So I am reading the construction of non-measurable set N . This is done so by considering an enumeration { r k } of Q ∩ [ − 1 , 1 ] I am stuck on seeing the obviousness in why (by monotonicity) The following is a contradiction. So we claim and show the following Inclusion [ 0 , 1 ] ⊂ ⋃ k N k ⊂ [ − 1 , 2 ]Where N k = N + r k I get that by monotonicity, we have 1 ≤ ∑ k ∈ N m ( N k ) ≤ 3But I do not see the contradiction as to why since neither m ( N ) = 0 nor m ( N ) &gt; 0 we are done by contradiction, can't the measure be strictly between 1 and 3? so 2?

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So I am reading the construction of non-measurable set       N  . This is done so by considering an enumeration   {      r    k    } of       Q    ∩  [  −  1  ,  1  ] I am stuck on seeing the obviousness in why (by monotonicity) The following is a contradiction. So we claim and show the following Inclusion  [  0  ,  1  ]  ⊂      ⋃          k                  N      k        ⊂  [  −  1  ,  2  ]Where             N        k    =      N    +      r    k  I get that by monotonicity, we have  1  ≤      ∑          k      ∈              N              m  (            N        k    )  ≤  3But I do not see the contradiction as to why since neither   m  (      N    )  =  0 nor   m  (      N    )  &amp;gt;  0 we are done by contradiction, can&#039;t the measure be strictly between 1 and 3? so 2?

Patricia Curry · Accepted Answer

Since   m  (            N        k    )  =  m  (      N    ) for each   k, both cases lead to a contradiction:1. If   m  (      N    )  =  0, then the infinite sum equals 0 as well, yielding   1  ≤  0  ≤  3, which is absurd.2. If   m  (      N    )  &amp;gt;  0, the infinite sum equals infinity, which gives   1  ≤  ∞  ≤  3, again a contradiction.

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