Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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?" –  William Feb 17 '12 at 1:25

1 Answer 1

Your attempt is good. The inverse image operation plays nicely with set operations.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.