Let λ and μ be the lebesgue measure and counting measure, respectfully on ( R...

EnvivyEvoxys6

EnvivyEvoxys6

Answered

2022-07-10

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-finite
I 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?

Answer & Explanation

Hayley Mccarthy

Hayley Mccarthy

Expert

2022-07-11Added 19 answers

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.

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