Can I define a measurable function using base of a topology?
$begingroup$
We know the more general definition of a measurable funnction: let $(X,mathcal{X})$ and $(Y,tau)$ a measurable and topological space, respectively. The function $f:Xto Y$ is a $mathcal{X}$-measurable function if $f^{-1}(V) in mathcal{X}$, $forall V in tau$.
I would like to know if it is true that we can define the measurability of the function $f$ using a base of the topology $tau$. That is, suposse that $beta$ is a base of $tau$. Then, I wold like to prove that
$$f hbox{ is } mathcal{X}hbox{-measurable function} Leftrightarrow f^{-1}(B) in mathcal{X},quad forall B in beta $$
We know that any $V in tau$ can be written as
$$V = bigcup_{A in mathcal{g}} A,quad mathcal{g} subset beta$$
with $g$ not necessarily enumerable. One implication of my goal is trivial. Any element of $beta$ is open, that is, it is a element of $tau$. Then, if $f$ is $mathcal{X}$-measurable function, then $f^{-1}(B) in mathcal{X}, forall B in beta$. But, because $g in beta$ is not necessarily enumerable, I can not conclude the other implication. That is, I can not conclude that
$$f^{-1}(V) = bigcup_{A in mathcal{g}} f^{-1}(A) in mathcal{X}$$
I would like to know if it is possible to circumvent this or in what cases this result is valid.
general-topology measure-theory
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
$endgroup$
|
show 1 more comment
$begingroup$
We know the more general definition of a measurable funnction: let $(X,mathcal{X})$ and $(Y,tau)$ a measurable and topological space, respectively. The function $f:Xto Y$ is a $mathcal{X}$-measurable function if $f^{-1}(V) in mathcal{X}$, $forall V in tau$.
I would like to know if it is true that we can define the measurability of the function $f$ using a base of the topology $tau$. That is, suposse that $beta$ is a base of $tau$. Then, I wold like to prove that
$$f hbox{ is } mathcal{X}hbox{-measurable function} Leftrightarrow f^{-1}(B) in mathcal{X},quad forall B in beta $$
We know that any $V in tau$ can be written as
$$V = bigcup_{A in mathcal{g}} A,quad mathcal{g} subset beta$$
with $g$ not necessarily enumerable. One implication of my goal is trivial. Any element of $beta$ is open, that is, it is a element of $tau$. Then, if $f$ is $mathcal{X}$-measurable function, then $f^{-1}(B) in mathcal{X}, forall B in beta$. But, because $g in beta$ is not necessarily enumerable, I can not conclude the other implication. That is, I can not conclude that
$$f^{-1}(V) = bigcup_{A in mathcal{g}} f^{-1}(A) in mathcal{X}$$
I would like to know if it is possible to circumvent this or in what cases this result is valid.
general-topology measure-theory
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
$endgroup$
$begingroup$
If the $Y$ is second countable the any open set is a countable union of sets from the base, so you get measurability. In general you cannot conclude that $f$ is measurable.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:31
$begingroup$
Tell me more about this property. Are metric spaces second countable?
$endgroup$
– orrillo
Dec 18 '18 at 5:48
1
$begingroup$
A metric space is second countable iff it is separable (in the sense there is a countable subset which is dense). A standard example of a metric space which is not separable is $ell ^{infty}$ the space of all bounded sequences of real numbers with the metric $d((a_n),(b_n))=sup{|a_n-b_n|:ngeq 1}$.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:52
$begingroup$
$mathbb{R}$ is separable because $mathbb{Q}$ is dense in $mathbb{R}$. Then $mathbb{R}$ has a enumerable base. Can I conclude that ${(q_1,q_2) , q_1 < q_2 , q_i in mathbb{Q}}$ is a base of the topology of $mathbb{R}$?
