Characterization of Sigma-Algebra Generated by Collection of Mappings

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Theorem

Let $\struct {X_i, \Sigma_i}$ be measurable spaces, with $i \in I$ for some index set $I$.

Let $X$ be a set, and let, for $i \in I$, $f_i: X \to X_i$ be a mapping.


Then:

$\map \sigma {f_i: i \in I} = \map \sigma {\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$

where:

$\map \sigma {f_i: i \in I}$ is the $\sigma$-algebra generated by $\family {f_i}_{i \mathop \in I}$
$\map \sigma {\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$ is the $\sigma$-algebra generated by $\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i}$
$\map {f_i^{-1} } {\Sigma_i}$ denotes the pre-image $\sigma$-algebra on $X$ by $f$


Proof

For each $i \in I$, one has by definition of generated $\sigma$-algebra:

$\ds \map {f_i^{-1} } {\Sigma_i} \subseteq \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$

which shows that each of the $f_i$ is measurable.


Next, suppose that $\Sigma$ is a $\sigma$-algebra such that each of the $f_i$ is $\Sigma \,/\, \Sigma_i$-measurable.

Then for all $i \in I$, one has:

$\map {f_i^{-1} } {\Sigma_i} \subseteq \Sigma$

and hence by Union is Smallest Superset: Family of Sets:

$\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \Sigma$


Finally, by Generated Sigma-Algebra Preserves Subset, it follows that:

$\ds \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} } \subseteq \Sigma$


Thus:

$\ds \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} } = \map \sigma {f_i : i \in I}$

by definition of the latter.

$\blacksquare$


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