Question

If \(\mu^{*}\) is an outer measure on \(X\) and \(\left\\{A_{j}\right\\}_{1}^{\infty}\) is a sequence of disjoint \(\mu^{*}\) measurable sets, then \(\mu^{*}\left(E \cap\left(\bigcup_{1}^{\infty} A_{j}\right)\right)=\sum_{1}^{\infty} \mu^{*}\left(E \cap A_{j}\right)\) for any \(E \subset X .\)

Short answer

The statement is proven using countable additivity of the outer measure over disjoint \(\mu^{*}\)-measurable sets.

Step-by-step solution

Understand Measurability

The set \( E \) is given as a subset of \( X \), equipped with an outer measure \( \mu^{*} \). We are given a sequence of disjoint \( \mu^{*} \)-measurable sets \( \{ A_j \}_{j=1}^{\infty} \). Recall that a set \( A \) is \( \mu^{*} \)-measurable if for every set \( E \subset X \), the equation \( \mu^{*}(E) = \mu^{*}(E \cap A) + \mu^{*}(E \cap A^c) \) holds, where \( A^c \) is the complement of \( A \). This property will be crucial for the proof.

Apply the Definition of Outer Measure

For each integer \( j \), define the set \( E \cap A_j \), which represents the intersection of the set \( E \) with each of the disjoint sets \( A_j \). The problem statement suggests using the sum of outer measures of these intersections. Using the countable additivity property of outer measures over these measurable sets will be leveraged to show equality.

Use Countable Additivity on Disjoint Sets

Since \( \{ A_j \} \) are disjoint and measurable, by definition of \( \mu^{*} \) being countably additive over disjoint measurable sets, we can assert that \[ \mu^{*}\left(\bigcup_{j=1}^{\infty} A_j\right) = \sum_{j=1}^{\infty} \mu^{*}(A_j). \] This property applies here.

Analyze the Intersection

Evaluate \( \mu^{*}(E \cap (\bigcup_{j=1}^{\infty} A_j)) \). This can be expressed using distributive properties of sets as: \[ \mu^{*}\left(\bigcup_{j=1}^{\infty} (E \cap A_j)\right). \] By the countability and disjointness of \( A_j \), and them being \( \mu^{*} \)-measurable, we apply additivity:

Apply Countable Additivity to the Intersection

The countable additivity implies that \[ \mu^{*}\left(\bigcup_{j=1}^{\infty} (E \cap A_j)\right) = \sum_{j=1}^{\infty} \mu^{*}(E \cap A_j) \] because each \( E \cap A_j \) is a measurable set. Thus, \[ \mu^{*}\left(E \cap\left(\bigcup_{j=1}^{\infty} A_j\right)\right) = \sum_{j=1}^{\infty} \mu^{*}(E \cap A_j). \] This completes the proof and establishes the required equality.

Key concepts

Measurable Sets

In the foundational topic of measure theory, we encounter measurable sets. These are crucial for defining and understanding measures as they provide the structural foundation needed by many mathematical concepts. Imagine you have a set, say set \(A\). For \(A\) to be termed as \(\mu^{*}\)-measurable, it must satisfy a specific condition. For every subset \(E\) within a larger set \(X\), the outer measure of \(E\) must equal the sum of the outer measures of two disjoint sets: \(E \cap A\) (the elements in both \(E\) and \(A\)) and \(E \cap A^c\) (the elements in \(E\) but not in \(A\)). In equation form, this can be written as \(\mu^{*}(E) = \mu^{*}(E \cap A) + \mu^{*}(E \cap A^c)\).
This essentially says that the measure of the entire set \(E\) can be broken down into parts: the part that overlaps with \(A\) and the part that doesn't. This is crucial because it allows measure theory to "see" sets as entities that can be churned through operations while maintaining important properties like measure-preserving compatibility.

Countable Additivity

Now, let's move on to a fundamental property in measure theory called countable additivity. This property ensures that when you combine an infinite sequence of disjoint measurable sets, the measure of the union of all these sets is equal to the sum of the measures of each set individually. In simpler words, if you have disjoint measurable sets \(\{A_j\}_{j=1}^{\infty}\), the measure of the whole union \(\mu^{*}(\bigcup_{j=1}^{\infty} A_j)\) is just the sum of the measures of each part: \(\sum_{j=1}^{\infty} \mu^{*}(A_j)\).
This principle is necessary for consistency and reliability in mathematics as it helps ensure that our concept of "size" or "measure" doesn't change unexpectedly when dealing with infinite collections of disjoint sets. Countable additivity is what binds operations within measure theory to work predictably over infinite cases.

Set Theory

Lastly, let's delve into the essential framework of set theory. Set theory is the study of collections of objects, considered as objects themselves. It provides a standardized way to describe collections using the terms like union, intersection, and complement. - **Union**: The set that contains all elements in either set, or both.- **Intersection**: The set containing elements common to both sets.- **Complement**: Elements not in the set but in the universal set.In measure theory and for the problem we are discussing, set theory is used to nest operations inside each other. For instance, we analyze \(E \cap (\bigcup_{j=1}^{\infty} A_j)\). This expression uses both intersection and union:- \(E\) intersects with every set in the union of \(\{A_j\}_{j=1}^{\infty}\),- Making them separable by countable additivity.Set theory languages provide the backbone for defining and manipulating these measurable entities in a rigorous yet versatile manner. Understanding these basic operations allows one to harness the full potential of measure theory efficiently.