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**Unformatted text preview: **Chapter 3: Lebesgue Measure Written by Men-Gen Tsai email: b89902089@ntu.edu.tw 1. Let m be a countably additive measure defined for all sets in a σ-algebra M . If A and B are two sets in M with A ⊂ B , then mA ≤ mB . This property is called monotonicity . Proof: B = A S ( A- B ). A and A- B are disjoint. Since m is a countably additive measure, mB = mA + m ( A- B ). Note that m is nonnegative, and m ( A- B ) ≥ 0. Hence mA ≤ mB . 2. Let < E n > be any sequence of sets in M . Then m ( S E n ) ≤ ∑ mE n . [Hint: Use Proposition 1.2.] This property of a measure is called count- able subadditivity . Proof: By Proposition 1.2 on page 17, since M is a σ-algebra, there is a sequence < B n > of sets in M such that B n T B m = φ for n 6 = m and ∞ [ i =1 B i = ∞ [ i =1 E i . Since m is a countably additive measure and B i ⊂ E i for all i , by Problem 3.1 I have that m ( [ E n ) = m ( [ B n ) = X mB n ≤ X mE n . 3. If there is a set A in M such that mA < ∞ , then mφ = 0. Proof: Note that A = A S φ and A and φ are disjoint, and thus mA = mA + mφ. Since mA < ∞ , mφ = 0 precisely. 1 4. 5. Let A be the set of rational numbers between 0 and 1, and let { I n } be a finite collection of open intervals covering A . Then ∑ l ( I n ) ≥ 1. Proof 1: (due to Meng-Gen Tsai) Since 0 ∈ A , there is an open interval J 1 in { I n } such that 0 ∈ J 1 . Let J 1 = ( a 1 ,b 1 ). Note that a 1 < 0 and b 1 > 0. If b 1 ≥ 1, then X l ( I n ) ≥ l ( J 1 ) = b 1- a 1 ≥ 1 . Suppose not. If a 1 is rational, then I can find an open interval J 2 ∈ { I n } such that a 1 ∈ J 2 . If a 1 is irrational, I consider the following cases....

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