TITLE: The measure of the diagonal of a unit square in an alternative measure.
QUESTION [8 upvotes]: Usually, we say that the measure of the diagonal of a unit square is 0, but that's with the preassumption the measure is Lebesgue measure in $\mathbb{R}^2$. But what if we are talking about a strange measure where its the product of a 1-dimensional Lebesgue measure and a counting measure? To be precise:

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 $X\times Y$, then what is $(\mu\times\nu)(D)$?

According to the definition, it should be the infimum of the measure sum of all the possible "rectangle" cover of $D$. But in this measure $\mu\times\nu$, the only rectangles that have finite measures are those with either finite "height", or with Lebesgue-null "width", but clearly you cannot cover the diagonal with those thin strips (or rather 1-d line segments in the former case). Does that mean this diagonal has a measure $\infty$?

REPLY [9 votes]: This is exercise 46 in Real Analysis by Folland. Yes, $(\mu\times \nu)(D)=\infty$. The reason is that any finite (or countable) cover of the diagonal by rectangles will contain a rectangle of nonzero height and width, and such a rectangle has infinite measure. Following the  construction of the product measure, we see that it's infinite on $D$.
The exercise has two other parts, concerning iterated integrals:
 $$\int \int \chi_D \,d\mu(x)\,d\nu(y) =\int  0 \,d\nu(y) = 0 $$
 $$\int \int \chi_D \,d\nu(y)\,d\mu(x) =\int  1 \,d\mu(x) = 1 $$
contrasted to 
 $$\iint   \chi_D \,d(\mu\times \nu) =(\mu\times \nu)(D)=\infty $$
The Fubini-Tonelli theorem does not apply, since $\nu$ is not $\sigma$-finite.