For each $x$, there is one and only one $y$ for which $f_x(y) = f(x, y) = 1$; namely $y = x$. Otherwise, $f_x(y) = f(x, y) = 0$. We have:
$$\int_{[0,1]} f(x,y) \, d\nu(y) = \int_{\{x\}} 1 \, d\nu(y) = \nu(\{x\}) = 1$$
Therefore:
$$\int_{[0,1]} d\mu(x) \int_{[0,1]} f(x,y) \, d\nu(y) = \mu([0,1]) = 1$$
On the other hand:
$$\int_{[0,1]} f(x,y) \, d\mu(x) = \int_{\{y\}} 1 \, d\mu(x) = \mu(\{y\}) = 0$$
Therefore:
$$\int_{[0,1]} d\nu(y) \int_{[0,1]} f(x,y) \, d\mu(x) = 0$$
As for (3), how do you define the product measure when the measures aren't both $\sigma$-finite? Some references only define it when the measures are both $\sigma$-finite, and it's uniquely determined only when so.
Being measurable w.r.t. $m\times\nu$ doesn't make sense, and furthermore you don't even use the product measure in this exercise.
Instead, you should have specified which sigma-algebra you equip $[0,1]$ with - both when you're speaking of $m$ and when you're speaking of $\nu$. You could for example consider the measure-space $([0,1],\mathcal{E},m)$ and $([0,1],\mathcal{F},\nu)$, where $\mathcal{E}=\mathcal{B}([0,1])$ is the Borel sigma-algebra on $[0,1]$ and $\mathcal{F}$ could be $\mathcal{B}([0,1])$ or even the power set $\mathcal{P}([0,1])$. But let us assume that $\mathcal{F}=\mathcal{B}(\mathbb{R})$ since this is the smallest of the two.
Now you should show that $$D\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})=\mathcal{B}(\mathbb{R}^2),$$
i.e. that $D$ belongs to the product-sigma-algebra of $[0,1]\times [0,1]$. One strategy for that is to show that $D$ is closed in $\mathbb{R}^2$. This ensures that the sections $D_x=\{y\in\mathbb{R}\mid (x,y)\in D\}$ and $D_y=\{x\in\mathbb{R}\mid (x,y)\in D\}$ belongs to $\mathcal{B}(\mathbb{R})$.
For (2) you just evaluate the inner integrals first: For a fixed $y\in [0,1]$ we have that $\chi_D(x,y)=\chi_{D_y}(x)=1$ if and only if $x=y$ and zero otherwise. Therefore,
$$
\int_{[0,1]}\chi_D(x,y)\,m(\mathrm dx)=\int_{[0,1]}\chi_{D_{y}}(x)\,m(\mathrm dx)=m(\{y\})=0,
$$
for all $y\in [0,1]$.
For the right-hand side we have that for a fixed $x\in [0,1]$:
$$
\int_{[0,1]}\chi_D(x,y)\,\nu(\mathrm dy)=\int_{[0,1]}\chi_{D_{x}}(y)\,\nu(\mathrm dy)=\nu(\{x\})=1.
$$
Is this a contradiction to Tonelli/Fubini's theorem? (This is probably the key point of the exercise).
Best Answer
Since $\Bbb N$ is countable and the only set for which $c\otimes c$ measure is $0$ is the emptyset, the $c\otimes c$ measurable sets of $\Bbb N\times \Bbb N$ are all the subsets, hence $f$ is measurable.
The measure $c$ is $\sigma$-finite, hence it show we cannot remove the assumption of non-negativeness or integrability.