$endgroup$
– orrillo
Dec 18 '18 at 6:04
1
$begingroup$
All Euclidean spaces are separable (hence second countable) so your definition of measurability works. Intervals with rational end points form a countable base in $mathbb R$, rectangles with rational coordinatesform a countable base in $mathbb R^{2}$, etc.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:08
|
show 1 more comment
$begingroup$
We know the more general definition of a measurable funnction: let $(X,mathcal{X})$ and $(Y,tau)$ a measurable and topological space, respectively. The function $f:Xto Y$ is a $mathcal{X}$-measurable function if $f^{-1}(V) in mathcal{X}$, $forall V in tau$.
I would like to know if it is true that we can define the measurability of the function $f$ using a base of the topology $tau$. That is, suposse that $beta$ is a base of $tau$. Then, I wold like to prove that
$$f hbox{ is } mathcal{X}hbox{-measurable function} Leftrightarrow f^{-1}(B) in mathcal{X},quad forall B in beta $$
We know that any $V in tau$ can be written as
$$V = bigcup_{A in mathcal{g}} A,quad mathcal{g} subset beta$$
with $g$ not necessarily enumerable. One implication of my goal is trivial. Any element of $beta$ is open, that is, it is a element of $tau$. Then, if $f$ is $mathcal{X}$-measurable function, then $f^{-1}(B) in mathcal{X}, forall B in beta$. But, because $g in beta$ is not necessarily enumerable, I can not conclude the other implication. That is, I can not conclude that
$$f^{-1}(V) = bigcup_{A in mathcal{g}} f^{-1}(A) in mathcal{X}$$
I would like to know if it is possible to circumvent this or in what cases this result is valid.
general-topology measure-theory
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
$endgroup$
We know the more general definition of a measurable funnction: let $(X,mathcal{X})$ and $(Y,tau)$ a measurable and topological space, respectively. The function $f:Xto Y$ is a $mathcal{X}$-measurable function if $f^{-1}(V) in mathcal{X}$, $forall V in tau$.
I would like to know if it is true that we can define the measurability of the function $f$ using a base of the topology $tau$. That is, suposse that $beta$ is a base of $tau$. Then, I wold like to prove that
$$f hbox{ is } mathcal{X}hbox{-measurable function} Leftrightarrow f^{-1}(B) in mathcal{X},quad forall B in beta $$
We know that any $V in tau$ can be written as
$$V = bigcup_{A in mathcal{g}} A,quad mathcal{g} subset beta$$
with $g$ not necessarily enumerable. One implication of my goal is trivial. Any element of $beta$ is open, that is, it is a element of $tau$. Then, if $f$ is $mathcal{X}$-measurable function, then $f^{-1}(B) in mathcal{X}, forall B in beta$. But, because $g in beta$ is not necessarily enumerable, I can not conclude the other implication. That is, I can not conclude that
$$f^{-1}(V) = bigcup_{A in mathcal{g}} f^{-1}(A) in mathcal{X}$$
I would like to know if it is possible to circumvent this or in what cases this result is valid.
general-topology measure-theory
general-topology measure-theory
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
asked Dec 18 '18 at 5:20
orrilloorrillo
315111
315111
$begingroup$
If the $Y$ is second countable the any open set is a countable union of sets from the base, so you get measurability. In general you cannot conclude that $f$ is measurable.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:31
$begingroup$
Tell me more about this property. Are metric spaces second countable?
$endgroup$
– orrillo
Dec 18 '18 at 5:48
1
$begingroup$
A metric space is second countable iff it is separable (in the sense there is a countable subset which is dense). A standard example of a metric space which is not separable is $ell ^{infty}$ the space of all bounded sequences of real numbers with the metric $d((a_n),(b_n))=sup{|a_n-b_n|:ngeq 1}$.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:52
$begingroup$
$mathbb{R}$ is separable because $mathbb{Q}$ is dense in $mathbb{R}$. Then $mathbb{R}$ has a enumerable base. Can I conclude that ${(q_1,q_2) , q_1 < q_2 , q_i in mathbb{Q}}$ is a base of the topology of $mathbb{R}$?
