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ntsivanidis
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Hey guys, below is a small question from introductory measure theory. Maybe be completely wrong on this, so if you could point me in the right direction I'd really appreciate it.
Claim: Let B=\mathbb{Q} \cap [0,1] and \{I_k\}_{k=1}^n be a finite open cover for B. Then \sum_{k=1}^n m^*(I_k) \geq 1
Proof: Let \ B = \{q_k\}_{k=1}^\infty.Since {I_k\}_{k=1}^n is a finite cover, there must be at least one j \in \{1,\dots,n\} s.t. I_j contains infinitely many elements of B.Fix \varepsilon > 0. WLOG, WMA I_k=(q_k - \frac{\varepsilon}{2(n-1)}, q_k + \frac{\varepsilon}{2(n-1)}) \ni q_k \ \forall \ k \neq j. Then \sum_{k=1}^n m^*(I_k) = \sum_{k\neq j} m^*(I_k) + m^*(I_j)=\varepsilon + m^*(I_j) \geq m^*(I_j)=1 since m^*([0,1]\backslash \mathbb{Q})=1 \ \Box
Claim: Let B=\mathbb{Q} \cap [0,1] and \{I_k\}_{k=1}^n be a finite open cover for B. Then \sum_{k=1}^n m^*(I_k) \geq 1
Proof: Let \ B = \{q_k\}_{k=1}^\infty.Since {I_k\}_{k=1}^n is a finite cover, there must be at least one j \in \{1,\dots,n\} s.t. I_j contains infinitely many elements of B.Fix \varepsilon > 0. WLOG, WMA I_k=(q_k - \frac{\varepsilon}{2(n-1)}, q_k + \frac{\varepsilon}{2(n-1)}) \ni q_k \ \forall \ k \neq j. Then \sum_{k=1}^n m^*(I_k) = \sum_{k\neq j} m^*(I_k) + m^*(I_j)=\varepsilon + m^*(I_j) \geq m^*(I_j)=1 since m^*([0,1]\backslash \mathbb{Q})=1 \ \Box
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