$(\mu\otimes \nu) (Z)=0$ implies $\nu(Z_x)=0$ for almost all $x$. Am reading the proof Lemma 8.3 in Lang's real and functional analysis book. The set-up is two $\sigma$-finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ and we have a
set $Z\in \mathcal{M}\otimes\mathcal{N}$ with $(\mu\otimes \nu)(Z)=0$. We want to show that this implies  $\nu(Z_x)=0$ for almost all $x$, where $Z_x$ denotes the $x$ section of $Z$. The proof reads like this:
For each positive integer $n$, let $S_n$ be the set of all $x$ such that $\nu(Z_x)\geq1/n$. Let $S=\cup_{n=1}^{\infty}S_n$. It suffices to show that $S$ is contained in a set of measure $0$. Given $\varepsilon$, let $\{R_k\}$ be a sequence of rectangles whose union contains $Z$ and such that
$$\sum_{k=1}^{\infty} (\mu\times\nu) (R_k)<\frac{\varepsilon}{n2^n}$$
Such $\{R_k\}$ exists by Hahn's extension theorem. The $Z_x\subset \cup_{k=1}^{\infty}R_{k,x}$. Let $T_n$ be the set of all $x$ such that $$\frac{1}{n}\leq\sum_{k=1}^{\infty} \nu (R_{k,x})$$ Then $T_n$ is measurable, and $S_n\subset T_n$. Furthermore, the expression on the right is integrable with respect to $x$, and we find that
$$\frac{1}{n}\mu(T_n)\leq \sum_{k=1}^{\infty} \int_X \nu (R_{k,x}) d\mu = \sum_{k=1}^{\infty} (\mu\times\nu) (R_k)<\frac{\varepsilon}{n2^n}$$
This shows that $\mu(T_n)<\frac{\varepsilon}{2^n}$ and whence $S$ is contained in a set of measure $0$.
Ok so am not sure I understand the structure of the proof. I think we showed that for any $\varepsilon>0$ there exist a sequence of sets $\{T_n\}$ in $\mathcal{M}$ such that $S\subset\cup_{n=1}^{\infty}T_n$ and $\sum_{k=1}^{\infty}\mu(T_n)<\varepsilon$. This implies that for any $\varepsilon>0$ there exist a measurable set $T$ in $\mathcal{M}$ such that $S\subset T$ and $\mu(T)<\varepsilon$.
Am also not sure about the measurability and integrability claims.
Any help on this is very appreciated.
 A: Your understarding of the conclusion is correct. Indeed, at the the end of the prove the author has proved that for every $\epsilon>0$ there exist a measurable $T$ such that $S\subseteq T$ and $\mu(T)<\epsilon$.
Now, by using this fact for $\epsilon=1/n$ you obtain a sequence of measurable sets $(T_n)$ such that $S\subseteq T_n$ and $\mu(T_n)<1/n$ for all $n$. Now, $T=\bigcap T_n$ is measurable with $S\subseteq T$ and $\mu(T)=0$, thereby $\mu(S)=0$.
Now, you first question is why the function $x\mapsto \sum_{k=1}^{\infty}\nu(R_{k,x})$ is measurable. This follows from the fact that every $R_{k}$ is of the form $A_k\times B_k$ for $A_k\in \mathcal{M},\,B_k\in \mathcal{N}$, this means that $R_k$ is a rectangle. Now, it's easy to see that
$$R_{k,x}=(A_k\times B_k)_x =
\begin{cases}
B_k,\,\, x\in A_k\\
\emptyset,\,\,\,\,\,\, x\notin A_k.
\end{cases}
$$
In either case $R_{k,x}$ belongs to $\mathcal{N}$. Hence,
$$\nu(R_{k,x})=\begin{cases}
\nu(B_k),\,\,x\in A_k\\
0,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x\notin A_k.
\end{cases}
$$
In other words, $\nu(R_{k,x})=\nu(B_k)\cdot 1_{A_{k}}$, where $1_{A_k}$ is the characteristic function on the set $A_k$. From the latter expression it follows that the function $x\mapsto \nu(R_{k,x})$ is an $\mathcal{M}$ - measurable function. Now you can view the function $x\mapsto \sum_{k=1}^{\infty}\nu(R_{k,x})$ as the limit of measurable functions $(x\mapsto \sum_{k=1}^{N}\nu(R_{k,x})$). I.e. if $g_N=\sum_{k=1}^{N}\nu(R_{k,x})$ then $\lim_{N}g_N=\sum_{k=1}^{\infty}\nu(R_{k,x})$ which is measurable.
Now for the integrability part using the fact that $g_N\nearrow g$, where $g(x)=\sum_{k=1}^{\infty}\nu(R_{k,x})$ and monotone convergence we obtain
\begin{align}
\int_{X}\sum_{k=1}^{\infty}\nu(R_{k,x})\,d\mu(x)&=\int_{X}g(x)\,d\mu(x)\\
&=\int_{X}\lim_{N\to \infty}g_N(x)\,d\mu(x)\\
&=\lim_{N\to \infty}\int_{X}g_N(x)\,d\mu(x)\\
&=\lim_{N\to \infty}\int_{X}\sum_{k=1}^{N}\nu(R_{k,x})\,d\mu(x)\\
&=\lim_{N\to \infty}\sum_{k=1}^{N}\int_{X}\nu(R_{k,x})\,d\mu(x)\\
&=\sum_{k=1}^{\infty}\int_{X}\nu(R_{k,x})\,d\mu(x).
\end{align}
Now, using the expression $\nu(R_{k,x})=\nu(B_k)\cdot 1_{A_k}$ (as functions) we see that
\begin{align}
\int_{X}\nu(R_{k,x})\,d\mu(x)&=\int_{X}\nu(B_k)\cdot 1_{A_k}(x)\,d\mu(x)=\nu(B_k)\mu(A_k)\\
&=(\mu\otimes \nu)(A_k\times B_k)=(\mu\otimes \nu)(R_k)
\end{align}
Hence, combining the above equalities with the latter expression we obtain
$$\int_{X}\sum_{k=1}^{\infty}\nu(R_{k,x})\,d\mu(x)=\sum_{k=1}^{\infty}\int_{X}\nu(R_{k,x})\,d\mu(x)=\sum_{k=1}^{\infty}(\mu\otimes \nu)(R_k).$$