$endgroup$
– orrillo
Dec 18 '18 at 6:04
1
$begingroup$
All Euclidean spaces are separable (hence second countable) so your definition of measurability works. Intervals with rational end points form a countable base in $mathbb R$, rectangles with rational coordinatesform a countable base in $mathbb R^{2}$, etc.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:08
|
show 1 more comment
$begingroup$
If the $Y$ is second countable the any open set is a countable union of sets from the base, so you get measurability. In general you cannot conclude that $f$ is measurable.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:31
$begingroup$
Tell me more about this property. Are metric spaces second countable?
$endgroup$
– orrillo
Dec 18 '18 at 5:48
1
$begingroup$
A metric space is second countable iff it is separable (in the sense there is a countable subset which is dense). A standard example of a metric space which is not separable is $ell ^{infty}$ the space of all bounded sequences of real numbers with the metric $d((a_n),(b_n))=sup{|a_n-b_n|:ngeq 1}$.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:52
$begingroup$
$mathbb{R}$ is separable because $mathbb{Q}$ is dense in $mathbb{R}$. Then $mathbb{R}$ has a enumerable base. Can I conclude that ${(q_1,q_2) , q_1 < q_2 , q_i in mathbb{Q}}$ is a base of the topology of $mathbb{R}$?
$endgroup$
– orrillo
Dec 18 '18 at 6:04
1
$begingroup$
All Euclidean spaces are separable (hence second countable) so your definition of measurability works. Intervals with rational end points form a countable base in $mathbb R$, rectangles with rational coordinatesform a countable base in $mathbb R^{2}$, etc.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:08
$begingroup$
If the $Y$ is second countable the any open set is a countable union of sets from the base, so you get measurability. In general you cannot conclude that $f$ is measurable.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:31
$begingroup$
If the $Y$ is second countable the any open set is a countable union of sets from the base, so you get measurability. In general you cannot conclude that $f$ is measurable.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:31
$begingroup$
Tell me more about this property. Are metric spaces second countable?
$endgroup$
– orrillo
Dec 18 '18 at 5:48
$begingroup$
Tell me more about this property. Are metric spaces second countable?
$endgroup$
– orrillo
Dec 18 '18 at 5:48
1
1
$begingroup$
A metric space is second countable iff it is separable (in the sense there is a countable subset which is dense). A standard example of a metric space which is not separable is $ell ^{infty}$ the space of all bounded sequences of real numbers with the metric $d((a_n),(b_n))=sup{|a_n-b_n|:ngeq 1}$.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:52
$begingroup$
A metric space is second countable iff it is separable (in the sense there is a countable subset which is dense). A standard example of a metric space which is not separable is $ell ^{infty}$ the space of all bounded sequences of real numbers with the metric $d((a_n),(b_n))=sup{|a_n-b_n|:ngeq 1}$.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:52
$begingroup$
$mathbb{R}$ is separable because $mathbb{Q}$ is dense in $mathbb{R}$. Then $mathbb{R}$ has a enumerable base. Can I conclude that ${(q_1,q_2) , q_1 < q_2 , q_i in mathbb{Q}}$ is a base of the topology of $mathbb{R}$?
$endgroup$
– orrillo
Dec 18 '18 at 6:04
$begingroup$
$mathbb{R}$ is separable because $mathbb{Q}$ is dense in $mathbb{R}$. Then $mathbb{R}$ has a enumerable base. Can I conclude that ${(q_1,q_2) , q_1 < q_2 , q_i in mathbb{Q}}$ is a base of the topology of $mathbb{R}$?
$endgroup$
– orrillo
Dec 18 '18 at 6:04
1
1
$begingroup$
All Euclidean spaces are separable (hence second countable) so your definition of measurability works. Intervals with rational end points form a countable base in $mathbb R$, rectangles with rational coordinatesform a countable base in $mathbb R^{2}$, etc.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:08
$begingroup$
All Euclidean spaces are separable (hence second countable) so your definition of measurability works. Intervals with rational end points form a countable base in $mathbb R$, rectangles with rational coordinatesform a countable base in $mathbb R^{2}$, etc.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:08
|
show 1 more comment
1 Answer
1
active
oldest
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$begingroup$
This is too long for the comment section.
