Let λ and μ be the lebesgue measure and counting measure, respectfully on ( R , B ( R ) ). Show that it holds that for A = { ( x , y ) ∈ R 2 | y = x } ∫ ∫ 1 A ( x , y ) d λ ( x ) d μ ( y ) = 0 ∫ ∫ 1 A ( x , y ) d μ ( y ) d λ ( x ) = ∞to further conclude that μ can't be Σ-finiteI know that if we have two Σ-finite measure spaces then their product measure is unique. Fubini then tells us that if the function f is integrable over the product measure then we can interchange the the order of integration. This is not the case.I realize that the point is in that the counting measure: The counting measure does exactly what it says: it counts the number of elements in a set. So any infinite set has counting measure ∞ , while the measure of any finite set is its cardinality. A would have measure ∞ wrt. the counting measure. The counting measure is just summation. The Lebesgue integral of the indicator would just be the Lebesgue measure of A.How do I actually show that the double integrals hold?

Question

Let   λ and   μ be the lebesgue measure and counting measure, respectfully on   (      R    ,  B  (      R    )  ). Show that it holds that for   A  =  {  (  x  ,  y  )  ∈            R        2        |    y  =  x  }  ∫  ∫      1    A    (  x  ,  y  )  d  λ  (  x  )  d  μ  (  y  )  =  0  ∫  ∫      1    A    (  x  ,  y  )  d  μ  (  y  )  d  λ  (  x  )  =  ∞to further conclude that   μ can&#039;t be   Σ-finiteI know that if we have two   Σ-finite measure spaces then their product measure is unique. Fubini then tells us that if the function f is integrable over the product measure then we can interchange the the order of integration. This is not the case.I realize that the point is in that the counting measure: The counting measure does exactly what it says: it counts the number of elements in a set. So any infinite set has counting measure       ∞  , while the measure of any finite set is its cardinality. A would have measure       ∞   wrt. the counting measure. The counting measure is just summation. The Lebesgue integral of the indicator would just be the Lebesgue measure of A.How do I actually show that the double integrals hold?

Hayley Mccarthy · Accepted Answer

Well, for each   y  ∈      R  ,  ∫      1    A    (  x  ,  y  )    d  λ  (  x  )  =  ∫      1          {      y      }        (  x  )    d  λ  (  x  )  =  0.On the other hand, for each   x  ∈      R  ,  ∫      1    A    (  x  ,  y  )    d  μ  (  y  )  =  ∫      1          {      x      }        (  y  )    d  μ  (  y  )  =  1.