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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

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 64E
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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

Suppose that f: RR" is locally Lipschitz. Show that if ECR has measure 0, then f(E) also has measure 0. Definition 0.1. A subset E of Rn has measure zero if for any e > 0 there exists a countable collection {R} of non-degenerate closed rectangles in Rn such that 1. ECUR; ₁R₁ < €. 2. Note that |R| = II1|ai - bi| for R = [a₁, b₁] × × × [an, bn]. A closed rectangle [a₁, b₁] × x [an, bn] being non-degenerate means that ai <bi for all i=1,2,..., n. Hint: Step 1: Show that AC Rn has measure 0 if and only if for any e > 0 there exists a countable collection of non-degenerate closed cubes {Di} such that Σ|D₂|<E. ∞ CU Di and i=1 Ac i=1 By definition, a cube D is a non-degenerate rectangle [a₁, b₁] × [an, bn] such that ai-bil X = |aj - bjl for all i, j = 1,..., n. The common value |ai - bil is called the width of D. Step 2: Show that if A₁, A2,... CR are all sets of measure 0, then U₁A has measure 0. Step 3: Let EC Rn have measure 0. Define Ek En B(0, k) for cach ke N. Prove that f(Ek) has measure 0 for any k N. Step 4: Use Steps 2 and 3 to show that f(E) has measure 0 since U₁f (Ek) = f(E).
Transcribed Image Text:Suppose that f: RR" is locally Lipschitz. Show that if ECR has measure 0, then f(E) also has measure 0. Definition 0.1. A subset E of Rn has measure zero if for any e > 0 there exists a countable collection {R} of non-degenerate closed rectangles in Rn such that 1. ECUR; ₁R₁ < €. 2. Note that |R| = II1|ai - bi| for R = [a₁, b₁] × × × [an, bn]. A closed rectangle [a₁, b₁] × x [an, bn] being non-degenerate means that ai <bi for all i=1,2,..., n. Hint: Step 1: Show that AC Rn has measure 0 if and only if for any e > 0 there exists a countable collection of non-degenerate closed cubes {Di} such that Σ|D₂|<E. ∞ CU Di and i=1 Ac i=1 By definition, a cube D is a non-degenerate rectangle [a₁, b₁] × [an, bn] such that ai-bil X = |aj - bjl for all i, j = 1,..., n. The common value |ai - bil is called the width of D. Step 2: Show that if A₁, A2,... CR are all sets of measure 0, then U₁A has measure 0. Step 3: Let EC Rn have measure 0. Define Ek En B(0, k) for cach ke N. Prove that f(Ek) has measure 0 for any k N. Step 4: Use Steps 2 and 3 to show that f(E) has measure 0 since U₁f (Ek) = f(E).
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