Real Analysis Suppose that f : Rn → Rn is locally Lipschitz. Show that if E ⊂Rn has measure 0, then f(E) also has measure 0. Please prove this with Steps
Real Analysis Suppose that f : Rn → Rn is locally Lipschitz. Show that if E ⊂Rn has measure 0, then f(E) also has measure 0. Please prove this with Steps
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Suppose that f : Rn → Rn is locally Lipschitz.
Show that if E ⊂Rn has measure 0, then f(E) also has measure 0.
Please prove this with Steps
![**Title: Understanding Measure Zero and Lipschitz Functions in \(\mathbb{R}^n\)**
**Introduction:**
This section explores the relationship between measure zero sets and Lipschitz functions, specifically examining how a locally Lipschitz function impacts the measure of a set.
**Theorem Statement:**
Suppose that \( f : \mathbb{R}^n \to \mathbb{R}^n \) is locally Lipschitz. Show that if \( E \subset \mathbb{R}^n \) has measure 0, then \( f(E) \) also has measure 0.
**Definition 0.1:**
A subset \( E \) of \(\mathbb{R}^n\) has measure zero if for any \(\epsilon > 0\), there exists a countable collection \(\{R_i\}_{i=1}^\infty\) of non-degenerate closed rectangles in \(\mathbb{R}^n\) such that:
1. \( E \subset \bigcup_{i=1}^\infty R_i \);
2. \(\sum_{i=1}^\infty |R_i| < \epsilon\).
*Note:* \(|R| = \prod_{i=1}^n |a_i - b_i|\) for \( R = [a_1, b_1] \times \cdots \times [a_n, b_n]\). A closed rectangle \([a_1, b_1] \times \cdots \times [a_n, b_n]\) is non-degenerate, meaning \( a_i < b_i \) for all \( i = 1, 2, \ldots, n \).
**Hint for Proof:**
**Step 1:**
Show that \( A \subset \mathbb{R}^n \) has measure 0 if and only if for any \(\epsilon > 0\), there exists a countable collection of non-degenerate closed cubes \(\{D_i\}_{i=1}^\infty\) such that:
\[ A \subset \bigcup_{i=1}^\infty D_i \quad \text{and} \quad \sum_{i=1}^\infty |D_i| < \epsilon. \]
*A cube \( D \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1187a7f3-21d6-4d03-a9a5-364417f249b2%2F96cb9d73-664b-48a0-85f0-3239db35d147%2Ffx235db_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Measure Zero and Lipschitz Functions in \(\mathbb{R}^n\)**
**Introduction:**
This section explores the relationship between measure zero sets and Lipschitz functions, specifically examining how a locally Lipschitz function impacts the measure of a set.
**Theorem Statement:**
Suppose that \( f : \mathbb{R}^n \to \mathbb{R}^n \) is locally Lipschitz. Show that if \( E \subset \mathbb{R}^n \) has measure 0, then \( f(E) \) also has measure 0.
**Definition 0.1:**
A subset \( E \) of \(\mathbb{R}^n\) has measure zero if for any \(\epsilon > 0\), there exists a countable collection \(\{R_i\}_{i=1}^\infty\) of non-degenerate closed rectangles in \(\mathbb{R}^n\) such that:
1. \( E \subset \bigcup_{i=1}^\infty R_i \);
2. \(\sum_{i=1}^\infty |R_i| < \epsilon\).
*Note:* \(|R| = \prod_{i=1}^n |a_i - b_i|\) for \( R = [a_1, b_1] \times \cdots \times [a_n, b_n]\). A closed rectangle \([a_1, b_1] \times \cdots \times [a_n, b_n]\) is non-degenerate, meaning \( a_i < b_i \) for all \( i = 1, 2, \ldots, n \).
**Hint for Proof:**
**Step 1:**
Show that \( A \subset \mathbb{R}^n \) has measure 0 if and only if for any \(\epsilon > 0\), there exists a countable collection of non-degenerate closed cubes \(\{D_i\}_{i=1}^\infty\) such that:
\[ A \subset \bigcup_{i=1}^\infty D_i \quad \text{and} \quad \sum_{i=1}^\infty |D_i| < \epsilon. \]
*A cube \( D \
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