SIGMA-ALGEBRAS AND MEASURES - FOUNDATIONS OF
MEASURE THEORY
1 NUMERICAL PROBLEMS ON SIGMA-ALGEBRAS AND MEASURES
1. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
a. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
b. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
2. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
3. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
4. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
a. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
b. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
c. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
5. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
a. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
b. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
c. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
6. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
7. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
8. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
9. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
10. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
11. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
12. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
13. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
14. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
15. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
16. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
17. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
18. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
19. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
20. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
21. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
22. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
23. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
24. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
25. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
26. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
27. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
28. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
29. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
30. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
31. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
32. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
33. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
34. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
35. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
36. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
37. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
38. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
39. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
40. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
41. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
42. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
43. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
44. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
45. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
46. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
47. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
48. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
49. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
50. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
51. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
52. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
53. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
54. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
55. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
56. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
57. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
58. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
59. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
60. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
61. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
62. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
63. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
64. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
65. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
66. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
67. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
68. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
69. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
70. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
71. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
72. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
73. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
74. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
75. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
76. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
77. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
78. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
79. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
80. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
81. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
82. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
83. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
84. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
85. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
86. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
87. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
88. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
89. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
90. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
91. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
92. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
93. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
94. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
95. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
96. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
97. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
98. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
99. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
100. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
101. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
102. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
103. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
104. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
105. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
106. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
107. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
108. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
109. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
110. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
111. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
112. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
113. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
114. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
115. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
116. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
117. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
118. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
119. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
120. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
121. Let 𝑋 = {1,2,3,4}. Consider the collection ℱ = {∅, {1,2}, {3,4}, 𝑋}.
a. Is ℱ a sigma-algebra on 𝑋? Justify your answer.
b. If not, find the smallest sigma-algebra containing ℱ.
Solution:
c. No, ℱ is not a sigma-algebra on 𝑋. While it contains 𝑋 and ∅, and is closed
under complements, it is not closed under countable unions. For example,
{1,2} ∪ {3,4} = 𝑋, but 𝑋 is already in ℱ.
d. The smallest sigma-algebra containing ℱ is the power set of 𝑋: 𝒫(𝑋)=
{∅, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, 𝑋}
122. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {𝑎, 𝑏, 𝑐}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1 Calculate 𝜇(𝐴) for all 𝐴 ∈ ℱ.
Solution: Using the additivity of measures:
– 𝜇(∅)= 0
– 𝜇({𝑎})= 2, 𝜇({𝑏})= 3, 𝜇({𝑐})= 1
– 𝜇({𝑎, 𝑏})= 𝜇({𝑎})+ 𝜇({𝑏})= 2 + 3 = 5
– 𝜇({𝑎, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑐})= 2 + 1 = 3
– 𝜇({𝑏, 𝑐})= 𝜇({𝑏})+ 𝜇({𝑐})= 3 + 1 = 4
– 𝜇(𝑋)= 𝜇({𝑎, 𝑏, 𝑐})= 𝜇({𝑎})+ 𝜇({𝑏})+ 𝜇({𝑐})= 2 + 3 + 1 = 6
123. Let 𝑋 = {1,2,3,4,5} and 𝒜 = {{1,2}, {3,4,5}}. Find the sigma-algebra generated by 𝒜.
Solution: The sigma-algebra generated by 𝒜, denoted 𝜎(𝒜), is:
𝜎(𝒜)= {∅, {1,2}, {3,4,5}, 𝑋}
This is the smallest collection containing 𝒜 that satisfies the properties of a sigma-
algebra.
124. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℕ, ℱ = 𝒫(ℕ), and 𝜇(𝐴)=∑1
2𝑛
𝑛∈𝐴 for all
𝐴 ∈ ℱ. Calculate:
a. 𝜇({1,2,3})
b. 𝜇(even numbers)
c. 𝜇(𝑋)
Solution:
d. 𝜇({1,2,3})=1
21+1
22+1
23=1
2+1
4+1
8=7
8
e. 𝜇(even numbers)=∑1
22𝑛
∞
𝑛=1 =1
4+1
16 +1
64 + ⋯ = 1/4
1−1/4 =1
3
f. 𝜇(𝑋)=∑1
2𝑛
∞
𝑛=1 = 1
125. Let 𝑋 = [0,1] and ℬ be the Borel sigma-algebra on 𝑋. Define 𝜇 on ℬ by 𝜇(𝐴)= 𝑚(𝐴)+
𝜒𝐴(1/2), where 𝑚 is the Lebesgue measure and 𝜒𝐴 is the indicator function of 𝐴.
