Question

In writing Eqn. (7.3.53) we assumed that $$ \int_{\mathbf{G}(S)} f(\mathbf{X}) d \mathbf{X}=\int_{\mathbf{G}\left(S_{1}\right)} f(\mathbf{X}) d \mathbf{X}+\int_{\mathbf{G}\left(S \cap S_{1}^{c}\right)} f(\mathbf{X}) d \mathbf{X}. $$ Justify this. HINT: Show that \(\mathbf{G}\left(S_{1}\right) \cap \mathbf{G}\left(S \cap S_{1}^{c}\right)\) has zero content.

Short answer

Short Answer: The given equation can be justified by showing that the intersection between the two regions \(\mathbf{G}(S_1)\) and \(\mathbf{G}(S \cap S_1^c)\) has zero content, making them disjoint sets. Since they are disjoint, their individual integrals can be summed together to yield the integral over the original region, \(\mathbf{G}(S)\).

Step-by-step solution

Define regions and content

From the given equation, we have three regions: \(\mathbf{G}(S)\), \(\mathbf{G}(S_1)\), and \(\mathbf{G}(S \cap S_1^c)\). These regions are formed by applying some functions or transformations to sets \(S\), \(S_1\), and their intersection and complement. Content refers to a measure of the size of a region similar to the concept of area or volume.

Rewrite the equation in terms of Sets

The equation can be rewritten in terms of sets as: $$\mathbf{G}(S) = \mathbf{G}(S_1) \cup \mathbf{G}(S \cap S_1^c).$$ We are supposed to prove that the equation represents two regions with no intersection, so that their sum is equal to the integral of the whole set.

Show intersection has zero content

We can rewrite the intersection as follows: $$\mathbf{G}(S_1) \cap \mathbf{G}(S \cap S_1^c) = \mathbf{G}(S_1 \cap (S \cap S_1^c)).$$ Now, using the hint, we will show that this intersection has zero content. Notice that: 1. \((S_1 \cap S) \subseteq S_1\) 2. \((S \cap S_1^c) \subseteq S_1^c\) From these two inclusions, we can conclude that the intersection of sets \((S_1 \cap S)\) and \((S \cap S_1^c)\) is empty, or they are disjoint sets: $$S_1 \cap S \cap S_1^c = \emptyset.$$ Since the intersection is empty, we can infer that the content of the intersection is zero: $$\mathrm{content}\left(\mathbf{G}(S_1) \cap \mathbf{G}(S \cap S_1^c)\right) = 0.$$

Justify the original equation

Now that we have shown that the intersection has zero content, we can justify the original equation. Recall the equation: $$ \int_{\mathbf{G}(S)} f(\mathbf{X}) d\mathbf{X}=\int_{\mathbf{G}\left(S_{1}\right)} f(\mathbf{X}) d\mathbf{X}+\int_{\mathbf{G}\left(S \cap S_{1}^{c}\right)} f(\mathbf{X}) d\mathbf{X}. $$ Since \(\mathbf{G}(S_1)\) and \(\mathbf{G}(S \cap S_1^c)\) have no intersection (zero content), the integrals over each region can be summed together to yield the integral over the original region, \(\mathbf{G}(S)\): $$\int_{\mathbf{G}(S)} f(\mathbf{X}) d\mathbf{X} = \int_{\mathbf{G}(S_1)} f(\mathbf{X}) d\mathbf{X} + \int_{\mathbf{G}(S \cap S_1^c)} f(\mathbf{X}) d\mathbf{X}. $$ This justifies the original equation.

Key concepts

Measure Theory

Measure theory is a branch of mathematics that deals with the concept of "measure," which generalizes the notion of lengths, areas, and volumes. This is particularly useful when dealing with more complex geometrical spaces that rise in mathematical analysis and probability. In measure theory, a measure is a systematic way of assigning a number to a subset of a given set, interpretable as its size. This is similar to generalizing the idea of how we understand dimensions and calculating exact sizes in geometric shapes like squares and circles. Measure theory, therefore, provides the tools required for integrating functions over complex regions.
When solving integrals, such as the one in the exercise, measure theory allows us to break down complex sets into smaller, more manageable pieces where standard arithmetic and algebraic operations can apply. This set-theoretic view of integration ensures that even if the space is fragmented into strange shapes, we can calculate larger integral sums accurately through these measures. Specifically, in the exercise, each part of the set can be measured individually and then added, as long as there’s no overlap between them.

Disjoint Sets

In set theory, disjoint sets are sets that have no elements in common. This is analogous to non-overlapping regions in geometry. In mathematical terms, two sets are disjoint if their intersection is an empty set, denoted as \(S_1 \cap S_2 = \emptyset\).
Disjoint sets are crucial in calculus and integration because when you consider the union of two sets where the sets have no common elements, measuring properties like their combined area or volume becomes straightforward. You can essentially sum the measures of the sets independently because there’s no overlapping region that’s included in the calculation twice.
In the context of the given exercise, showing that \(\mathbf{G}(S_1)\) and \(\mathbf{G}(S \cap S_1^c)\) are disjoint ensures that when taking integrals over these regions, the results from each are accurately added without repetition, reaffirming the given integral equation. The crucial part here is proving their intersection has zero content, confirming they are indeed disjoint.

Region Content

Region content often refers to the measure or "size" of a region, similar to how we think about the area of a two-dimensional plane or the volume of a three-dimensional space. It's essential to understand that in mathematics, especially in scenarios involving measure and integration, region content does not always align with our intuitive notions of size, particularly when we deal with more complex or abstract spaces.
In integrals, the content of a region is used to compute the integral of a function over that space. For instance, in the solved exercise, we dealt with function values summed over specific regions. Each region's content is crucial because it dictates how much "weight" or "influence" each part of the space has in the entire calculation.
When asserting that a region has zero content, we claim that this particular segment of space contributes nothing to the integral's sum. Therefore, recognizing and proving that some intersections between regions have zero content allows one to conclude safely that the function values make no additional contribution from that area, simplifying the calculation by removing any unnecessary components.