You are probably after the following result. On any set $mathrm{Z},$ denote by $sigma(mathscr{A})$ the minimal sigma algebra containing the subset $mathscr{A}$ of the power set of $mathrm{Z}$. If $varphi$ is a function with values in $mathrm{Z},$ we also denote by $varphi(mathscr{A})$ the set of $varphi^{-1}(mathrm{A})$ as $mathrm{A}$ runs through $mathscr{A}.$
Proposition. Let $f$ be any function $mathrm{X} to mathrm{Y}$ and let $mathscr{Y}$ be any subset of the power set of $mathrm{Y}.$ Then, $f^{-1}(sigma(mathscr{Y})) = sigma(f^{-1}(mathscr{Y})).$
Proof. You probably already know that the set of preimages of a sigma algebra conforms a sigma algebra, this proves at once the $supset.$ For the other side, consider the set $mathscr{Y}'$ of sets $mathrm{E}$ in $sigma(mathscr{Y})$ such that $f^{-1}(mathrm{E}) in sigma(f^{-1}(mathscr{Y})).$ Straightforwardly, $mathscr{Y}'$ is a sigma field containing $mathscr{Y},$ hence, it also contains $mathscr{Y},$ this proves $subset.$ Q.E.D.
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
$endgroup$
add a comment |
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$begingroup$
This is too long for the comment section.
You are probably after the following result. On any set $mathrm{Z},$ denote by $sigma(mathscr{A})$ the minimal sigma algebra containing the subset $mathscr{A}$ of the power set of $mathrm{Z}$. If $varphi$ is a function with values in $mathrm{Z},$ we also denote by $varphi(mathscr{A})$ the set of $varphi^{-1}(mathrm{A})$ as $mathrm{A}$ runs through $mathscr{A}.$
Proposition. Let $f$ be any function $mathrm{X} to mathrm{Y}$ and let $mathscr{Y}$ be any subset of the power set of $mathrm{Y}.$ Then, $f^{-1}(sigma(mathscr{Y})) = sigma(f^{-1}(mathscr{Y})).$
Proof. You probably already know that the set of preimages of a sigma algebra conforms a sigma algebra, this proves at once the $supset.$ For the other side, consider the set $mathscr{Y}'$ of sets $mathrm{E}$ in $sigma(mathscr{Y})$ such that $f^{-1}(mathrm{E}) in sigma(f^{-1}(mathscr{Y})).$ Straightforwardly, $mathscr{Y}'$ is a sigma field containing $mathscr{Y},$ hence, it also contains $mathscr{Y},$ this proves $subset.$ Q.E.D.
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
$endgroup$
add a comment |
$begingroup$
This is too long for the comment section.
You are probably after the following result. On any set $mathrm{Z},$ denote by $sigma(mathscr{A})$ the minimal sigma algebra containing the subset $mathscr{A}$ of the power set of $mathrm{Z}$. If $varphi$ is a function with values in $mathrm{Z},$ we also denote by $varphi(mathscr{A})$ the set of $varphi^{-1}(mathrm{A})$ as $mathrm{A}$ runs through $mathscr{A}.$
Proposition. Let $f$ be any function $mathrm{X} to mathrm{Y}$ and let $mathscr{Y}$ be any subset of the power set of $mathrm{Y}.$ Then, $f^{-1}(sigma(mathscr{Y})) = sigma(f^{-1}(mathscr{Y})).$
Proof. You probably already know that the set of preimages of a sigma algebra conforms a sigma algebra, this proves at once the $supset.$ For the other side, consider the set $mathscr{Y}'$ of sets $mathrm{E}$ in $sigma(mathscr{Y})$ such that $f^{-1}(mathrm{E}) in sigma(f^{-1}(mathscr{Y})).$ Straightforwardly, $mathscr{Y}'$ is a sigma field containing $mathscr{Y},$ hence, it also contains $mathscr{Y},$ this proves $subset.$ Q.E.D.