Compute:
a. 𝜇([0,1/4])
b. 𝜇([1/4,3/4])
c. 𝜇([0,1])
Solution:
d. 𝜇([0,1/4])= 𝑚([0,1/4])+ 𝜒[0,1/4](1/2)= 1/4 + 0 = 1/4
e. 𝜇([1/4,3/4])= 𝑚([1/4,3/4])+ 𝜒[1/4,3/4](1/2)= 1/2 + 1 = 3/2
f. 𝜇([0,1])= 𝑚([0,1])+ 𝜒[0,1](1/2)= 1 + 1 = 2
126. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = {1,2,3,4,5}, ℱ = 𝒫(𝑋), and 𝜇 is defined by:
𝜇(𝐴)=|𝐴|2 for all 𝐴 ∈ ℱ, where |𝐴| denotes the number of elements in 𝐴. Is 𝜇 a
measure? If not, explain why.
Solution: No, 𝜇 is not a measure. While it satisfies 𝜇(∅)= 0, it fails the additivity
property. For example:
𝜇({1} ∪ {2})= 𝜇({1,2})= 22= 4
But
𝜇({1})+ 𝜇({2})= 12+ 12= 2
Since 4 ≠ 2, 𝜇 is not additive and therefore not a measure.
127. Let 𝑋 = {1,2,3,4} and ℱ = {∅, {1,2}, {3,4}, 𝑋}. Define 𝜇 on ℱ by: 𝜇(∅)= 0, 𝜇({1,2})= 3,
𝜇({3,4})= 2, 𝜇(𝑋)= 5 Is (𝑋, ℱ, 𝜇) a measure space? Justify your answer.
Solution: Yes, (𝑋, ℱ, 𝜇) is a measure space.
– ℱ is a sigma-algebra on 𝑋: it contains 𝑋 and ∅, is closed under complements,
and is closed under countable unions (since it’s finite).
– 𝜇 is a measure: - 𝜇(∅)= 0 - 𝜇 is non-negative - 𝜇 is countably additive (which
reduces to finite additivity in this case): 𝜇({1,2})+ 𝜇({3,4})= 3 + 2 = 5 = 𝜇(𝑋)
128. Let (𝑋, ℱ, 𝜇) be a measure space where 𝑋 = ℝ, ℱ is the Borel sigma-algebra, and 𝜇 is
the Lebesgue measure. Define a new set function 𝜈 on ℱ by: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1]) for all
𝐴 ∈ ℱ. Prove that 𝜈 is a measure and compute 𝜈(ℝ).
Solution: To prove 𝜈 is a measure:
– 𝜈(∅)= 𝜇(∅ ∩ [0,1])= 𝜇(∅)= 0
– Non-negativity: 𝜈(𝐴)= 𝜇(𝐴 ∩ [0,1])≥ 0 for all 𝐴 ∈ ℱ
– Countable additivity: For disjoint 𝐴1, 𝐴2, … ∈ ℱ, 𝜈(⋃𝐴𝑖
∞
𝑖=1 )= 𝜇((⋃𝐴𝑖
∞
𝑖=1 )∩[0,1])
= 𝜇(⋃(𝐴𝑖∩[0,1])
∞
𝑖=1 )=∑𝜇
∞
𝑖=1 (𝐴𝑖∩[0,1])=∑𝜈
∞
𝑖=1 (𝐴𝑖)
Therefore, 𝜈 is a measure.
To compute 𝜈(ℝ): 𝜈(ℝ)= 𝜇(ℝ ∩ [0,1])= 𝜇([0,1])= 1
