Let A = { ( x , y ) : x ∈ Q , y ∈ R }. Show that m ( A ) = 0.We notice that A = Q × R . Now since Q is countable we can denote it as { x 1 , x 2 , … }. Now consider the intervals { ( x n − ε 2 k , x n + ε 2 k ) × ( k − 1 , k + 1 ) } .I can show that the sum of the lengths of these intervals is zero as ℓ ( I k ) = 4 ε 2 k so ∑ k = 1 ∞ ℓ ( I k ) = 4 εbut I don't know how can I show that the intervals cover Q × R ??

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Let   A  =  {  (  x  ,  y  )  :  x  ∈      Q    ,  y  ∈      R    }. Show that   m  (  A  )  =  0.We notice that   A  =      Q    ×      R  . Now since       Q   is countable we can denote it as   {      x    1    ,      x    2    ,  …  }. Now consider the intervals   {  (      x    n    −      ε          2      k        ,      x    n    +      ε          2      k        )  ×  (  k  −  1  ,  k  +  1  )  }  .I can show that the sum of the lengths of these intervals is zero as  ℓ  (      I    k    )  =            4      ε              2      k      so      ∑          k      =      1        ∞    ℓ  (      I    k    )  =  4  εbut I don&#039;t know how can I show that the intervals cover       Q    ×      R    ??

Xavier Vargas · Accepted Answer

You have to conisder   (      x    n    −      ϵ          2              k        +        n              ,      x    n    +      ϵ          2              k        +        n              )  ×  (  k  −  1  ,  k  +  1  ). Both   n and   k are varying.If   q  ∈      Q   and   y  ∈      R   then there exist k such that   y  ∈  (  k  −  1  ,  k  +  1  ) and   q  =      x    n   for some   n. Hence,   (  q  ,  y  )  ∈  (      x    n    −      ϵ          2              k        +        n              ,      x    n    +      ϵ          2              k        +        n              )  ×  (  k  −  1  ,  k  +  1  ). Now check that the total measure of all these intervals is less than   4  ϵ.

kincirrboh7 · Accepted Answer

Your set is a countable union of copies of the real line (indexed by       Q  ). The two-dimensional measure of a line in             R        2   is zero so your total measure is zero by countable additivity

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