Understanding Lang's proof of product $\sigma$-algebras We have two $\sigma$-finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ and we let $\mathcal{A}$ and $\mathcal{B}$ denote the rings of sets of finite measure in $\mathcal{M}$ and $\mathcal{N}$ respectively. We also let $\mathcal{A}\times \mathcal{B}$ denote the collection of all finite disjoint unions of rectangles $A\times B$ with $A\in \mathcal{A}$ and $B\in \mathcal{B}$. On page 158 of his book Lang proves the following lemma:
Lemma. Denote by $\mathcal{A}\otimes \mathcal{B}$ the $\sigma$-algebra generated by $\mathcal{A}\times \mathcal{B}$ in $X\times Y$ and denote by $\mathcal{A}^\sigma$ the $\sigma$-algebra generated by $\mathcal{A}$ in $X$. Then
$$\mathcal{A}^\sigma\otimes \mathcal{B}^\sigma=(\mathcal{A}\times \mathcal{B})^{\sigma}$$
Proof. Since $\mathcal{A}\times \mathcal{B}\subset \mathcal{A}^\sigma\times \mathcal{B}^\sigma \subset \mathcal{A}^\sigma\otimes \mathcal{B}^\sigma$ it follows that $(\mathcal{A}\times \mathcal{B})^{\sigma}\subset \mathcal{A}^\sigma\otimes \mathcal{B}^\sigma$ and we must show the reverse inclusion. For each $B\in \mathcal{B}$ consider the $\sigma$-algebra in $X\times B$ generated by all sets $A\times B$ with $A\in\mathcal{A}$. It is contained in $(\mathcal{A}\times \mathcal{B})^{\sigma}$, which therefore contains $\mathcal{A}^{\sigma}\times\{B\}$ for all $B\in \mathcal{B}$. Now for any $A\in\mathcal{A}^{\sigma}$, it follows that $\{A\}\times \mathcal{B}^{\sigma}$ is contained in $(\mathcal{A}\times \mathcal{B})^{\sigma}$. Thus finally $ \mathcal{A}^{\sigma}\times  \mathcal{B}^{\sigma}\subset (\mathcal{A}\times \mathcal{B})^{\sigma}$, whence the reverse inclusion $\mathcal{A}^\sigma\otimes \mathcal{B}^\sigma\subset  (\mathcal{A}\times \mathcal{B})^{\sigma}$.
I understand that the $\sigma$-algebra in $X\times B$ generated by all sets $A\times B$ with $A\in\mathcal{A}$ is in fact the collection $\mathcal{C}:=\{A\times B:A\in \mathcal{A}^\sigma\}$, but I don't see why this collection must be contained in $(\mathcal{A}\times \mathcal{B})^{\sigma}$.
Any help on this is very appreciated.
 A: The quoted proof is sloppy, probably you will need some previous result to handle the assertions on the proof, or some work to make every assertion clear.
This is a way to see it: let $f: (X,\mathcal{X})\to Y$ a function from a measurable space to some set $Y$, then we can define a $\sigma $-algebra in $Y$ as $f_*(\mathcal{X}):=\{C\subset Y: f^{-1}(C)\in \mathcal{X}\}$ (it can be shown, using set theoretic properties of $f^{-1}$ that $f_*(\mathcal{X})$ is indeed a $\sigma $-algebra in $Y$, namely the push-forward $\sigma $-algebra induced by $f$).
The proof below need the fact that $X\in \mathcal{A}$ and $Y\in \mathcal{B}$, so I will assume it from now.
Let $\tilde {\mathcal{A}}:=\operatorname{proj_1}_*((\mathcal{A}\times \mathcal{B})^{\sigma })$ where $\operatorname{proj_1}:X\times Y\to X,\, (x,y)\mapsto x$ is the canonical projection in the first coordinate. From the definition of the push-forward $\sigma $-algebra its easy to see that $\mathcal{A}\subset \tilde {\mathcal{A}}$ so $\mathcal{A}^{\sigma }\subset \tilde {\mathcal{A}}$, and consequently $A\times Y= \operatorname{proj_1}^{-1}(A)\in (\mathcal{A}\times \mathcal{B})^{\sigma}$ for any chosen $A\in \mathcal{A}^{\sigma }$.
Finally it follows that $A\times B=(A\times Y)\cap (X\times B)\in (\mathcal{A}\times \mathcal{B})^{\sigma}$ for any $A\times B\in \mathcal{A}^{\sigma }\times \mathcal{B}^{\sigma }$.∎
A: It seems to me that Lang omitted an important assumption needed for the lemma, which is that $X$ and $Y$ must be countable unions of sets in $\mathcal{A}$ and  $\mathcal{B}$ respectively. In particular it holds if $\mathcal{A}$ and  $\mathcal{B}$ are algebras of subsets in $X$ and $Y$ respectively.
