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Let $\{(X_k, \mathcal{X}_k) : k \in K \}$ be a nonempty family of measurable spaces. In my measure theory class, we defined $\otimes_{k \in K}\mathcal{X}_k$ (product of $\sigma$-algebras) to be the smallest $\sigma$-algebra for which all the projections $\pi_k$ are measurable.

So, does this mean that combining measurable and unmeasurable sets gives you a measurable set? To be precise, let's say we are on $(\mathbb{R}^2 , \mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R}))$. Is $V \times [0, 1]$ measurable in $\mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})$, where V is the Vitali set (or any other $\mathcal{B}(\mathbb{R})$ - unmeasurable set)?

$\pi_2(V \times [0, 1]) = [0,1]$, and $\pi_2$ is measurable on product $\sigma$-algebra... so the answer should be yes? Or am I making a mistake? Or $\mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})$ isn't equal with $\mathcal{B}(\mathbb{R}^2)$?

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That $\pi_2$ is measurable with respect to $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$ means that $$ \pi_2^{-1}(A)\in \mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}),\quad A\in\mathcal{B}(\mathbb{R}) $$ but not that if $B\subseteq \mathbb{R}^2$ is a set such that $\pi_2(B)\in\mathcal{B}(\mathbb{R})$, then $B\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$. Note that in your example $\pi_2^{-1}([0,1])=\mathbb{R}\times [0,1]$.

For example, consider the constant function $f:\mathbb{R}^2\to \mathbb{R}$ given by $f(x,y)=0$. This is clearly measurable with respect to any sigma-algebra you may put on $\mathbb{R}^2$, but for any non-measurable set $B\subseteq\mathbb{R}^2$ you have $f(B)=\{0\}\in\mathcal{B}(\mathbb{R})$.

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