EnvivyEvoxys6

Answered

2022-07-10

Let $\lambda $ and $\mu $ be the lebesgue measure and counting measure, respectfully on $(\mathbb{R},B(\mathbb{R}))$. Show that it holds that for $A=\{(x,y)\in {\mathbb{R}}^{2}|y=x\}$

$\int \int {1}_{A}(x,y)d\lambda (x)d\mu (y)=0$

$\int \int {1}_{A}(x,y)d\mu (y)d\lambda (x)=\mathrm{\infty}$

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?

$\int \int {1}_{A}(x,y)d\lambda (x)d\mu (y)=0$

$\int \int {1}_{A}(x,y)d\mu (y)d\lambda (x)=\mathrm{\infty}$

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?

Answer & Explanation

Hayley Mccarthy

Expert

2022-07-11Added 19 answers

Well, for each $y\in \mathbb{R}$,

$\int {1}_{A}(x,y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)=\int {1}_{\{y\}}(x)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)=0.$

On the other hand, for each $x\in \mathbb{R}$,

$\int {1}_{A}(x,y)\phantom{\rule{thinmathspace}{0ex}}d\mu (y)=\int {1}_{\{x\}}(y)\phantom{\rule{thinmathspace}{0ex}}d\mu (y)=1.$

$\int {1}_{A}(x,y)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)=\int {1}_{\{y\}}(x)\phantom{\rule{thinmathspace}{0ex}}d\lambda (x)=0.$

On the other hand, for each $x\in \mathbb{R}$,

$\int {1}_{A}(x,y)\phantom{\rule{thinmathspace}{0ex}}d\mu (y)=\int {1}_{\{x\}}(y)\phantom{\rule{thinmathspace}{0ex}}d\mu (y)=1.$

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