2. Let ([0, 1], L, m) be a Lebesgue measure space, and let A be a nonempty measurable subset of [0, 1]. Let {Ek}_1 ≤ [0, 1] be a countable disjoint collection of Lebesgue measurable sets. Show that a. m(AnŪR) - Σm(AMB). Ek). k=1 k=1 b. Let f [0, 1] → (0, 1] be a measurable function. Show that for every € > 0, there is a natural number N and a set C such that m(C₂) < € and < f(x) ≤ Ne+1 for all x ECE. Ne
2. Let ([0, 1], L, m) be a Lebesgue measure space, and let A be a nonempty measurable subset of [0, 1]. Let {Ek}_1 ≤ [0, 1] be a countable disjoint collection of Lebesgue measurable sets. Show that a. m(AnŪR) - Σm(AMB). Ek). k=1 k=1 b. Let f [0, 1] → (0, 1] be a measurable function. Show that for every € > 0, there is a natural number N and a set C such that m(C₂) < € and < f(x) ≤ Ne+1 for all x ECE. Ne
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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Question
Answer b only
![2. Let ([0, 1], L, m) be a Lebesgue measure space, and let A be a nonempty measurable
[0, 1] be a countable disjoint collection of Lebesgue
subset of [0, 1]. Let {E}
measurable sets.
a.
Show that
m
¹ (ADŮ Ek) = [m(An Ek).
-
k=1
k=1
b.
Let f [0, 1] → (0, 1] be a measurable function. Show that for every € > 0,
there is a natural number N, and a set Ce such that m(Ce) < € and < f(x) ≤
Ne+1
1
NE
for all x € CE.
∞](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4228d8f-8dd1-458e-9871-4eecf6e388c5%2Fbefadd4a-1345-4969-b064-b03ce401b5e8%2F3aghbx_processed.png&w=3840&q=75)
Transcribed Image Text:2. Let ([0, 1], L, m) be a Lebesgue measure space, and let A be a nonempty measurable
[0, 1] be a countable disjoint collection of Lebesgue
subset of [0, 1]. Let {E}
measurable sets.
a.
Show that
m
¹ (ADŮ Ek) = [m(An Ek).
-
k=1
k=1
b.
Let f [0, 1] → (0, 1] be a measurable function. Show that for every € > 0,
there is a natural number N, and a set Ce such that m(Ce) < € and < f(x) ≤
Ne+1
1
NE
for all x € CE.
∞
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