TITLE: Borel $\sigma$-algebra of the product is the product of the Borel $\sigma$-algebras with $\sigma$-compactness
QUESTION [1 upvotes]: Suppose $X_1$ and $X_2$ are Hausdorff, locally compact, $\sigma$-compact spaces. 
Clearly the same holds for their product $X=X_1\times X_2$. 
We know that in general the Borel $\sigma$-algebra of the product, $B(X)$, includes $B(X_1)\times B(X_2)$ (but strict inclusion can happen). Now the question: 
Is it true that $B(X)=B(X_1)\times B(X_2)$ holds under the stated hypotheses on the spaces? 
What I observed so far is that it would suffice to show that $X$ is strongly Lindelof..

REPLY [2 votes]: No, not even if $X_1 = X_2 = X$ is compact Hausdorff.  Take a compact Hausdorff space with cardinal strictly greater than the continuum.  Then the diagonal
$$
\Delta = \{(x,x): x \in X\}
$$
is a closed set in $X \times X$, therefore $\Delta$ is a Borel set.  But $\Delta$ does not belong to the product sigma-algebra.  It does not even belong to the product sigma-algebra $\mathcal P(X) \otimes \mathcal P(X)$.
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