A question on Borel measurabilityCardinality of Borel sigma algebraProve that the graph of a measurable function is measurable.Cartesian Product of Borel Sets is Borel AgainIs $A$ a Borel set?Is the measure evaluation map measurable?Measurability of the pushforward operator on measuresBorel sets and measurabilityBorel MeasurabilityWeak measurability of a set-valued maptotally disconnected and non Borel set.Classifying measurability of functions for $mathcalL(mathbbR)$Borel class of multivalued mappingBorel set of large measure in product space contains a product set of large measure
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A question on Borel measurability
Cardinality of Borel sigma algebraProve that the graph of a measurable function is measurable.Cartesian Product of Borel Sets is Borel AgainIs $A$ a Borel set?Is the measure evaluation map measurable?Measurability of the pushforward operator on measuresBorel sets and measurabilityBorel MeasurabilityWeak measurability of a set-valued maptotally disconnected and non Borel set.Classifying measurability of functions for $mathcalL(mathbbR)$Borel class of multivalued mappingBorel set of large measure in product space contains a product set of large measure
$begingroup$
Let $X$ be a compact metric space. Given a map $xmapsto E_x$ where $E_xsubset X$ is a Borel set in X. What can be said about the Borel measurability of the set $F_E:=(x,y)in Xtimes X:yin E_x$? What are some (possibly useful) sufficient conditions for $F_E$ to be Borel?
measure-theory borel-sets
asked Mar 26 at 15:35
debdeb
1345
$endgroup$
|
show 1 more comment
$begingroup$
Let $X$ be a compact metric space. Given a map $xmapsto E_x$ where $E_xsubset X$ is a Borel set in X. What can be said about the Borel measurability of the set $F_E:=(x,y)in Xtimes X:yin E_x$? What are some (possibly useful) sufficient conditions for $F_E$ to be Borel?
measure-theory borel-sets
asked Mar 26 at 15:35
debdeb
1345
$endgroup$
$begingroup$
I think $F_E = (x, y) in X times X mid y in E = X times E$ is always Borel in $X times X$ for any topological space $X$ and $E$ Borel in $X$. The reason is that (finite) product of Borel sets is Borel in the product space. A hint of the proof can be found in Cartesian Product of Borel Sets is Borel Again. Feel free to ask for more details.
$endgroup$
– Alex Vong
Mar 26 at 16:42
$begingroup$
I meant $xmapsto E_x$ describes a map $XtomathcalB(X)$, from $X$ to the collection of Borel subsets of X (the sets $E_x$ vary with $xin X$).
$endgroup$
– deb
Mar 26 at 16:49
$begingroup$
I don't understand, we are talking about the Borel measurability of each $E_x$ in $X$ instead of the Borel measurability of the map $x mapsto E_x$, right? If so, then each $E_x$ being Borel in $X$ implies that $F_E = X times E_x$ being Borel in $X times X$, right?
$endgroup$
– Alex Vong
Mar 26 at 16:59
$begingroup$
Yes, the set $Xtimes E_x$ is a Borel set. But the set that I denoted $F_E$ (defined above) is a different set. Are you saying that $Xtimes E_x$ being Borel, implies $F_E$ is Borel? How, if so?
$endgroup$
– deb
Mar 26 at 17:02
$begingroup$
The Borel measurability of $E_x$ is given, yes. It is not clear what you mean by the Borel measurability of the map $xmapsto E_x$ and $F_E$ is not equal to $Xtimes E_x$.
