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Measure and Integration

The document provides proofs for several theorems related to measure and integration, including the monotone convergence theorem, layer cake representation, bathtub principle, completeness of a measure, and regularity of Lebesgue measure. Each proof is presented succinctly, demonstrating key properties of measurable functions and sets. The results emphasize the relationships between integrals, measurable functions, and the structure of measure theory.

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14 views2 pages

Measure and Integration

The document provides proofs for several theorems related to measure and integration, including the monotone convergence theorem, layer cake representation, bathtub principle, completeness of a measure, and regularity of Lebesgue measure. Each proof is presented succinctly, demonstrating key properties of measurable functions and sets. The results emphasize the relationships between integrals, measurable functions, and the structure of measure theory.

Uploaded by

coolsirmohanmaths
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Measure and Integration - Selected Solutions

Q2. Prove equation (2) in Theorem 1.6 (monotone convergence).


Theorem 1.6 states that if (f_n) is a sequence of non-negative measurable
functions increasing pointwise to f, then:
lim ∫ f_n dµ = ∫ f dµ.

Proof: Since f_n ≤ f, by monotonicity of the integral, we have ∫ f_n dµ ≤ ∫ f dµ for all
n. Define L = lim ∫ f_n dµ. By monotone convergence theorem (Beppo Levi), f_n ↑ f
implies ∫ f_n dµ ↑ ∫ f dµ. Thus, L = ∫ f dµ. Hence proved.

Q3. Alternative proof of the layer cake representation (without Fubini’s


theorem).
For a non-negative measurable function f, the layer cake representation is:
f(x) = ∫0∞ 1_{f(x) > t} dt.

Proof: For fixed x, let a = f(x). Then the indicator function 1_{f(x) > t} equals 1
when 0 ≤ t < a and 0 otherwise. Hence, ∫0∞ 1_{f(x) > t} dt = ∫0a 1 dt = a = f(x). Thus,
the formula holds pointwise without invoking Fubini’s theorem.

Q4. Prove Theorem 1.14 (Bathtub Principle).


The bathtub principle asserts that among all measurable functions g with 0 ≤ g ≤ 1
satisfying ∫ g dµ = m, the function that minimizes ∫ fg dµ is the indicator of a level
set of f.

Proof: Arrange values of f decreasingly. Define g = 1 on the set where f is largest


until the measure constraint ∫ g dµ = m is satisfied, and 0 elsewhere. Then, ∫ fg dµ
is minimized due to the rearrangement inequality. Hence the optimal g is an
indicator of a sublevel set.

Q6. Prove that the measure constructed in Theorem 1.15 is complete.


Proof: Let N be a measurable set with µ(N) = 0. For any subset A ⊆ N, we show A
is measurable. Given ε > 0, there exists a measurable set B ⊇ N with µ(B) < ε.
Since A ⊆ B, outer measure µ*(A) ≤ µ(B) < ε. Thus µ*(A) = 0. As A ⊆ measurable
null set, Carathéodory’s criterion gives µ*(E) = µ*(E ∩ A) + µ*(E ∩ Ac) for all E.
Hence A is measurable with µ(A) = 0. Therefore, the measure is complete.

Q10. Using monotone class theorem, prove Lebesgue measure is inner and
outer regular.
Proof: Outer regularity: For measurable E, by definition µ(E) = inf{ µ(O): O ⊇ E, O
open }. This follows from Carathéodory’s construction.

Inner regularity: By the monotone class theorem, the class of sets for which µ(E) =
sup{ µ(K): K ⊆ E, K compact } is a monotone class containing open intervals,
hence all Borel sets. Extending via completion gives all Lebesgue sets. Thus
Lebesgue measure is both inner and outer regular.

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