EnvivyEvoxys6

2022-07-10

Let $\lambda$ and $\mu$ be the lebesgue measure and counting measure, respectfully on $\left(\mathbb{R},B\left(\mathbb{R}\right)\right)$. Show that it holds that for $A=\left\{\left(x,y\right)\in {\mathbb{R}}^{2}|y=x\right\}$
$\int \int {1}_{A}\left(x,y\right)d\lambda \left(x\right)d\mu \left(y\right)=0$
$\int \int {1}_{A}\left(x,y\right)d\mu \left(y\right)d\lambda \left(x\right)=\mathrm{\infty }$
to further conclude that $\mu$ can't be $\mathrm{\Sigma }$-finite
I know that if we have two $\mathrm{\Sigma }$-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 $\infty$, while the measure of any finite set is its cardinality. A would have measure $\infty$ 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

Expert

Well, for each $y\in \mathbb{R}$,
$\int {1}_{A}\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(x\right)=\int {1}_{\left\{y\right\}}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}d\lambda \left(x\right)=0.$
On the other hand, for each $x\in \mathbb{R}$,
$\int {1}_{A}\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}d\mu \left(y\right)=\int {1}_{\left\{x\right\}}\left(y\right)\phantom{\rule{thinmathspace}{0ex}}d\mu \left(y\right)=1.$

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