$endgroup$
– deb
Mar 26 at 17:07
|
show 1 more comment
$begingroup$
Let $X$ be a compact metric space. Given a map $xmapsto E_x$ where $E_xsubset X$ is a Borel set in X. What can be said about the Borel measurability of the set $F_E:=(x,y)in Xtimes X:yin E_x$? What are some (possibly useful) sufficient conditions for $F_E$ to be Borel?
measure-theory borel-sets
asked Mar 26 at 15:35
debdeb
1345
$endgroup$
Let $X$ be a compact metric space. Given a map $xmapsto E_x$ where $E_xsubset X$ is a Borel set in X. What can be said about the Borel measurability of the set $F_E:=(x,y)in Xtimes X:yin E_x$? What are some (possibly useful) sufficient conditions for $F_E$ to be Borel?
measure-theory borel-sets
measure-theory borel-sets
asked Mar 26 at 15:35
debdeb
1345
asked Mar 26 at 15:35
debdeb
1345
asked Mar 26 at 15:35
debdeb
1345
asked Mar 26 at 15:35
debdeb
1345
asked Mar 26 at 15:35
debdeb
1345
1345
$begingroup$
I think $F_E = (x, y) in X times X mid y in E = X times E$ is always Borel in $X times X$ for any topological space $X$ and $E$ Borel in $X$. The reason is that (finite) product of Borel sets is Borel in the product space. A hint of the proof can be found in Cartesian Product of Borel Sets is Borel Again. Feel free to ask for more details.
$endgroup$
– Alex Vong
Mar 26 at 16:42
$begingroup$
I meant $xmapsto E_x$ describes a map $XtomathcalB(X)$, from $X$ to the collection of Borel subsets of X (the sets $E_x$ vary with $xin X$).
$endgroup$
– deb
Mar 26 at 16:49
$begingroup$
I don't understand, we are talking about the Borel measurability of each $E_x$ in $X$ instead of the Borel measurability of the map $x mapsto E_x$, right? If so, then each $E_x$ being Borel in $X$ implies that $F_E = X times E_x$ being Borel in $X times X$, right?
$endgroup$
– Alex Vong
Mar 26 at 16:59
$begingroup$
Yes, the set $Xtimes E_x$ is a Borel set. But the set that I denoted $F_E$ (defined above) is a different set. Are you saying that $Xtimes E_x$ being Borel, implies $F_E$ is Borel? How, if so?
$endgroup$
– deb
Mar 26 at 17:02
$begingroup$
The Borel measurability of $E_x$ is given, yes. It is not clear what you mean by the Borel measurability of the map $xmapsto E_x$ and $F_E$ is not equal to $Xtimes E_x$.
$endgroup$
– deb
Mar 26 at 17:07
|
show 1 more comment
$begingroup$
I think $F_E = (x, y) in X times X mid y in E = X times E$ is always Borel in $X times X$ for any topological space $X$ and $E$ Borel in $X$. The reason is that (finite) product of Borel sets is Borel in the product space. A hint of the proof can be found in Cartesian Product of Borel Sets is Borel Again. Feel free to ask for more details.
$endgroup$
– Alex Vong
Mar 26 at 16:42
$begingroup$
I meant $xmapsto E_x$ describes a map $XtomathcalB(X)$, from $X$ to the collection of Borel subsets of X (the sets $E_x$ vary with $xin X$).
$endgroup$
– deb
Mar 26 at 16:49
$begingroup$
I don't understand, we are talking about the Borel measurability of each $E_x$ in $X$ instead of the Borel measurability of the map $x mapsto E_x$, right? If so, then each $E_x$ being Borel in $X$ implies that $F_E = X times E_x$ being Borel in $X times X$, right?
$endgroup$
– Alex Vong
Mar 26 at 16:59
$begingroup$
Yes, the set $Xtimes E_x$ is a Borel set. But the set that I denoted $F_E$ (defined above) is a different set. Are you saying that $Xtimes E_x$ being Borel, implies $F_E$ is Borel? How, if so?
$endgroup$
– deb
Mar 26 at 17:02
$begingroup$
The Borel measurability of $E_x$ is given, yes. It is not clear what you mean by the Borel measurability of the map $xmapsto E_x$ and $F_E$ is not equal to $Xtimes E_x$.
$endgroup$
– deb
Mar 26 at 17:07
$begingroup$
I think $F_E = (x, y) in X times X mid y in E = X times E$ is always Borel in $X times X$ for any topological space $X$ and $E$ Borel in $X$. The reason is that (finite) product of Borel sets is Borel in the product space. A hint of the proof can be found in Cartesian Product of Borel Sets is Borel Again. Feel free to ask for more details.