To see why this assumption is necessary consider an example where $X\neq \emptyset \neq Y$ and $B\subset Y$ with $\emptyset\neq B\neq Y$. Let $\mathcal{A}=\{\emptyset\}$ and $\mathcal{B}=\{\emptyset,B,Y\setminus B,Y\}$. Then  $\mathcal{A}$ and  $\mathcal{B}$ are rings of subsets in $X$ and $Y$ respectively. We have $\mathcal{A}\times \mathcal{B}=\{\emptyset\}$ and so $(\mathcal{A}\times \mathcal{B})^{\sigma }=\{\emptyset,X\times Y\}$. Other the other hand we have $\mathcal{A}^{\sigma}\times\mathcal{B}^{\sigma}=\{\emptyset,X\}\times \{\emptyset,B,Y\setminus B,Y\}$. It follows that $X\times B\in \mathcal{A}^{\sigma}\times\mathcal{B}^{\sigma}\subset \mathcal{A}^\sigma\otimes \mathcal{B}^\sigma$ but $X\times B\notin (\mathcal{A}\times \mathcal{B})^{\sigma }$ so the lemma dosen't hold.
To see why the assumption is also sufficient lets rewrite the proof in details. Since $\mathcal{A}\times \mathcal{B}\subset \mathcal{A}^\sigma\times \mathcal{B}^\sigma \subset \mathcal{A}^\sigma\otimes \mathcal{B}^\sigma$ it follows that $(\mathcal{A}\times \mathcal{B})^{\sigma}\subset \mathcal{A}^\sigma\otimes \mathcal{B}^\sigma$ and we must show the reverse inclusion.
For each $B\in\mathcal{B}$ consider the $\sigma$-algebra $\Sigma_B$ in $X\times B$ generated by all sets $A\times B$ with $A\in\mathcal{A}$. One checks that we have $\Sigma_B=\{A\times B:A\in \mathcal{A}^{\sigma}\}$. By assumption there exist a sequence $(A_n)\subset \mathcal{A}$ with $X=\cup_{n}A_n$ which implies $X\times B=\cup_n (A_n\times B) \in (\mathcal{A}\times \mathcal{B})^{\sigma}$. It follows that $(\mathcal{A}\times \mathcal{B})^{\sigma}\restriction_{X\times B} \subset (\mathcal{A}\times \mathcal{B})^{\sigma}$, where $(\mathcal{A}\times \mathcal{B})^{\sigma}\restriction_{X\times B}$ denotes the trace of $(\mathcal{A}\times \mathcal{B})^{\sigma}$ on $X\times B$. Moreover, $(\mathcal{A}\times \mathcal{B})^{\sigma}\restriction_{X\times B}$ is a $\sigma$-algebra in $X\times B$ containing all sets  $A\times B$ with $A\in\mathcal{A}$ since $A\times B=(A\times B)\cap (X\times B)$ with $A\times B\in (\mathcal{A}\times \mathcal{B})^{\sigma}$. It follows that $\Sigma_B\subset (\mathcal{A}\times \mathcal{B})^{\sigma}\restriction_{X\times B} \subset (\mathcal{A}\times \mathcal{B})^{\sigma}$ for all $B\in \mathcal{B}$. This implies $\mathcal{A}^{\sigma}\times\{B\} \subset (\mathcal{A}\times \mathcal{B})^{\sigma}$ for all $B\in \mathcal{B}$.
For each $A\in \mathcal{A}^{\sigma}$ consider the $\sigma$-algebra $\Sigma_A$ in $A\times Y$ generated by all sets $A\times B$ with $B\in\mathcal{B}$. One checks that we have $\Sigma_A=\{A\times B:B\in \mathcal{B}^{\sigma}\}$. By assumption there exist a sequence $(B_n)\subset \mathcal{B}$ with $Y=\cup_{n}B_n$ which implies $A\times Y=\cup_n (A\times B_n) \in (\mathcal{A}\times \mathcal{B})^{\sigma}$ since $A\times B_n\in \mathcal{A}^{\sigma}\times\{B_n\}$ for each $n$. It follows that $(\mathcal{A}\times \mathcal{B})^{\sigma}\restriction_{A\times Y} \subset (\mathcal{A}\times \mathcal{B})^{\sigma}$. Moreover, $(\mathcal{A}\times \mathcal{B})^{\sigma}\restriction_{A\times Y} $ is a $\sigma$-algebra in $A\times Y$ which contains all sets $A\times B$ with $B\in\mathcal{B}$, since $A\times B=(A\times B)\cap (A\times Y)$ with $A\times B\in \mathcal{A}^{\sigma}\times\{B\} \subset (\mathcal{A}\times \mathcal{B})^{\sigma}$. It follows that $\Sigma_A\subset (\mathcal{A}\times \mathcal{B})^{\sigma}\restriction_{A\times Y} \subset (\mathcal{A}\times \mathcal{B})^{\sigma}$ for all $A\in \mathcal{A}^{\sigma}$. This implies $A\times B \in  (\mathcal{A}\times \mathcal{B})^{\sigma}$ for all $A\in \mathcal{A}^{\sigma}$ and $B\in \mathcal{B}^{\sigma}$ and whence $\mathcal{A}^{\sigma}\times \mathcal{B}^{\sigma}\subset (\mathcal{A}\times \mathcal{B})^{\sigma}$. We conclude that $\mathcal{A}^\sigma\otimes \mathcal{B}^\sigma\subset(\mathcal{A}\times \mathcal{B})^{\sigma}$.
A: Since $\mathcal{A}\times B \subset \mathcal{A}\times \mathcal{B} \subset (\mathcal{A}\times \mathcal{B})^{\sigma}$, we have that the $\sigma-$algebra generated by $\mathcal{A}\times B$ is contained in $(\mathcal{A}\times \mathcal{B})^{\sigma}$ as well.
