Question

Prove the additivity of the volume of intervals, namely, if \(A\) is subdivided, in any manner, into \(m\) mutually disjoint subintervals \(A_{1}, A_{2}, \ldots, A_{m}\) in \(E^{n}\), then $$ v A=\sum_{i=1}^{m} v A_{i} $$ (This is true also if some \(A_{i}\) contain common faces). [Proof outline: For \(m=2\), use Problem 8 Then by induction, suppose additivity holds for any number of intervals smaller than a certain \(m\) \((m>1) .\) Now let $$ A=\bigcup_{i=1}^{m} A_{i} \quad\left(A_{i} \text { disjoint }\right) $$ One of the \(A_{i}\) (say, \(\left.A_{1}=[\bar{a}, \bar{p}]\right)\) must have some edge-length smaller than the corresponding edge-length of \(A\left(\right.\) say \(\left., \ell_{1}\right) .\) Now cut all of \(A\) into $$ P=[\bar{a}, \bar{d}] \text { and } Q=A-P \text { (Figure 4) } $$ by the plane \(x_{1}=c\left(c=p_{1}\right)\) so that \(A_{1} \subseteq P\) while \(A_{2} \subseteq Q\). For simplicity, assume that the plane cuts each \(A_{i}\) into two subintervals \(A_{i}^{\prime}\) and \(A_{i}^{\prime \prime}\). (One of them may be empty.) Then $$ P=\bigcup_{i=1}^{m} A_{i}^{\prime} \text { and } Q=\bigcup_{i=1}^{m} A_{i}^{\prime \prime} $$ Actually, however, \(P\) and \(Q\) are split into fewer than \(m\) (nonempty) intervals since \(A_{1}^{\prime \prime}=\emptyset=A_{2}^{\prime}\) by construction. Thus, by our inductive assumption, $$ v P=\sum_{i=1}^{m} v A_{i}^{\prime} \text { and } v Q=\sum_{i=1}^{m} v A_{i}^{\prime \prime} $$ where \(v A_{1}^{\prime \prime}=0=v A_{2}^{\prime},\) and \(v A_{i}=v A_{i}^{\prime}+v A_{i}^{\prime \prime}\) by Problem \(8 .\) Complete the inductive proof by showing that $$ \left.v A=v P+v Q=\sum_{i=1}^{m} v A_{i} .\right] $$

Short answer

The additivity of volume for intervals, \( vA = \sum_{i=1}^{m} vA_i \), holds for any disjoint subintervals.

Step-by-step solution

Base Case: Additivity for Two Intervals

Let's prove the additivity of the volume for the case where the interval \( A \) is divided into two mutually disjoint subintervals, \( A_1 \) and \( A_2 \). For \( m=2 \), we have \( A = A_1 \cup A_2 \) with \( A_1 \cap A_2 = \emptyset \). Based on Problem 8, if intervals \( A_1 \) and \( A_2 \) are disjoint, the volume of \( A \), denoted as \( vA \), is given by \( vA = vA_1 + vA_2 \). Hence, the base case holds: the additivity of the volume for 2 intervals.

Inductive Hypothesis

Assume that the additivity of volumes holds for any collection of \( m-1 \) or fewer disjoint subintervals. This means that for any subdivision of an interval \( A \) into \( m-1 \) subintervals \( A_1, A_2, \ldots, A_{m-1} \), the volume \( vA = \sum_{i=1}^{m-1} vA_i \) holds true by assumption.

Inductive Step for m Intervals

Suppose we subdivide \( A \) into \( m \) disjoint subintervals \( A_1, A_2, \ldots, A_m \). We need to show \( vA = \sum_{i=1}^{m} vA_i \). Choose one subinterval, say \( A_1 \), with an edge shorter than the corresponding edge of \( A \). Cut \( A \) into \( P \) and \( Q \) by the plane such that \( P=A_1 \cup A_1' \) and \( Q=A_2 \cup \cdots \cup A_m \).

