Generated Sigma-Algebra Preserves Subset

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Theorem

Let $X$ be a set.

Let $\mathcal F, \mathcal G \subseteq \mathcal P \left({X}\right)$ be collections of subsets of $X$.


Suppose that $\mathcal F \subseteq \mathcal G$.

Then $\sigma \left({\mathcal F}\right) \subseteq \sigma \left({\mathcal G}\right)$, where $\sigma \left({\mathcal G}\right)$ denotes the $\sigma$-algebra generated by $\mathcal G$


Proof

By definition of $\sigma \left({\mathcal G}\right)$, $\mathcal G \subseteq \sigma \left({\mathcal G}\right)$.

It follows that also $\mathcal F \subseteq \sigma \left({\mathcal G}\right)$.

Hence, by definition of $\sigma \left({\mathcal F}\right)$, $\sigma \left({\mathcal F}\right) \subseteq \sigma \left({\mathcal G}\right)$.

$\blacksquare$


Sources