Any continuous function is Baire function?
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Afirmation. All continuous function is Baire function.
Let $f:Xto mathbb{R}$ continuous function. X compact hausdorff with $mathcal{B} sigma$-algebra of Baire.
Let $bigcap_{n=1}^{infty} O_ninmathcal{B}(mathbb{R})$ compact set. (sigma-algebra of Baire in $mathbb{R}$)
Now, $f^{-1}(bigcap_{n=1}^{infty} O_n)=bigcap_{n=1}^{infty}f^{-1}(O_n)$ compact set, and $f^{-1}(O_n)$ open sets. Therefore, $f^{-1}(bigcap_{n=1}^{infty} O_n)in mathcal{B}(X)$. Therefore $f$ is Baire function.
It is correct?
functional-analysis analysis measure-theory
asked Nov 18 at 16:33
eraldcoil
25119
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up vote
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A have doubt.
Afirmation. All continuous function is Baire function.
Let $f:Xto mathbb{R}$ continuous function. X compact hausdorff with $mathcal{B} sigma$-algebra of Baire.
Let $bigcap_{n=1}^{infty} O_ninmathcal{B}(mathbb{R})$ compact set. (sigma-algebra of Baire in $mathbb{R}$)
Now, $f^{-1}(bigcap_{n=1}^{infty} O_n)=bigcap_{n=1}^{infty}f^{-1}(O_n)$ compact set, and $f^{-1}(O_n)$ open sets. Therefore, $f^{-1}(bigcap_{n=1}^{infty} O_n)in mathcal{B}(X)$. Therefore $f$ is Baire function.
It is correct?
functional-analysis analysis measure-theory
asked Nov 18 at 16:33
eraldcoil
25119
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A have doubt.
Afirmation. All continuous function is Baire function.
Let $f:Xto mathbb{R}$ continuous function. X compact hausdorff with $mathcal{B} sigma$-algebra of Baire.
Let $bigcap_{n=1}^{infty} O_ninmathcal{B}(mathbb{R})$ compact set. (sigma-algebra of Baire in $mathbb{R}$)
Now, $f^{-1}(bigcap_{n=1}^{infty} O_n)=bigcap_{n=1}^{infty}f^{-1}(O_n)$ compact set, and $f^{-1}(O_n)$ open sets. Therefore, $f^{-1}(bigcap_{n=1}^{infty} O_n)in mathcal{B}(X)$. Therefore $f$ is Baire function.
It is correct?
functional-analysis analysis measure-theory
asked Nov 18 at 16:33
eraldcoil
25119
A have doubt.
Afirmation. All continuous function is Baire function.
Let $f:Xto mathbb{R}$ continuous function. X compact hausdorff with $mathcal{B} sigma$-algebra of Baire.
Let $bigcap_{n=1}^{infty} O_ninmathcal{B}(mathbb{R})$ compact set. (sigma-algebra of Baire in $mathbb{R}$)
Now, $f^{-1}(bigcap_{n=1}^{infty} O_n)=bigcap_{n=1}^{infty}f^{-1}(O_n)$ compact set, and $f^{-1}(O_n)$ open sets. Therefore, $f^{-1}(bigcap_{n=1}^{infty} O_n)in mathcal{B}(X)$. Therefore $f$ is Baire function.
It is correct?
functional-analysis analysis measure-theory
functional-analysis analysis measure-theory
asked Nov 18 at 16:33
eraldcoil
25119
asked Nov 18 at 16:33
eraldcoil
25119
edited Nov 18 at 16:40
asked Nov 18 at 16:33
eraldcoil
25119
asked Nov 18 at 16:33
eraldcoil
25119
asked Nov 18 at 16:33
eraldcoil
25119
25119
add a comment |
add a comment |
1 Answer
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The Baire-$sigma$-algebra is generated by closed $G_delta$-sets. If $bigcap_{i=1}^infty O_i$ is closed with open $O_n$, then also $f^{-1}(bigcap_{i=1}^infty O_i)$ is closed. Moreover $f^{-1}(bigcap_{i=1}^infty O_i) = bigcap_{i=1}^infty f^{-1}(O_i)$. Again using the continuity we see that $f^{-1}(O_i)$ is open. Thus $f^{-1}(bigcap_{i=1}^infty O_i)$ is Baire-measurable.
Since we only need to prove measurability on a generating set-system, we get that $f$ is measurable according to the Baire-$sigma$-algebras.
answered Nov 22 at 15:55
p4sch
3,765216
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The Baire-$sigma$-algebra is generated by closed $G_delta$-sets. If $bigcap_{i=1}^infty O_i$ is closed with open $O_n$, then also $f^{-1}(bigcap_{i=1}^infty O_i)$ is closed. Moreover $f^{-1}(bigcap_{i=1}^infty O_i) = bigcap_{i=1}^infty f^{-1}(O_i)$. Again using the continuity we see that $f^{-1}(O_i)$ is open. Thus $f^{-1}(bigcap_{i=1}^infty O_i)$ is Baire-measurable.
Since we only need to prove measurability on a generating set-system, we get that $f$ is measurable according to the Baire-$sigma$-algebras.
answered Nov 22 at 15:55
p4sch
3,765216
add a comment |
up vote
1
down vote
accepted
The Baire-$sigma$-algebra is generated by closed $G_delta$-sets. If $bigcap_{i=1}^infty O_i$ is closed with open $O_n$, then also $f^{-1}(bigcap_{i=1}^infty O_i)$ is closed. Moreover $f^{-1}(bigcap_{i=1}^infty O_i) = bigcap_{i=1}^infty f^{-1}(O_i)$. Again using the continuity we see that $f^{-1}(O_i)$ is open. Thus $f^{-1}(bigcap_{i=1}^infty O_i)$ is Baire-measurable.
Since we only need to prove measurability on a generating set-system, we get that $f$ is measurable according to the Baire-$sigma$-algebras.
answered Nov 22 at 15:55
p4sch
3,765216
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The Baire-$sigma$-algebra is generated by closed $G_delta$-sets. If $bigcap_{i=1}^infty O_i$ is closed with open $O_n$, then also $f^{-1}(bigcap_{i=1}^infty O_i)$ is closed. Moreover $f^{-1}(bigcap_{i=1}^infty O_i) = bigcap_{i=1}^infty f^{-1}(O_i)$. Again using the continuity we see that $f^{-1}(O_i)$ is open. Thus $f^{-1}(bigcap_{i=1}^infty O_i)$ is Baire-measurable.
Since we only need to prove measurability on a generating set-system, we get that $f$ is measurable according to the Baire-$sigma$-algebras.
answered Nov 22 at 15:55
p4sch
3,765216
The Baire-$sigma$-algebra is generated by closed $G_delta$-sets. If $bigcap_{i=1}^infty O_i$ is closed with open $O_n$, then also $f^{-1}(bigcap_{i=1}^infty O_i)$ is closed. Moreover $f^{-1}(bigcap_{i=1}^infty O_i) = bigcap_{i=1}^infty f^{-1}(O_i)$. Again using the continuity we see that $f^{-1}(O_i)$ is open. Thus $f^{-1}(bigcap_{i=1}^infty O_i)$ is Baire-measurable.
Since we only need to prove measurability on a generating set-system, we get that $f$ is measurable according to the Baire-$sigma$-algebras.
answered Nov 22 at 15:55
p4sch
3,765216
answered Nov 22 at 15:55
p4sch
3,765216
answered Nov 22 at 15:55
p4sch
3,765216
answered Nov 22 at 15:55
p4sch
3,765216
3,765216
add a comment |
add a comment |
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