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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..

Mizar
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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 $\mathcal P(X) \otimes \mathcal P(X)$.

GEdgar
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    An interesting follow-up question would be: is it even possible for the equality to hold in an arbitrary product of (necessarily distinct) Hausdorff spaces of cardinality larger than continuum? – tomasz Dec 14 '13 at 14:21