# How to show that the pull back of a sigma algebra is a sigma algebra.

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?

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Yes. The point being that $f^{-1}$ commutes with set operations. Which you should show...! – john w. Feb 17 '12 at 1:23
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?" – user12918723509187 Feb 17 '12 at 1:25

@Jacob: You should probably call your construction a "pushforward" rather than a "pullback" since you started with a $\sigma$-algebra on $X$ and built a $\sigma$-algebra on $Y$. – Mike F Sep 21 '15 at 17:14