$endgroup$
– Alex Vong
Mar 26 at 16:42
$begingroup$
I think $F_E = (x, y) in X times X mid y in E = X times E$ is always Borel in $X times X$ for any topological space $X$ and $E$ Borel in $X$. The reason is that (finite) product of Borel sets is Borel in the product space. A hint of the proof can be found in Cartesian Product of Borel Sets is Borel Again. Feel free to ask for more details.
$endgroup$
– Alex Vong
Mar 26 at 16:42
$begingroup$
I meant $xmapsto E_x$ describes a map $XtomathcalB(X)$, from $X$ to the collection of Borel subsets of X (the sets $E_x$ vary with $xin X$).
$endgroup$
– deb
Mar 26 at 16:49
$begingroup$
I meant $xmapsto E_x$ describes a map $XtomathcalB(X)$, from $X$ to the collection of Borel subsets of X (the sets $E_x$ vary with $xin X$).
$endgroup$
– deb
Mar 26 at 16:49
$begingroup$
I don't understand, we are talking about the Borel measurability of each $E_x$ in $X$ instead of the Borel measurability of the map $x mapsto E_x$, right? If so, then each $E_x$ being Borel in $X$ implies that $F_E = X times E_x$ being Borel in $X times X$, right?
$endgroup$
– Alex Vong
Mar 26 at 16:59
$begingroup$
I don't understand, we are talking about the Borel measurability of each $E_x$ in $X$ instead of the Borel measurability of the map $x mapsto E_x$, right? If so, then each $E_x$ being Borel in $X$ implies that $F_E = X times E_x$ being Borel in $X times X$, right?
$endgroup$
– Alex Vong
Mar 26 at 16:59
$begingroup$
Yes, the set $Xtimes E_x$ is a Borel set. But the set that I denoted $F_E$ (defined above) is a different set. Are you saying that $Xtimes E_x$ being Borel, implies $F_E$ is Borel? How, if so?
$endgroup$
– deb
Mar 26 at 17:02
$begingroup$
Yes, the set $Xtimes E_x$ is a Borel set. But the set that I denoted $F_E$ (defined above) is a different set. Are you saying that $Xtimes E_x$ being Borel, implies $F_E$ is Borel? How, if so?
$endgroup$
– deb
Mar 26 at 17:02
$begingroup$
The Borel measurability of $E_x$ is given, yes. It is not clear what you mean by the Borel measurability of the map $xmapsto E_x$ and $F_E$ is not equal to $Xtimes E_x$.
$endgroup$
– deb
Mar 26 at 17:07
$begingroup$
The Borel measurability of $E_x$ is given, yes. It is not clear what you mean by the Borel measurability of the map $xmapsto E_x$ and $F_E$ is not equal to $Xtimes E_x$.
$endgroup$
– deb
Mar 26 at 17:07
|
show 1 more comment
1 Answer
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$begingroup$
Since this is quite open-ended and have not received any answer yet, I'll give it a go. Please correct me if I get something wrong!
It seems there are quite a number of non-examples. Say each $E_x$ is a singleton, then the map $x mapsto E_x$ becomes a function, call it $f$. Observe that $F_E$ is the graph of $f$. Also, each $E_x$ is measurable since singleton is closed in $T_1$ space (in particular metric space).
From this question, any Borel sigma algebra has cardinality at most continuum. In particular, $|B(X times X)| le mathfrakc$
Suppose $X$ has cardinality $|X| ge mathfrakc$ (e.g. $X = [0, 1]$), then there are $|X^X| = |X|^ ge 2^ ge 2^mathfrakc$ many $f: X to X$ functions.
We thus conclude $|X^X| ge 2^mathfrakc > mathfrakc ge |B(X times X)|$ which means that there are more $f: X to X$ functions than Borel sets in $X times X$. Since different functions corresponds to different graphs, there must be a $f: X to X$ function with non-Borel graph (i.e. $F_E$ is not Borel).
