Product measure of this set I want to compute the measure of $\mu_1 \times \mu_2 (A)$, where $A = \{ (x, x): x\in [0, 1]\}$, $\mu_1$ is the Lebesgue measure and $\mu_2$ the counting measure on $[0, 1]$.
So far, I have been trying to write $A = A_1 \times A_2$ so I can compute $\mu_1 \times \mu_2 (A) = \mu_1(A_1)\mu_2(A_2)$, but I don't know how to write $A$ as a cartesian product.
 A: I know it probably already too late to answer this, but I'll do it just in case someone else needs it.
Let's start by remembering that product measures are defined as induced by an outer measure. Therefore, if our two measure spaces are $(\mathbb{X},\mathcal{M},\mu)$ and $(\mathbb{Y},\mathcal{N},\nu)$ and $A\in\mathcal{M}\otimes\mathcal{N}$ (the product $\sigma$-algebra) then
$$ (\mu\times\nu)(A) = \inf\left\{\sum(\mu\times\nu)(A_n) : A_n\in\mathcal{M}\times\mathcal{N},\,A\subset\bigcup A_n\right\}.$$
And $\mu\times\nu$ is the only measure in the product space such that for all $A\times B\in\mathcal{M}\times\mathcal{N}$ satisfies $(\mu\times\nu)(A\times B) = \mu(A)\nu(B)$, which gives us:
$$ (\mu\times\nu)(A) = \inf\left\{\sum \mu(B_n)\nu(C_n) : B_n\in\mathcal{M},C_n\in\mathcal{N},\,A\subset\bigcup (B_n\times C_n)\right\}.$$
With all this in mind we can start the exercise. We want to calulate $(\mu\times\nu)(A)$, where $A = \{(x,x) : x\in[0,1]\}$. Let's start by taking a cover of $A\in\mathcal{M}\otimes\mathcal{N}$ in $\mathcal{M}\times\mathcal{N}$.
Let $\{B_n\}\subset\mathcal{M}, \{C_n\}\subset\mathcal{N}$ be such that $A\subset\bigcup(B_n\times C_n)$, that is, for all $(x,x)\in A$ there exists $n\in\mathbb{N}$ such that $(x,x)\in (B_n\times C_n)$. So $x\in B_n$ and $x\in C_n$, then $x\in B_n\cap C_n$. So every time we find a cover of $A$ we can turn it into a cover of $[0,1]$ of the form $\bigcup (B_n\cap C_n)$.
Now remember that $\mu([0,1])=1$ ($\mu$ is Lebesgue), so there must exist $n_0$ such that $\mu(B_{n_0}\cap C_{n_0})>0$ then $\mu(B_{n_0})>\mu(B_{n_0}\cap C_{n_0})>0$ and $\mu(C_{n_0})>\mu(B_{n_0}\cap C_{n_0})>0$.
For any measurable set $D$ of $[0,1]$, if $\mu(D)>0$ then $\nu(D)=\infty$ ($\nu$ is the counting measure). Then it must be $\nu(B_{n_0})=\nu(C_{n_0})=\infty$.
As the cover $\bigcup B_n\times C_n$ was arbitrary then
$$(\mu\times\nu)(A)\leq(\mu\times\nu)\left(\bigcup B_n\times C_n\right)\leq\sum\mu(B_n)\nu(C_n) = \infty.$$
As $([0,1],\mathcal{B}_{[0,1]},\nu)$ is not $\sigma$-finite, so Fubini's theorem is not applicable, let's check this by calculating the iterated integrals.
$$A_x = \{x\},\quad\quad A^y = \{y\},$$
then
$$\nu(A_x) = 1,\quad\quad \mu(A^y)=0$$
Which result in
$$\int\int\chi_A\,d\mu d\nu = \int_{[0,1]}\mu(A^y)\,d\nu = 0,$$
$$\int\int\chi_A\,d\nu d\mu = \int_{[0,1]}\nu(A_x)\,d\mu = 1,$$
$$\int\int\chi_A\,d(\mu\times\nu) = \infty.$$
