Product of Borel sigma algebra Consider two generic topological spaces $(X,\mathcal{T}_1),(Y,\mathcal{T}_2)$.
We can then define the Borel measurable spaces $(X,\sigma(\mathcal{T}_1)),(Y,\sigma(\mathcal{T}_2))$, where $\sigma(\mathcal{T})$ is the smaller sigma algebra that cointains $\mathcal{T}$.
Now consider the product topological space $(X \times Y,\mathcal{T})$, where $\mathcal{T}$ is the product topology, namely the topology which has $\mathcal{B}=\{A_1 \times A_2 \mid A_1 \in \mathcal{T}_1, A_2 \in \mathcal{T}_2\}$ as a base.
We can now define a sigma algebra on $X \times Y$ in two different ways:

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*We can define the Borel product sigma algebra $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2):=\sigma(\{A_1 \times A_2 \mid A_1 \in \sigma(\mathcal{T}_1), A_2 \in \sigma(\mathcal{T}_2)\})$;


*We can instead define the Borel sigma algebra on product topology $\sigma(\mathcal{T})$.
I have read that it is always true that $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \subseteq \sigma(\mathcal{T})$.
The converse ($\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \supseteq \sigma(\mathcal{T})$) is true for example when $(X,\mathcal{T}_1)$ and $(Y,\mathcal{T}_2)$ are second countable (namely they own a countable basis).
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I want to proof these two relations. The second one is quite easy, and here is my attempt:
$X,Y$ are second countable, so there exist two countable basis $\mathcal{B}_1$ and $\mathcal{B}_2$ of these topological spaces.
We have that $\mathcal{B}:=\{B_1 \times B_2 \mid B_1 \in \mathcal{B}_1, B_2 \in \mathcal{B}_2\}$ is a countable base for $\mathcal{T}$, and then $\sigma(\mathcal{T})=\sigma(\mathcal{B})$.
Moreover $\mathcal{B} \subseteq \{A_1 \times A_2 \mid A_1 \in \mathcal{T}_1, A_2 \in \mathcal{T}_2\}=\{A_1 \times A_2 \mid A_1 \in \sigma(\mathcal{T}_1), A_2 \in \sigma(\mathcal{T}_2)\}$.
So $\sigma(\mathcal{T})=\sigma(\mathcal{B}) \subseteq \sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$.
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My question now is: how can I prove the first relation? (namely $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \subseteq \sigma(\mathcal{T})$) I have searched a lot, but I didn't find a decent proof.
My attempt is the following:
$X,Y$ now are generic topological spaces (not second countable).
It is enough to prove that $\{A_1 \times A_2 \mid A_1 \in \sigma(\mathcal{T}_1), A_2 \in \sigma(\mathcal{T}_2)\} \subseteq \sigma(\mathcal{T})$.
However, I'm pretty stuck here. Any hint? Thank you so much!
 A: If $X,Y$ are second countable, so is $X \times Y$ and this implies that every open subset $O$ of $X \times Y$ is a countable union of basic open subsets (of the form $O_1 \times O_2$) which are by definition in $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$. So all open subsets of the product are in $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$, which is a $\sigma$-algebra and so contains $\sigma(\mathcal{T})$ by minimality.
OTOH, define $\mathcal{A} = \{A \subseteq X\mid: A \times Y \in \sigma(\mathcal{T})\}$ and show this is a $\sigma$-algebra that contains all open subsets of $X$, so $\sigma(\mathcal{T_1}) \subseteq \mathcal{A}$. Similarly, show that $$\sigma(\mathcal{T_2}) \subseteq \mathcal{B}:= \{B \subseteq Y\mid X \times B \in \sigma({T})\}$$ and note that from these facts it follows that for $A \in \sigma(\mathcal{T_1})$ and $B \in \sigma(\mathcal{T_2})$ we have that $A \times B = (A \times Y) \cap (X \times B) \in \sigma(\mathcal{T})$ and so the whole generating set of $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$ is in $\sigma(\mathcal{T})$ which implies what you need:
$$\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \subseteq \sigma(\mathcal{T}) = \textrm{Bor}(X \times Y)$$
It can be shown that for a discrete uncountable space $X$ of size $|\Bbb R|$ we have that the diagonal $\Delta_X = \{(x,x) \mid x \in X\}$ is Borel in $X \times X$ but not in the product Borel algebra as defined under 1. Proofs can be found on this site.