In short, $F_E$ can be non-Borel even when each $E_x$ is a singleton! (assuming $|X| ge mathfrakc$)
On the other hand, if $f$ is Borel measurable, then $F_E$ being the graph of $f$ is Borel (I use the hint by saz). But this is probably too restrictive to be useful. (since we have to assume each $E_x$ is a singleton)
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
$endgroup$
add a comment |
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$begingroup$
Since this is quite open-ended and have not received any answer yet, I'll give it a go. Please correct me if I get something wrong!
It seems there are quite a number of non-examples. Say each $E_x$ is a singleton, then the map $x mapsto E_x$ becomes a function, call it $f$. Observe that $F_E$ is the graph of $f$. Also, each $E_x$ is measurable since singleton is closed in $T_1$ space (in particular metric space).
From this question, any Borel sigma algebra has cardinality at most continuum. In particular, $|B(X times X)| le mathfrakc$
Suppose $X$ has cardinality $|X| ge mathfrakc$ (e.g. $X = [0, 1]$), then there are $|X^X| = |X|^ ge 2^ ge 2^mathfrakc$ many $f: X to X$ functions.
We thus conclude $|X^X| ge 2^mathfrakc > mathfrakc ge |B(X times X)|$ which means that there are more $f: X to X$ functions than Borel sets in $X times X$. Since different functions corresponds to different graphs, there must be a $f: X to X$ function with non-Borel graph (i.e. $F_E$ is not Borel).
In short, $F_E$ can be non-Borel even when each $E_x$ is a singleton! (assuming $|X| ge mathfrakc$)
On the other hand, if $f$ is Borel measurable, then $F_E$ being the graph of $f$ is Borel (I use the hint by saz). But this is probably too restrictive to be useful. (since we have to assume each $E_x$ is a singleton)
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
$endgroup$
add a comment |
$begingroup$
Since this is quite open-ended and have not received any answer yet, I'll give it a go. Please correct me if I get something wrong!
It seems there are quite a number of non-examples. Say each $E_x$ is a singleton, then the map $x mapsto E_x$ becomes a function, call it $f$. Observe that $F_E$ is the graph of $f$. Also, each $E_x$ is measurable since singleton is closed in $T_1$ space (in particular metric space).
From this question, any Borel sigma algebra has cardinality at most continuum. In particular, $|B(X times X)| le mathfrakc$
Suppose $X$ has cardinality $|X| ge mathfrakc$ (e.g. $X = [0, 1]$), then there are $|X^X| = |X|^ ge 2^ ge 2^mathfrakc$ many $f: X to X$ functions.
We thus conclude $|X^X| ge 2^mathfrakc > mathfrakc ge |B(X times X)|$ which means that there are more $f: X to X$ functions than Borel sets in $X times X$. Since different functions corresponds to different graphs, there must be a $f: X to X$ function with non-Borel graph (i.e. $F_E$ is not Borel).
In short, $F_E$ can be non-Borel even when each $E_x$ is a singleton! (assuming $|X| ge mathfrakc$)
On the other hand, if $f$ is Borel measurable, then $F_E$ being the graph of $f$ is Borel (I use the hint by saz). But this is probably too restrictive to be useful. (since we have to assume each $E_x$ is a singleton)
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
$endgroup$
add a comment |
$begingroup$
Since this is quite open-ended and have not received any answer yet, I'll give it a go. Please correct me if I get something wrong!
It seems there are quite a number of non-examples. Say each $E_x$ is a singleton, then the map $x mapsto E_x$ becomes a function, call it $f$. Observe that $F_E$ is the graph of $f$. Also, each $E_x$ is measurable since singleton is closed in $T_1$ space (in particular metric space).
From this question, any Borel sigma algebra has cardinality at most continuum. In particular, $|B(X times X)| le mathfrakc$
Suppose $X$ has cardinality $|X| ge mathfrakc$ (e.g. $X = [0, 1]$), then there are $|X^X| = |X|^ ge 2^ ge 2^mathfrakc$ many $f: X to X$ functions.
