Showing $B$ Borel set, then $f^{-1}(B)$ is a Borel set with $f$ continuous function.
$begingroup$
Let $f:mathbb{R}to [-infty,infty]$ continuous function.
(a) Let $Omega=left{E:f^{-1}(E) text{ is a Borel set } right}$. Show that $Omega$ is $sigma$-algebra.
I already proved this. :)
(b) Let $B$ be a Borel set. Show that $f^{-1}(B)$ is a Borel set.
For (b) I have this.
$f$ continuous then $f$ is measurable, because $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set.
It is correct?...
real-analysis measure-theory lebesgue-measure
edited Dec 3 '18 at 23:34
Bernard
119k740113
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}to [-infty,infty]$ continuous function.
(a) Let $Omega=left{E:f^{-1}(E) text{ is a Borel set } right}$. Show that $Omega$ is $sigma$-algebra.
I already proved this. :)
(b) Let $B$ be a Borel set. Show that $f^{-1}(B)$ is a Borel set.
For (b) I have this.
$f$ continuous then $f$ is measurable, because $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set.
It is correct?...
real-analysis measure-theory lebesgue-measure
edited Dec 3 '18 at 23:34
Bernard
119k740113
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
$endgroup$
$begingroup$
No, your logic is assailable. Since $f$ is continuous, $Omega$ contains the open sets, and since it is a $sigma$-field, it contains all the Borel sets.
$endgroup$
– copper.hat
Dec 4 '18 at 0:00
add a comment |
$begingroup$
Let $f:mathbb{R}to [-infty,infty]$ continuous function.
(a) Let $Omega=left{E:f^{-1}(E) text{ is a Borel set } right}$. Show that $Omega$ is $sigma$-algebra.
I already proved this. :)
(b) Let $B$ be a Borel set. Show that $f^{-1}(B)$ is a Borel set.
For (b) I have this.
$f$ continuous then $f$ is measurable, because $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set.
It is correct?...
real-analysis measure-theory lebesgue-measure
edited Dec 3 '18 at 23:34
Bernard
119k740113
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
$endgroup$
Let $f:mathbb{R}to [-infty,infty]$ continuous function.
(a) Let $Omega=left{E:f^{-1}(E) text{ is a Borel set } right}$. Show that $Omega$ is $sigma$-algebra.
I already proved this. :)
(b) Let $B$ be a Borel set. Show that $f^{-1}(B)$ is a Borel set.
For (b) I have this.
$f$ continuous then $f$ is measurable, because $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set.
It is correct?...
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
edited Dec 3 '18 at 23:34
Bernard
119k740113
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
edited Dec 3 '18 at 23:34
Bernard
119k740113
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
edited Dec 3 '18 at 23:34
Bernard
119k740113
edited Dec 3 '18 at 23:34
Bernard
119k740113
edited Dec 3 '18 at 23:34
Bernard
119k740113
119k740113
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
asked Dec 3 '18 at 23:31
eraldcoileraldcoil
385211
385211
$begingroup$
No, your logic is assailable. Since $f$ is continuous, $Omega$ contains the open sets, and since it is a $sigma$-field, it contains all the Borel sets.
$endgroup$
– copper.hat
Dec 4 '18 at 0:00
add a comment |
$begingroup$
No, your logic is assailable. Since $f$ is continuous, $Omega$ contains the open sets, and since it is a $sigma$-field, it contains all the Borel sets.
$endgroup$
– copper.hat
Dec 4 '18 at 0:00
$begingroup$
No, your logic is assailable. Since $f$ is continuous, $Omega$ contains the open sets, and since it is a $sigma$-field, it contains all the Borel sets.
$endgroup$
– copper.hat
Dec 4 '18 at 0:00
$begingroup$
No, your logic is assailable. Since $f$ is continuous, $Omega$ contains the open sets, and since it is a $sigma$-field, it contains all the Borel sets.
$endgroup$
– copper.hat
Dec 4 '18 at 0:00
add a comment |
1 Answer
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$begingroup$
It appears that you are assuming what you are supposed to prove. The purpose of this exercise is to prove that a continuous function is Borel measurable. The correct argument is to use a): the sigma algebra in a) contains all open sets because $f^{-1}(U)$ is open for any open set $U$; if a sigma algebra contains all open sets it contains all Borel sets because Borel sigma algebra is the smallest sigma algebra containing all open sets.
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It appears that you are assuming what you are supposed to prove. The purpose of this exercise is to prove that a continuous function is Borel measurable. The correct argument is to use a): the sigma algebra in a) contains all open sets because $f^{-1}(U)$ is open for any open set $U$; if a sigma algebra contains all open sets it contains all Borel sets because Borel sigma algebra is the smallest sigma algebra containing all open sets.
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
$begingroup$
It appears that you are assuming what you are supposed to prove. The purpose of this exercise is to prove that a continuous function is Borel measurable. The correct argument is to use a): the sigma algebra in a) contains all open sets because $f^{-1}(U)$ is open for any open set $U$; if a sigma algebra contains all open sets it contains all Borel sets because Borel sigma algebra is the smallest sigma algebra containing all open sets.
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
add a comment |
$begingroup$
It appears that you are assuming what you are supposed to prove. The purpose of this exercise is to prove that a continuous function is Borel measurable. The correct argument is to use a): the sigma algebra in a) contains all open sets because $f^{-1}(U)$ is open for any open set $U$; if a sigma algebra contains all open sets it contains all Borel sets because Borel sigma algebra is the smallest sigma algebra containing all open sets.
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
It appears that you are assuming what you are supposed to prove. The purpose of this exercise is to prove that a continuous function is Borel measurable. The correct argument is to use a): the sigma algebra in a) contains all open sets because $f^{-1}(U)$ is open for any open set $U$; if a sigma algebra contains all open sets it contains all Borel sets because Borel sigma algebra is the smallest sigma algebra containing all open sets.
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 4 '18 at 0:01
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
55.7k42158
add a comment |
add a comment |
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No, your logic is assailable. Since $f$ is continuous, $Omega$ contains the open sets, and since it is a $sigma$-field, it contains all the Borel sets.
$endgroup$
– copper.hat
Dec 4 '18 at 0:00