# Pre-Image of Generated Sigma-Algebra

There is very well known lemma: Let $$\mathit f$$: $$X \mapsto Y$$ be a mapping.

Let $$\mathcal G \subseteq \mathcal P(Y)$$ be a collection of subsets of $$Y$$. $$\\$$

then $$\mathit f^{-1}(\sigma(\mathcal G)) = \sigma(\mathit f^{-1}(\mathcal G))$$. $$\\$$

I do not understand this lemma. The domain of the function $$\mathit f$$ is $$X$$, and the codomain is $$Y$$. $$\\$$

Thus, the domain of the function $$\mathit f^{-1}$$ is $$Y$$, not $$2^{Y}$$. The function $$\mathit f$$ is just not defined on $$2^{Y}$$. But obviously the elements of $$\mathcal G$$ and $$\sigma(\mathcal G)$$ consist of the elements of $$2^{Y}$$? Where is the mistake in my arguments? $$\\$$

E.g.

$$\left(\begin{matrix} 1 \\ 2 \\ 3 \\ 4 \\ \end{matrix}\right) \begin{matrix} \to \\ \to \\ \to \\ \to \\ \end{matrix} \left(\begin{matrix} 5 \\ 6 \\ 7 \\ 8 \\ \end{matrix}\right)$$

But this mapping is not defined on e.g. {5,7}, which is a subset of $$2^{Y}$$.

• It’s simply notation. By $f^{-1}(G)$ we mean the set of preimages $f^{-1}(G)$ for all $G$ in $\mathcal{G}$.
– user512346
Jun 18, 2019 at 21:13
• I did not get it. $\mathit f^{-1}$ is defined on $2^{Y}$? Do you mean that the domain of $\mathit f$ is $2^X$ and the codomain is $2^Y$? May be you meant to write $\mathit f^{-1}(\mathcal G)$, not $\mathit f^{-1}(G)$? In any case you provided no definition of notation which I suppose I do not inderstand. Jun 18, 2019 at 21:32
• For $G \subset Y$, $f^{-1}(G) = \{x \in X : f(x) \in G\}$.
– snar
Jun 18, 2019 at 21:35
• Yes, I understand the definition of $\mathit f^{-1}(G)$. But the lemma is about $\mathit f^{-1}(\mathcal G)$, not about $\mathit f^{-1}(G)$. If $\mathcal G=\{\{5,7\},\{5\}\}$ in my example, what is $\mathit f^{-1}(\mathcal G)$? Jun 18, 2019 at 21:42
• @snar here is the link of the lemma proofwiki.org/wiki/… Jun 18, 2019 at 22:07

OP agrees that $$f^{-1}(G) = \{x \in X : f(x) \in G\}$$ for $$G \subset Y$$, and the definition of $$f^{-1}(\mathcal{G})$$ was given in words in the comments as "the set of preimages $$f^{-1}(G)$$ for all $$G \in \mathcal{G}$$ ". Written verbatim, $$f^{-1}(\mathcal{G}) = \{f^{-1}(G) : G \in \mathcal{G}\} \subset \mathcal{P}(X).$$ The first step of the linked proof, $$f^{-1}(\mathcal{G}) \subset f^{-1}(\sigma(\mathcal{G}))$$ is spelled out as \begin{align*} f^{-1}(\mathcal{G}) &= \{f^{-1}(G) : G \in \mathcal{G}\} \\ &\subset \{f^{-1}(G) : G \in \sigma(\mathcal{G})\} \\ &= f^{-1}(\sigma(\mathcal{G})). \end{align*}