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Nov 12 at 10:05 vote accept guest1
Nov 11 at 23:42 answer added Ramiro timeline score: 2
Nov 11 at 12:43 history edited guest1 CC BY-SA 4.0
added 295 characters in body
Nov 11 at 12:17 history reopened guest1
amWhy
Dean Miller
Another User
Dermot Craddock
Nov 11 at 10:49 history edited Dean Miller CC BY-SA 4.0
Fixed some further typos.
Nov 11 at 10:42 history edited guest1 CC BY-SA 4.0
I have made the question more precise and corrected some mistakes.
Nov 11 at 9:01 review Reopen votes
Nov 11 at 12:17
Nov 11 at 9:00 history left closed in review Harish Chandra Rajpoot
José Carlos Santos
Another User
Original close reason(s) were not resolved
Nov 11 at 8:19 comment added guest1 @KaviRamaMurthy Thanks for your comments. Yes I did indeed forgot to mention that the spaces need to be Polish. Sorry for this inconvenience
S Nov 11 at 8:18 review Reopen votes
Nov 11 at 9:00
S Nov 11 at 8:18 history edited guest1 CC BY-SA 4.0
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Nov 11 at 5:01 history closed Kavi Rama Murthy
Harish Chandra Rajpoot
Leucippus
Bowei Tang
Afntu
Not suitable for this site
Nov 10 at 23:20 review Close votes
Nov 11 at 5:01
Nov 10 at 23:12 answer added Ramiro timeline score: 2
Nov 10 at 22:56 comment added Kavi Rama Murthy Consider the identity map from $\mathbb R$ with power set to $\mathbb R$ with $\{\emptyset, \mathbb R\}$. This is injective and measurable, but does not map measurable sets to measurable sets. Such things have been pointed out to you many times in the past but you keep ignoring them. I am downvoting your question.
Nov 10 at 22:43 comment added Kavi Rama Murthy According to which theorem $A\in\mathcal{B}(X)$, implies that $f(A)\in\mathcal{B}(Y)$ if $f$ is measurable and injective ? I don't think this is true for general measurable spaces.
Nov 10 at 14:03 history asked guest1 CC BY-SA 4.0