For a second-countable space $X$, we have $${\rm Bor}(X\times X) = {\rm Bor}\, X \otimes {\rm Bor}\, X,$$ that is the Borel $\sigma$-algebra of the product is the product $\sigma$-algebra. Some [counterexamples to this statement can be produced for uncountable discrete spaces](https://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras), *etc*. However, I was wondering what happens in this particular example of 


$$X = \beta \mathbb N,$$


the [Stone–Čech compactification](https://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification) of the integers? Does this formula still hold?



Borel $\sigma$-algebra in $\beta \mathbb N \times \beta \mathbb N$
The answer is no.


Jiří Nedoma proved that if $(X,\Sigma)$ is a measurable space $|X| > 2^{\aleph\_0}$, then the diagonal is not a measurable subset of $(X\times X, \Sigma \otimes \Sigma)$. (The article is called *Note on Generalized Random Variables*, the result is Lemma 2, a proof can also be found in Schechter's *Handbook of Analysis and Its Foundations* section 21.8, which I found out about from David McIver on this very website).


Now, $|\beta(\mathbb{N})| = 2^{2^{\aleph\_0}} > 2^{\aleph\_0}$, so the diagonal is not an element of $\mathrm{Bor}(\beta(\mathbb{N})) \otimes \mathrm{Bor}(\beta(\mathbb{N}))$. But, as $\beta(\mathbb{N})$ is Hausdorff, the diagonal is closed, and therefore Borel, in $\beta(\mathbb{N}) \times \beta(\mathbb{N})$. 


