Let $\lambda$ and $\mu$ be the lebesgue measure and counting measure, respectfully on $(\mathbb{R}, B(\mathbb{R}))$. Show that it holds that for $A=\{(x,y)\in \mathbb{R}^2 | y=x\}$
$$\int \int 1_A(x,y)d\lambda (x) d\mu(y) =0$$$$\int \int 1_A(x,y)d\mu(y) d\lambda (x) =\infty$$
to further conclude that $\mu$ can't be $\Sigma$-finite
I know that if we have two $\Sigma$-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 $∞$, 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?
Best Answer
Well, for each $y\in\mathbb{R}$, $$ \int 1_A(x,y)\,d\lambda(x)=\int 1_{\{y\}}(x)\,d\lambda(x)=0. $$ On the other hand, for each $x\in\mathbb{R}$, $$ \int 1_A(x,y)\,d\mu(y)=\int 1_{\{x\}}(y)\,d\mu(y)=1. $$