Measure theory problem from Folland Let $X=Y=[0,1]$, $\mathcal M=\mathcal N=\mathcal B_{[0,1]}$, $\mu=$ Lebesgue measure, and $\nu=$counting measure. If $D={(x,x):x\in [0,1]}$ is the diagonal in $X$x$Y$, then $\int \int \mathcal X_D d\mu d\nu $, $\int \int \mathcal X_D d\nu d\mu $ and $ \int \mathcal X_D d(\mu \times \nu) $ are all unequal.
 A: Ok, well $\int \limits_{[0,1]} \int \limits_{[0,1]} \chi_{D} \, d\nu \,d\mu$ is integration with respect to $nu$, counting measure, first.  For fixed $x$, we are counting how many $y$ satisfy that $(x,y) \in D = \{(x,x) : x \in [0,1] \}$.  But if $x$ is fixed, and ordered pairs in $D$ have both the $x$ and $y$ coordinates equal, then there is only one $y$.  So $\nu(D) = 1$.  Then we are integrating $\int \limits_{[0,1]} 1 \, d\mu = \mu([0,1]) = 1$ since the Lebesgue measure of the unit interval is $1$.
Now, for the other order: $\int \limits_{[0,1]} \int \limits_{[0,1]} \chi_{D} \, d\mu \,d\nu$
First, the inner integral: for fixed $y$, we want to know the Lebesgue measure of the set of $x$ such that $(x,y) \in \{(y,y) : y \in [0,1] \}$.  But $y$ is fixed, and since $x$ must equal $y$, then there is only one $x$, and the Lebesgue measure of a set with a single element is $0$.  Thus, the inner integral evaluates to $0$.  But that means the outer integral is an integral of $0$, so it equals $0$.
This shows that $\int \limits_{[0,1]} \int \limits_{[0,1]} \chi_{D} \, d\nu \,d\mu \neq \int \limits_{[0,1]} \int \limits_{[0,1]} \chi_{D} \, d\mu \,d\nu$.  
Then, we know by Fubini's theorem that if our original function $\chi_{D}$ in $X \times Y$ were in $L^{1}(d(\mu \times \nu))$ (i.e., if it had finite integral), then the iterated integrals must be equal.  But we found that the iterated integrals were not equal, so that means the original function is not in $L^{1}$, which means its integral is infinite.  That shows that none of the three integrals are equal.
Note that the end of my argument used the fact that the statement $p \to q$ is equivalent to $\neg q \to \neg p$.
