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Let $\mathcal{M}$ be a sigma algebra on $X$. Let $f:X\to Y$. Define $$ \mathcal{A}=\{B\subset Y : f^{-1}(B)\in \mathcal{M}\}. $$ The problem is to show that $\mathcal{A}$ is a sigma algebra on $Y$.

This is my attempt.

Clearly, $\emptyset \in \mathcal{A}$.
Let $\{B_k\}_{k=1}^\infty$ be a countable collection of sets such that $f^{-1}({B_k})\in \mathcal{M}$. Then $f^{-1}(\cup B_k)=\cup f^{-1}(B_k)\in \mathcal{M}$. So $\cup B_k \subset Y$ and hence $\cup B_k\in\mathcal{A}$.
Also, $f^{-1}(B^c)=(f^{-1}(B))^c\in \mathcal{M}$ and so $B^c\in \mathcal{A}$.

Thus $\mathcal{A}$ is a sigma algebra on $Y$.

Please, is what I have done right?

  • 1
    Yes. The point being that $f^{-1}$ commutes with set operations. Which you should show...!2012-02-17
  • 0
    Looks fine to me, but I have a question: what motivated you to ask this question? No no, don't get me wrong, I am just trying to help: is there anything specific that is bothering you about your solution? Something that feels "not right?"2012-02-17

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