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
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.2: Mappings
Problem 11E: For the given f:ZZ, decide whether f is onto and whether it is one-to-one. Prove that your decisions...
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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
![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).](/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: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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