Timeline for Understanding the theorem from Kechris' book that Borel sets are mapped to Borel sets under injective functions
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 10 at 10:26 | comment | added | guest1 | @AlessandroCodenotti Thank you again for your answer! Is it true (and if so how can I see this) that all sets in $\mathcal{B}(f(X))$ are of the form $f(A)$, for some A in $\mathcal{B}(X)$? | |
| Nov 7 at 10:12 | comment | added | Alessandro Codenotti | No, if $B\not\in\mathcal A$ there is no reason for $\mathcal A\mid B$ to be contained in $\mathcal A$. In particular if $B$ is Borel (in $Y$) but $f(X)$ is not, then $f(X)\cap B$ will be Borel in $f(X)$ but it doesn't have to be Borel in $Y$ | |
| Nov 7 at 10:00 | comment | added | guest1 | @AlessandroCodenotti thanks again for your answer! What I am wondering now is, if not every trace sigma algebra $\mathcal{A}|B$ is a sub sigma algebra of $\mathcal{A}$? I.e., do we in our problem here even need the condition that $f(X)\in\mathcal{B}(Y)$ to say that $\mathcal{B}(f(X))\subseteq \mathcal{B}(Y)$? since we already know that $\mathcal{B}(f(X))$ is a trace sigma algebra of $\mathcal{B}(Y)$... | |
| Nov 5 at 12:56 | comment | added | Alessandro Codenotti | Every element $B'$ of $\mathcal B(f(X))$ is of the form $B'=f(X)\cap B$ for some $B\in \mathcal B(Y)$. If $f$ is injective then $f(X)\in\mathcal B(Y)$ and so $B'\in\mathcal B(Y)$, since $\mathcal B(Y)$ is closed under finite intersections. | |
| Nov 5 at 12:44 | comment | added | guest1 | @AlessandroCodenotti thanks again! So re-reading your first comment I actually dont understand why $f(X)\in\mathcal{Y})$ implies that $\mathcal{B}(f(X))\subseteq \mathcal{B}(Y)$. Could you briefly explain the logic behind this? So usually one would need to show that an arbitrary element from $\mathcal{B}(f(X))$ is contained in $\mathcal{B}(Y)$, to show that it is a sub set | |
| Nov 5 at 12:40 | comment | added | Alessandro Codenotti | Yes. But in most applications you care about $f(A)$ being in $\mathcal B(Y)$ which is why the result is stated this way I'd guess | |
| Nov 5 at 11:40 | comment | added | guest1 | @AlessandroCodenotti thanks for your comment! So does that mean though that the statement (i) could be made more precise to say that $f(A)$ is actually in the borel sigma algebra on $f(X))$? | |
| Nov 5 at 11:18 | comment | added | Alessandro Codenotti | $\mathcal B(f(X))=\mathcal B(Y)\mid f(X)=\{f(X)\cap B\mid B\in\mathcal B(Y)$ by the remark at the beginning of 11.A, so, since $f(X)\in B(Y)$, we have $\mathcal B(f(X))\subseteq\mathcal B(Y)$. | |
| Nov 5 at 11:09 | history | asked | guest1 | CC BY-SA 4.0 |