Problem 25
Question
3-25. Use induction on \(n\) to show that \(\left[a_{1}, b_{1}\right] \times
\times\left[a_{n}, b_{n}\right]\) is not a set of measure 0 (or content 0\()\) if
\(a_{i}
Step-by-Step Solution
Verified Answer
By induction, \( [a_{1}, b_{1}] \times \, \cdots \, \times [a_{n}, b_{n}] \) is not of measure 0 if \( a_{i} < b_{i} \) for each \( i \).
1Step 1: Base Case
Consider the base case where \( n = 1 \). We need to show that \( [a_{1}, b_{1}] \) is not a set of measure 0 if \( a_{1} < b_{1} \). A set consisting of a single interval \( [a_{1}, b_{1}] \) has measure \( b_{1} - a_{1} \) which is positive because \( a_{1} < b_{1} \). Thus, for \( n = 1 \), the set \( [a_{1}, b_{1}] \) is not of measure 0.
2Step 2: Inductive Step
Assume that the statement is true for \( n = k \). This means that the product set \( [a_{1}, b_{1}] \times [a_{2}, b_{2}] \times \, \cdots \, \times [a_{k}, b_{k}] \) is not of measure 0 if \( a_{i} < b_{i} \) for each \( i \) from \( 1 \) to \( k \).
3Step 3: Consider \( n = k+1 \)
We need to show that the set \( [a_{1}, b_{1}] \times [a_{2}, b_{2}] \times \, \cdots \, \times [a_{k}, b_{k}] \times [a_{k+1}, b_{k+1}] \) is not of measure 0 if \( a_{i} < b_{i} \) for each \( i \) from \( 1 \) to \( k+1 \).
4Step 4: Compute the Measure
The measure of the product set \( [a_{1}, b_{1}] \times [a_{2}, b_{2}] \times \, \cdots \, \times [a_{k}, b_{k}] \times [a_{k+1}, b_{k+1}] \) is the product of the lengths of each interval: \[ (b_{1} - a_{1}) \times (b_{2} - a_{2}) \times \, \cdots \, \times (b_{k} - a_{k}) \times (b_{k+1} - a_{k+1}) \] Since each \( b_{i} - a_{i} > 0 \), the product is also greater than 0, indicating that the set is not of measure 0.
5Step 5: Complete the Induction
By the principle of mathematical induction, since the base case \( n = 1 \) is true and the inductive step \( n = k \rightarrow n = k+1 \) holds, the statement is true for all \( n \geq 1 \). Therefore, \( [a_{1}, b_{1}] \times [a_{2}, b_{2}] \times \, \cdots \, \times [a_{n}, b_{n}] \) is not a set of measure 0 if \( a_{i} < b_{i} \) for each \( i \).
Key Concepts
Measure TheoryLebesgue MeasureCartesian Product
Measure Theory
Measure theory is a mathematical framework aimed at generalizing and extending the concept of length, area, and volume. It deals with the quantification of size or measure of sets in a rigorous way. This is essential for many advanced topics in analysis and probability. When working with measure theory:
- Measurable Sets: Not all sets are measurable. Only certain sets qualify under the rules of measure theory.
- Measure Functions: Measures assign a non-negative value to a set, representing its size. Examples include counting measures, Lebesgue measures, and probability measures.
Lebesgue Measure
Lebesgue measure is a specific type of measure in measure theory used to generalize the concept of length, area, and volume to more complex sets. It's a core concept in real analysis and its properties include inclusivity and comprehensiveness.
Unlike other measures that may only work well with simpler sets (like intervals), the Lebesgue measure applies to a larger variety, including disjoint unions and more intricate constructs.
Important Features:
Unlike other measures that may only work well with simpler sets (like intervals), the Lebesgue measure applies to a larger variety, including disjoint unions and more intricate constructs.
Important Features:
- Additivity: If you have non-overlapping sets, the total measure is the sum of individual measures.
- Translation Invariance: Lebesgue measure doesn't change even if you move the set throughout your space.
Cartesian Product
The Cartesian product is a mathematical operation that returns a set from multiple sets. It can be visualized as combining elements from each set into ordered pairs or tuples.
For example, if you have two sets, A and B, the Cartesian product, denoted as A × B, will be the set of all ordered pairs \( (a, b) \) you can form where \( a \) is from A and \( b \) is from B.
Critical Points to Note:
For example, if you have two sets, A and B, the Cartesian product, denoted as A × B, will be the set of all ordered pairs \( (a, b) \) you can form where \( a \) is from A and \( b \) is from B.
Critical Points to Note:
- Order Matters: In a Cartesian product, the pairs are ordered, meaning \( (a, b) \) is different from \( (b, a) \).
- Higher Dimensions: Cartesian products can be extended to multiple sets, forming triplets, quadruplets, and beyond.
Other exercises in this chapter
Problem 18
3-18. If \(f: A \rightarrow \mathbf{R}\) is non-negative and \(\int_{A} f=0\), show that \(\\{x: f(x) \neq 0\\}\) has measure 0. Hint: Prove that \(\\{x: f(x)>1
View solution Problem 23
3-23. Let \(C \subset A \times B\) be a set of content \(0 .\) Let \(A^{\prime} \subset A\) be the set of all \(x \in A\) such that \(\\{y \in B:(x, y) \in C\\}
View solution Problem 26
3-26. Let \(f:[a, b] \rightarrow \mathbf{R}\) be integrable and non-negative and let \(A_{f}=\) \(\\{(x, y): a \leq x \leq b\) and \(0 \leq y \leq f(x)\\} .\) S
View solution Problem 27
3-27. If \(f:[a, b] \times[a, b] \rightarrow \mathbf{R}\) is continuous, show that $$ \int_{a}^{b} \int_{a}^{y} f(x, y) d x d y=\int_{a} \int_{x} f(x, y) d y d
View solution