We thus conclude $|X^X| ge 2^mathfrakc > mathfrakc ge |B(X times X)|$ which means that there are more $f: X to X$ functions than Borel sets in $X times X$. Since different functions corresponds to different graphs, there must be a $f: X to X$ function with non-Borel graph (i.e. $F_E$ is not Borel).
In short, $F_E$ can be non-Borel even when each $E_x$ is a singleton! (assuming $|X| ge mathfrakc$)
On the other hand, if $f$ is Borel measurable, then $F_E$ being the graph of $f$ is Borel (I use the hint by saz). But this is probably too restrictive to be useful. (since we have to assume each $E_x$ is a singleton)
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
$endgroup$
Since this is quite open-ended and have not received any answer yet, I'll give it a go. Please correct me if I get something wrong!
It seems there are quite a number of non-examples. Say each $E_x$ is a singleton, then the map $x mapsto E_x$ becomes a function, call it $f$. Observe that $F_E$ is the graph of $f$. Also, each $E_x$ is measurable since singleton is closed in $T_1$ space (in particular metric space).
From this question, any Borel sigma algebra has cardinality at most continuum. In particular, $|B(X times X)| le mathfrakc$
Suppose $X$ has cardinality $|X| ge mathfrakc$ (e.g. $X = [0, 1]$), then there are $|X^X| = |X|^ ge 2^ ge 2^mathfrakc$ many $f: X to X$ functions.
We thus conclude $|X^X| ge 2^mathfrakc > mathfrakc ge |B(X times X)|$ which means that there are more $f: X to X$ functions than Borel sets in $X times X$. Since different functions corresponds to different graphs, there must be a $f: X to X$ function with non-Borel graph (i.e. $F_E$ is not Borel).
In short, $F_E$ can be non-Borel even when each $E_x$ is a singleton! (assuming $|X| ge mathfrakc$)
On the other hand, if $f$ is Borel measurable, then $F_E$ being the graph of $f$ is Borel (I use the hint by saz). But this is probably too restrictive to be useful. (since we have to assume each $E_x$ is a singleton)
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
edited Mar 28 at 20:09
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
answered Mar 28 at 19:54
Alex VongAlex Vong
1,342819
1,342819
add a comment |
add a comment |
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$begingroup$
I think $F_E = (x, y) in X times X mid y in E = X times E$ is always Borel in $X times X$ for any topological space $X$ and $E$ Borel in $X$. The reason is that (finite) product of Borel sets is Borel in the product space. A hint of the proof can be found in Cartesian Product of Borel Sets is Borel Again. Feel free to ask for more details.
$endgroup$
– Alex Vong
Mar 26 at 16:42
$begingroup$
I meant $xmapsto E_x$ describes a map $XtomathcalB(X)$, from $X$ to the collection of Borel subsets of X (the sets $E_x$ vary with $xin X$).
$endgroup$
– deb
Mar 26 at 16:49
$begingroup$
I don't understand, we are talking about the Borel measurability of each $E_x$ in $X$ instead of the Borel measurability of the map $x mapsto E_x$, right? If so, then each $E_x$ being Borel in $X$ implies that $F_E = X times E_x$ being Borel in $X times X$, right?
$endgroup$
– Alex Vong
Mar 26 at 16:59
$begingroup$
Yes, the set $Xtimes E_x$ is a Borel set. But the set that I denoted $F_E$ (defined above) is a different set. Are you saying that $Xtimes E_x$ being Borel, implies $F_E$ is Borel? How, if so?
$endgroup$
– deb
Mar 26 at 17:02
$begingroup$
The Borel measurability of $E_x$ is given, yes. It is not clear what you mean by the Borel measurability of the map $xmapsto E_x$ and $F_E$ is not equal to $Xtimes E_x$.
$endgroup$
– deb
Mar 26 at 17:07