Proof of Inductiveness in m Intervals

For the partition \( A \), it's known that \( P \) and \( Q \) are now fewer than \( m \) intervals. By the inductive hypothesis, the volumes are \( vP = vA_1 + vA_1' \) and \( vQ = \, \sum_{i=2}^{m} vA_i \). Since \( A_A'' \) and \( A_2' \) are empty, it is simplified to: \( vP + vQ = \sum_{i=1}^{m} vA_i \), thus completing the induction.

Conclusion of Proof

Finally, by proving the base case for two intervals and the general case for \( m \) using the inductive hypothesis, the additivity of the volume for any \( m \) disjoint subintervals \( A_1, A_2, \ldots, A_m \) is demonstrated: \( vA = \sum_{i=1}^{m} vA_i \).

Key concepts

Measure Theory

Measure theory is a fundamental area of mathematics that deals with the generalization and study of concepts related to size, length, area, and volume. One of its central aims is to define a systematic way of assigning a measure to subsets of a given space, which in this context can be intervals within one or any number of dimensions. Measure theory helps establish a formal mathematical framework for talking about the "volume" of complex geometric objects, whether simple intervals or more complex structures.
This theory gives us the tools to develop, validate, and apply the essential principles of integration and probability. Within measure theory, an interval's "volume" can be seen as a measure, and the additivity principle asserts that the total volume of a set is the sum of the volumes of its mutually disjoint parts. This rule is pivotal in ensuring consistent results, such as confirming that sub-volumes add up correctly without overlap, which is crucial in both theoretical and practical applications like calculus and probability theory.

Inductive Proof

Inductive proof is a powerful mathematical technique used to prove a statement for all natural numbers by proving it for a starting point, typically 1 or 2, and then showing that if it holds for a number \( n \), it also holds for \( n+1 \). This step-by-step logical construction helps affirm that a property, like the additivity of volume, is universally true across all conditions defined in the hypothesis.
In proving the additivity of volume, induction begins with a **base case** where the statement is shown to be true for a small number of subintervals, often two. With this base case established, the method shifts to the **inductive step**. Here, the proof assumes that the property is true for \( m-1 \) intervals and demonstrates that it must then also be true for \( m \) intervals.
This approach builds on smaller, easily verifiable cases to prove larger, more complex ones, providing a comprehensive method to validate the property beyond initial examples. Through this structured proof by induction, we achieve a rigorous and reliable understanding of mathematical properties like the additivity of volume.

Subintervals

Subintervals are smaller segments or partitions within a larger interval that are separated along the same continuum. If we consider an interval as a piece of line or space, subintervals can be visually imagined as distinct non-overlapping pieces of that line. By examining subintervals, we can break down complex problems into simpler, more manageable parts.
In the context of the problem, understanding subintervals assumes dividing a given interval into smaller portions in such a way that each part independently contributes to the whole. The relationship between the sum of the measures of all subintervals and the measure of the initial interval is a focal point in the proof of volume additivity.
Using subintervals allows mathematicians to explore and demonstrate properties such as continuity and differentiability fabricating bridges between theory and application. By systematically examining each subinterval, we can ensure completeness and accurate measurements across the entire interval.

Disjoint Intervals

Disjoint intervals are intervals that do not overlap; each interval is separate and distinct from its companions within the same set. This is a crucial concept in measure theory as it ensures that when summing up the volumes of such intervals, there is no "double-counting" of common parts, which aligns with the principle of additivity.
When breaking down a broader interval into smaller subintervals, ensuring these subintervals are disjoint allows for a straightforward summation of their individual volumes to obtain the total volume.
In our proof, each subinterval must be meticulously constructed to be disjoint, allowing us to leverage the additivity property accurately. By maintaining these conditions, we guarantee that the integrity of the measurement process holds, affirming the total volume as merely the sum of its disjunct, individual parts.
Understanding how and why intervals are disjoint plays a crucial role in various applications of measure theory, particularly when dealing with integration and the comprehensive analysis of real-valued functions.