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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 [tex]B=\mathbb{Q} \cap [0,1][/tex] and [tex]\{I_k\}_{k=1}^n[/tex] be a finite open cover for [tex]B[/tex]. Then [tex]\sum_{k=1}^n m^*(I_k) \geq 1[/tex]

Proof: Let [tex]\ B = \{q_k\}_{k=1}^\infty[/tex].Since [tex]{I_k\}_{k=1}^n[/tex] is a finite cover, there must be at least one [tex] j \in \{1,\dots,n\}[/tex] s.t. [tex]I_j[/tex] contains infinitely many elements of [tex]B[/tex].Fix [tex]\varepsilon > 0[/tex]. WLOG, WMA [tex]I_k=(q_k - \frac{\varepsilon}{2(n-1)}, q_k + \frac{\varepsilon}{2(n-1)}) \ni q_k \ \forall \ k \neq j[/tex]. Then [tex]\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[/tex] since [tex]m^*([0,1]\backslash \mathbb{Q})=1 \ \Box[/tex]

Claim: Let [tex]B=\mathbb{Q} \cap [0,1][/tex] and [tex]\{I_k\}_{k=1}^n[/tex] be a finite open cover for [tex]B[/tex]. Then [tex]\sum_{k=1}^n m^*(I_k) \geq 1[/tex]

Proof: Let [tex]\ B = \{q_k\}_{k=1}^\infty[/tex].Since [tex]{I_k\}_{k=1}^n[/tex] is a finite cover, there must be at least one [tex] j \in \{1,\dots,n\}[/tex] s.t. [tex]I_j[/tex] contains infinitely many elements of [tex]B[/tex].Fix [tex]\varepsilon > 0[/tex]. WLOG, WMA [tex]I_k=(q_k - \frac{\varepsilon}{2(n-1)}, q_k + \frac{\varepsilon}{2(n-1)}) \ni q_k \ \forall \ k \neq j[/tex]. Then [tex]\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[/tex] since [tex]m^*([0,1]\backslash \mathbb{Q})=1 \ \Box[/tex]

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