Section of a set with measure 0 in the product space is measurable almost everywhere for a complete space Let $(X,\mathcal S, \mu)$ and $(Y,\mathcal T, \nu)$ be complete positive measure spaces, and let $(X\times Y,\mathcal V, w)$ be the product space. And let $N$ be a set such that $w(N)=0$.
How can I prove that:
$$N_x := \{y\in Y \mid (x, y)\in N\}$$
is measurable for almost all $x\in X$ in the measure space $(Y,\mathcal T, \nu)$?
 A: It is clear that $\mathbb{1}_N$ is lebesgue measurable in the product space $(X\times Y,\mathcal V, w)$.
Additionally, $$\int\limits_{X \times Y} \, \mathbb{1}_N (x,y) \, \mathrm{d} w = w(N)= 0$$ so $\mathbb{1}_N$ is integrable $\mathrm{d} w$.
Now notice that both $(X,\mathcal S, \mu)$ and $(Y,\mathcal T, \nu)$ are complete positive measure spaces, so we may apply fubini's theorem:
$$\int\limits_{X} \, \left(\int\limits_{Y} \, \mathbb{1}_{N_x }(x,y) \, \nu(\mathrm{d} y)\right)\, \mu(\mathrm{d} x) = \int\limits_{X} \, \nu \left(N_x (x,y) \right)\, \mu(\mathrm{d} x )= 0$$
But $\nu \left(N_x (x,y) \right) \geq 0 \; a.e$ and $\int\limits_X \nu \left(N_x (x,y) \right)\, \mathrm{d} x = 0$ and thus it follows that $\nu \left(N_x (x,y) \right) = 0 \Rightarrow N_x$  is measurable for almost all $x\in X$ in the measure space $(Y,\mathcal T, \nu)$
Edit:
This may also be proven without the use of fubini’s theorem.
To do so define $\Sigma := \{E \in \mathcal S \times \mathcal T \mid \forall x \in X:E_x \in \mathcal T\}$. Notice that if $E=A\times B$ then $\forall x \in A: E_x=B$ and $\forall x \in A^c: E_x=\emptyset$.
Therefore every measurable rectangle in the product space belongs to $\Sigma$.
Now since $\mathcal T$ is a sigma-algebra it is simple to prove that $\Sigma$ is also a sigma-algebra in the product space $X \times Y \Rightarrow \Sigma = \mathcal S \times \mathcal T$.
Now we know that $w(N) = 0$ and so $N$ is measurable in the product space. But then according to according to what was proven above: $\forall x \in X: N_x \in \mathcal T$
q.e.d.
