# Induced $\sigma$-algebra vs. product $\sigma$-algebra

Let $(E,\mathscr E)$ be a measurable space and $(S,2^S)$ be a finite set. Let $$\xi:(E,\mathscr E)\to(S,2^S)$$ be a mesaurable function, i.e. $\xi^{-1}(s)\in \mathscr E$ for any $s\in S$. Now, let us denote by $\Omega = E^{\mathbb N_0}$ and by $\mathscr F$ its product $\sigma$-algebra. Also, let $\Sigma = S^{\mathbb N_0}$ and let $\mathscr S$ be the correspondent product $\sigma$-algebra.

Let $\eta:\Omega\to\Sigma$ be the element-wise extension of $\xi$, i.e. $$\eta(\omega_0,\omega_1,\dots) = (\xi(\omega_0),\xi(\omega_1),\dots).$$

I wonder if $\mathscr S$ is different from $$\mathscr C = \{A\subseteq \Sigma:\eta^{-1}(A)\in \mathscr F\}.$$ It is clear that $\mathscr S\subseteq \mathscr C$ - but can the strict inclusion actually happen?

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Yes, strict inclusion can happen. If $S$ has more than one element, $(\Sigma,\mathscr{S})$ will be an uncountable compact metric space with the Borel $\sigma$-algebra. By standard arguments, there will be $\mathfrak{c}$ many measurable sets, there are as many measurable sets as real numbers. Also, $|\Sigma|=\mathfrak{c}$.

Now let $S=\{0,1,2\}$ and let $\xi$ be the constant function that maps everything to $2$. There are $2^\mathfrak{c}$ subsets of $\{0,1\}^\mathbb{N}$. Take any subset $A$ of $\{0,1\}^\mathbb{N}$ and let $A'=A\cup\{(2,2,2,\ldots)\}$. There are $2^\mathfrak{c}$ sets of this form and since $\eta^{-1}(A')=\Omega$ for all of them, they are all in $\mathscr{C}$.

So for cardinality reasons, the inclusion will be strict. Note that $\mathscr{C}$ is the largest $\sigma$-algebra on $\Sigma$ that is compatible with $\eta$ being measurable. It shouldn't be surprising that it is quite large.

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