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I need some help with subparagraph b) of the question.

$X$ and $Y$ are sets and $f$ a mapping from $X$ to $Y$. Let $\mathcal{A}$ be a sigma-algebra on X.

A) Show that the collection $\mathcal{B} = \{ B\subseteq Y: f^{-1}(B)\in \mathcal{A} \}$ is a $\sigma$- algebra on $Y$.

b) Let $C$ be the $\sigma$-algebra generated in $Y$ by $f(\mathcal{A}) =\{ f (A): A \in \mathcal{A} \} $. Show that if f is injective, then $C \subseteq \mathcal{B}$ and if $f$ is surjective, then $\mathcal{B} \subseteq C$.

I dont know how work with the set generated by $f(\mathcal{A})$ and the inyectivity, surjectivity of the preimage... THX

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2 Answers 2

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If $B\in \mathcal B$ then $f^{-1}(B^c)=f^{-1}(B)^c\in \mathcal A$ hence $B^c\in\mathcal B$. Now suppose $(B_n)_n\subseteq \mathcal B$ then $f^{-1}(\bigcup B_n)=\bigcup f^{-1}(B_n)\in\mathcal A$ hence $\bigcup B_n\in\mathcal B$. Therefore $\mathcal B$ is a $\sigma$-algebra.

For part (b): If $f$ is surjective and $B\in\mathcal B$ then $A=f^{-1}(B)\in\mathcal A$. Note that $f(A)=B$ as $f$ is surjective hence $\mathcal B\subseteq \mathcal C$. Now, let us consider the case of $f$ injective: Observe that it suffices to show that each generator $f(A)\in \mathcal B$ for all $A\in\mathcal A$. Since $f$ is injective $f^{-1}(f(A))=A\in\mathcal A$. This finishes the proof.

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THX, your answer will be very helpfull –  kEoz Apr 14 '12 at 2:12

Here are two general results you can use. Let $f:X\to Y$ be any function.

  1. Let $S\subseteq X$. If $f$ is injective, then $S=f^{-1}\big(f(S)\big)$.

  2. Let $T\subseteq Y$. If $f$ is surjective, then $f\big(f^{-1}(T)\big)=T$.

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