)\)

By sub-additivity of an outer measure \(\mu^{*}\), we have \(\mu^{*}(E \cup N) \leq \mu^{*}(E) + \mu^{*}(N)\). Since \(\mu^{*}(N) = 0\), we get \(\mu^{*}(E \cup N) \leq \mu^{*}(E)\). We also know that \(E \subseteq E \cup N\) so \(\mu^{*}(E) \leq \mu^{*}(E \cup N)\) by monotonicity of the outer measure. Hence, \(\mu^{*}(E \cup N) = \mu^{*}(E)\).

Step 2: Show that \(\mu^{*}(E-N) = \mu^{*}(E)\)

We know that \(E = (E-N) \cup (E \cap N)\). By sub-additivity, \(\mu^{*}(E) \leq \mu^{*}(E-N) + \mu^{*}(E \cap N)\). Since \(E \cap N\) is subset of \(N\) which is a null set, \(\mu^{*}(E \cap N) = 0\). Therefore, \(\mu^{*}(E) \leq \mu^{*}(E-N)\). By monotonicity, since \(E-N \subseteq E\), \(\mu^{*}(E-N) \leq \mu^{*}(E)\). Hence, \(\mu^{*}(E-N) = \mu^{*}(E)\).

Step 3: Prove the final assertion

Now, E is measurable if and only if for all \(A \subset \mathbb{R}\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^c)\). Replace \(A\) with \(A \cup N\) and \(A-N\) in this expression and use the results from step 1 and 2 to show that this condition is also fulfilled for \(E \cup N\) and \(E-N\) if \(E\) is measurable.

Key Concepts

Sub-additivity

Sub-additivity is a fundamental property of measures, particularly when dealing with the concept of an outer measure. It states that for any two sets, say \( A \) and \( B \), the measure of their union is less than or equal to the sum of their individual measures:

• \( \mu^*(A \cup B) \leq \mu^*(A) + \mu^*(B) \)

This property is particularly useful when dealing with complex sets in mathematics, as it provides a method to estimate the measure of the union of multiple sets.
In the given exercise, this concept helps show that \( \mu^*(E \cup N) \leq \mu^*(E) + \mu^*(N) \). Given \( \mu^*(N) = 0 \), it simplifies the inequality to \( \mu^*(E \cup N) \leq \mu^*(E) \). This is a powerful tool in measure theory that helps in handling cases involving null sets and the uniqueness of set measures.

Monotonicity

Monotonicity is another crucial characteristic of measure theory. It indicates that if a set \( A \) is a subset of another set \( B \) (meaning every element of A is also in B), the measure of \( A \) should be less than or equal to the measure of \( B \):

• \( \mu^*(A) \leq \mu^*(B) \) if \( A \subseteq B \)

This property assures that the measure does not increase when moving to a subset.
Within the exercise, monotonicity helps us establish that due to \( E \subseteq E \cup N \), we have \( \mu^*(E) \leq \mu^*(E \cup N) \). This relation, together with the result from sub-additivity, confirms that \( \mu^*(E \cup N) = \mu^*(E) \). It assures the consistency of measures when considering subsets and unions of sets.

Outer Measure

The concept of outer measure, denoted often as \( \mu^* \), is a way of assigning a measure to a set in a broader context than traditional measures allow.
Outer measures help to quantify the size of any subset of any space, even those that are not directly measurable through standard means. This is pivotal when dealing with sets that may not exhibit clear boundaries or definable characteristics.

• Outer measure includes the concepts of both inner and covering capabilities, providing more flexible criteria for set measurement.
• It forms the basis of the Lebesgue measure, allowing for the extension of measure to more irregular sets.

In the exercise at hand, the outer measure is employed to demonstrate that both \( \mu^*(E \cup N) \) and \( \mu^*(E-N) \) are equivalent to \( \mu^*(E) \). Thus, even when dealing with a null set \( N \), we can apply the outer measure reliably.

Null Set

Null sets, often also called zero measure sets, are crucial elements in measure theory. A null set is a set that has no 'size' in terms of measure. More formally, for a set \( N \) if \( \mu^*(N) = 0 \), then \( N \) is a null set.
In practical terms, this means that from the perspective of measure theory, null sets do not contribute any size or value, regardless of their elements.

• They can contain any number of elements but have zero measure because they are so "small", even if they contain infinitely many points.
• Null sets are critical in various proofs and theorems, as their property of adding zero measure plays an important role in simplifying equations and relationships like unions.

The original exercise hinges on the idea that because \( \mu^*(N) = 0 \), the null set \( N \) does not alter the measure of any set \( E \) when performing operations like union or set difference. This reflects the robust nature of measurement and its ability to handle sets of negligible size efficiently.