The following is a proposition in Folland's Real Analysis about product sigma algebra:

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Here $\mathcal{B}_X$ denotes the Borel $\sigma$-algebra on $X$.

Could anyone come up with an example that $$ \mathcal{B}_X\setminus\otimes_{1}^n\mathcal{B}_{X_j} $$ is nonempty?


Let $X_1$ and $X_2$ have the discrete topology. Then $X=X_1\times X_2$ has the discrete topology, so $\mathcal{B}_X=P(X)$. But if $X_1$ and $X_2$ are large enough $\mathcal{B}_{X_1}\otimes\mathcal{B}_{X_2}$ will not be all of $P(X)$. Here's one way to prove this. For $S\subset X$ and $x\in X_1$, let $S_x=\{y\in X_2:(x,y)\in S\}$. Now let $$\mathcal{A}=\{S:|\{S_x:x\in X_1\}|\leq 2^{\aleph_0}\}.$$ Then it is straightforward to check that $\mathcal{A}$ is a $\sigma$-algebra (the key point is that $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$). Since $\mathcal{A}$ contains all rectangles, it follows that $\mathcal{B}_{X_1}\otimes \mathcal{B}_{X_2}\subseteq\mathcal{A}$. But if $|X_1|>2^{\aleph_0}$ and $2^{|X_2|}>2^{\aleph_0}$, then $\mathcal{A}$ is not all of $P(X)$, since there are more than $2^{\aleph_0}$ different choices for $S_x$ and there are enough points $x\in X_1$ to build a set $S$ that realizes more than $2^{\aleph_0}$ of them.


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