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.
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 $$\int_{X}\sum_{k=1}^{\infty}\nu(R_{k,x})\,d\mu(x)=\sum_{k=1}^{\infty}\int_{X}\nu(R_{k,x})\,d\mu(x)=\sum_{k=1}^{\infty}(\mu\otimes \nu)(R_k).$$