How do you show Tonelli’s theorem can fail in the σ-nonfinite case, and also that product measure does not need to be unique. For this, I showed that
If we have the set $I=[0,1]$ equipped on one side with the Borel set and Lebesgue measure λ for X and on the other side the same set $I=[0,1]$ equipped with the power set and the counting measure m for Y.
The diagonal set Δ={$(x,x) : x∈[0,1]$} is a closed subset of $I_2$. Hence
Δ$\in(I_2)=\mathcal{B}(I)×\mathcal{B}(I)⊂\mathcal{B}(I)×\mathcal{B}(I)$.
Therefore, the characteristic map $1_Δ$ is measurable for the measure induced on $X×Y$. However, we have on one side
$\int_{I}(\int_{I}1_Δ(x,y)dm(y))dλ(x)=\int_{I}(\int_{I}1_{x}(y)dm(y))dλ(x)=\int_{I}m({x})dλ(x)=λ(I)=1$
and on the other side
$\int_{I}(\int_{I}1_Δ(x,y)dλ(x))dm(y)=\int_{I}(
\int_{I}1_{y}(x)dλ(x))dm(y)=\int_{I}λ({y})dm(y)=\int_{I}0 dm(y)=0$
These are not equal and Tonelli's Theorem failed. Is this right?
Also, in the case for uniqueness. I need to show that there is more than one measure $\mu$ on $\mathcal{L}[0,1]\times 2^{[0,1]}$ with the property that $\mu(E\times F)=\mu(E)$#$(F)$ (where # is the counting measure of F) for all $E\in \mathcal{L}([0,1])$ and $F\in 2^{[0,1]}$. How should I start with this?
 A: $1$. Yes, your proof that Tonelli's theorem conclusion fails in this case is correct. The only minor correction is:

The diagonal set $\Delta=\{(x,x) : x∈[0,1]\}$ is a closed subset of $I_2$. Hence
$\Delta\in\mathcal{B}(I_2)=\mathcal{B}(I)×\mathcal{B}(I)\subset \mathcal{B}(I)×2^I$.

$2$. Now, let us show that there is more than one measure $\mu$ on $\mathcal{B}([0,1])\times 2^{[0,1]}$ with the property that $\mu(E\times F)=\lambda(E)\#(F)$ (where $\lambda$ is the Lebesgue measure and  $\#$ is the counting measure) for all $E\in \mathcal{B}([0,1])$ and $F\in 2^{[0,1]}$.
Let $\mu_1$ be the the product measure $\lambda \times \#$. It is immediate that for all $E\in \mathcal{B}([0,1])$ and $F\in 2^{[0,1]}$,
$$\mu_1(E\times F) = (\lambda \times \#) (E\times F)= \lambda(E)\#(F)$$
Let $\mu_2$ be a measure defined on $\mathcal{B}([0,1])\times 2^{[0,1]}$ by, for all $A \in \mathcal{B}([0,1])\times 2^{[0,1]}$,
$$\mu_2(A) = \sum_{y\in[0,1]} \lambda(([0,1]\times\{y\}) \cap A)$$
It is easy to see that $\mu_2$ is in fact a measure. Moreover, for all $E\in \mathcal{B}([0,1])$ and $F\in 2^{[0,1]}$,
$$\mu_2(E\times F) = \sum_{y\in[0,1]} \lambda(([0,1]\times\{y\}) \cap (E\times F))=\sum_{y\in F}\lambda(E)=\lambda(E)\#(F)$$
Lets us prove that $\mu_1\neq \mu2$.
Consider the diagonal set $\Delta=\{(x,x) : x∈[0,1]\}$ again.
Note that, if $E\times F$ is a rectangle such that $\mu_1(E\times F)$ is finite, then $\lambda(E)=0$ or $F$ is a finite set.  So $\Delta$ can not be covered by any countable collection of rectangles $E\times F$ with $\mu_1(E\times F)$ finite.
So $\mu_1(\Delta)=+\infty$ (in fact, we proved more: $\Delta$ is not even $\mu_1$-$\sigma$-finite).
On the other hand,
$$\mu_2(\Delta) = \sum_{y\in[0,1]} \lambda(([0,1]\times\{y\}) \cap \Delta)= \sum_{y\in[0,1]} \lambda(\{y\})=0$$
A: The measurability of $\Delta$ is actually due to the fact that one can write
$$\Delta = \bigcap_{n=1}^\infty \left(\bigcup_{k=1}^{2^n}\left[\frac{k-1}{2^n},\frac{k}{2^n}\right]^2\right)$$
and each of the $\left[\frac{k-1}{2^n},\frac{k}{2^n}\right]^2\in \mathcal{B}(I)\times 2^I$ (note that this also works in the case that you have $\mathcal{B}(I)$ as your $\sigma$-algebra for $Y$). The reason the closedness argument you gave is a bit tricky is that it requires one to show that the product $\sigma$-algebra on $I^2$ is equivalent to the Borel $\sigma$-algebra on $I^2$ that's generated by open sets.
Your iterated integrals are correct. Further, notice that (I'll adopt your notation and set $\mu = \lambda\times m$, where $\lambda$ is the Lebesgue measure and $m$ is the counting measure)
$$\int_{X\times Y}1_{\Delta}\ \mathrm{d}\mu = \mu(\Delta) = \infty$$
with the last equality coming from the fact that any covering of the diagonal will include infinitely many elements in $Y$.
For uniqueness of the measure, you should take a look here. I'll briefly go over the idea behind it. Let $\varpi$ be a measure on $\mathcal{B}(I)\otimes2^I$ such that
$$\varpi(E) = \sup\{\mu(E\cap (M\times N)):M\in \mathcal{B}(I),N\in 2^I, \lambda(M),m(N)<\infty\}.$$
The link I provided shows that such definition indeed gives us a measure on $\mathcal{B}(I)\otimes2^I$, the product $\sigma$-algebra.
We first show that $\varpi=\mu$ on $\mathcal{B}(I)\times2^I$. Let $A\in\mathcal{B}(I)$ and $B\in 2^I$. If $\#B<\infty$, then $\varpi(A\times B) = \mu(A\times B)$ is straightforward. On the other hand, if $\#B=\infty$, we know that $\mu(A\times B)=\infty$. Further, observe that, by our above definition, we have
$$\varpi(A\times B)\geq \lambda(A)(\sup\{m(N):N\subseteq B,m(N)<\infty\}) = \infty$$
since $m(B) = \#B = \infty$.
We will now show that $\varpi(\Delta)=0$ and thus $\varpi\ne\mu$ on $\mathcal{B}(I)\otimes2^I$. Consider some $M\in\mathcal{B}(I)$ and $N\in 2^I$ such that $m(N)=\#N<\infty$. To show that $\varpi(\Delta)=0$, note that it's sufficient to consider $M=[0,1]$ since $\lambda([0,1]) = 1<\infty$. Further, let $N=\{n_1,\ldots,n_k\}$ so we have
$$\mu(\Delta\cap ([0,1]\times\{n_1,\ldots,n_k\})) \leq \sum_{i=1}^k\mu(\Delta\cap ([0,1]\times\{n_i\})) = \sum_{i=1}^k\mu(\{n_i\}^2) = 0. $$
Since $N$ was chosen arbitrarily, we then have $\varpi(\Delta)=0$ as it's a supremum of a set of zeros.
