I'm trying to evaluate this integral:
$$\int_A\int_B1_{x\neq y} \, d\mu(x) \, d\mu(y)$$
where $\mu$ is the Lebesgue measure.
I think the answer should be $\mu(A)\mu(B)$ but I'm not sure how to prove or disprove it. What is the value of this integral?
Assuming that $A$ and $B$ have finite measure, the set $E=\{(x,y)\in\mathbb R^2:x\ne y\}$ has measure $\mu(A)\mu(B)$, as $E^c$ is a subset of the image of $\mathbb R$ under the Lipschitz map $x\mapsto (x,x)$ and therefore has measure zero. It follows then that $$\iint_{A\times B}\chi_E(x,y)\ \mathsf d\mu_2(x\times y) = \mu_2(A\times B)=\mu(A)\mu(B). $$ Since $\chi_E$ is nonnegative and $\mu_2$-measurable (where $\mu_2$ denotes $2$-dimensional Lebesgue measure), Tonelli's theorem yields $$\iint_{A\times B}\chi_E(x,y)\ \mathsf d\mu_2(x\times y)=\int_A\int_B\chi_E(x,y)\ \mathsf dy\ \mathsf dx=\int_B\int_A\chi_E(x,y)\ \mathsf dx\ \mathsf dy.$$