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
$endgroup$
add a comment |
$begingroup$
This is too long for the comment section.
You are probably after the following result. On any set $mathrm{Z},$ denote by $sigma(mathscr{A})$ the minimal sigma algebra containing the subset $mathscr{A}$ of the power set of $mathrm{Z}$. If $varphi$ is a function with values in $mathrm{Z},$ we also denote by $varphi(mathscr{A})$ the set of $varphi^{-1}(mathrm{A})$ as $mathrm{A}$ runs through $mathscr{A}.$
Proposition. Let $f$ be any function $mathrm{X} to mathrm{Y}$ and let $mathscr{Y}$ be any subset of the power set of $mathrm{Y}.$ Then, $f^{-1}(sigma(mathscr{Y})) = sigma(f^{-1}(mathscr{Y})).$
Proof. You probably already know that the set of preimages of a sigma algebra conforms a sigma algebra, this proves at once the $supset.$ For the other side, consider the set $mathscr{Y}'$ of sets $mathrm{E}$ in $sigma(mathscr{Y})$ such that $f^{-1}(mathrm{E}) in sigma(f^{-1}(mathscr{Y})).$ Straightforwardly, $mathscr{Y}'$ is a sigma field containing $mathscr{Y},$ hence, it also contains $mathscr{Y},$ this proves $subset.$ Q.E.D.
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
$endgroup$
This is too long for the comment section.
You are probably after the following result. On any set $mathrm{Z},$ denote by $sigma(mathscr{A})$ the minimal sigma algebra containing the subset $mathscr{A}$ of the power set of $mathrm{Z}$. If $varphi$ is a function with values in $mathrm{Z},$ we also denote by $varphi(mathscr{A})$ the set of $varphi^{-1}(mathrm{A})$ as $mathrm{A}$ runs through $mathscr{A}.$
Proposition. Let $f$ be any function $mathrm{X} to mathrm{Y}$ and let $mathscr{Y}$ be any subset of the power set of $mathrm{Y}.$ Then, $f^{-1}(sigma(mathscr{Y})) = sigma(f^{-1}(mathscr{Y})).$
Proof. You probably already know that the set of preimages of a sigma algebra conforms a sigma algebra, this proves at once the $supset.$ For the other side, consider the set $mathscr{Y}'$ of sets $mathrm{E}$ in $sigma(mathscr{Y})$ such that $f^{-1}(mathrm{E}) in sigma(f^{-1}(mathscr{Y})).$ Straightforwardly, $mathscr{Y}'$ is a sigma field containing $mathscr{Y},$ hence, it also contains $mathscr{Y},$ this proves $subset.$ Q.E.D.
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
answered Dec 18 '18 at 7:00
Will M.Will M.
2,865315
2,865315
add a comment |
add a comment |
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$begingroup$
If the $Y$ is second countable the any open set is a countable union of sets from the base, so you get measurability. In general you cannot conclude that $f$ is measurable.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:31
$begingroup$
Tell me more about this property. Are metric spaces second countable?
$endgroup$
– orrillo
Dec 18 '18 at 5:48
1
$begingroup$
A metric space is second countable iff it is separable (in the sense there is a countable subset which is dense). A standard example of a metric space which is not separable is $ell ^{infty}$ the space of all bounded sequences of real numbers with the metric $d((a_n),(b_n))=sup{|a_n-b_n|:ngeq 1}$.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 5:52
$begingroup$
$mathbb{R}$ is separable because $mathbb{Q}$ is dense in $mathbb{R}$. Then $mathbb{R}$ has a enumerable base. Can I conclude that ${(q_1,q_2) , q_1 < q_2 , q_i in mathbb{Q}}$ is a base of the topology of $mathbb{R}$?
$endgroup$
– orrillo
Dec 18 '18 at 6:04
1
$begingroup$
All Euclidean spaces are separable (hence second countable) so your definition of measurability works. Intervals with rational end points form a countable base in $mathbb R$, rectangles with rational coordinatesform a countable base in $mathbb R^{2}$, etc.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 6:08