Is a simply connected subset of $mathbb{R}^n$ necessarily measurable?
$begingroup$
Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.
Is $S$ necessarily Lebesgue measurable?
Is it necessarily Borel measurable?
What's the proof?
(Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)
general-topology measure-theory lebesgue-measure
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
$endgroup$
add a comment |
$begingroup$
Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.
Is $S$ necessarily Lebesgue measurable?
Is it necessarily Borel measurable?
What's the proof?
(Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)
general-topology measure-theory lebesgue-measure
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
$endgroup$
add a comment |
$begingroup$
Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.
Is $S$ necessarily Lebesgue measurable?
Is it necessarily Borel measurable?
What's the proof?
(Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)
general-topology measure-theory lebesgue-measure
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
$endgroup$
Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.
Is $S$ necessarily Lebesgue measurable?
Is it necessarily Borel measurable?
What's the proof?
(Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)
general-topology measure-theory lebesgue-measure
general-topology measure-theory lebesgue-measure
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
asked Jan 16 at 8:09
Ben Blum-SmithBen Blum-Smith
10.1k23086
10.1k23086
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.
To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.
To see that this set is simply connected is best to just draw it.
edited Jan 16 at 23:10
DanielWainfleet
35k31648
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
$endgroup$
2
$begingroup$
Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
$endgroup$
– stressed out
Jan 16 at 8:24
3
$begingroup$
the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
$endgroup$
– postmortes
Jan 16 at 9:33
2
$begingroup$
My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
$endgroup$
– DanielWainfleet
Jan 16 at 23:11
1
$begingroup$
@stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:02
$begingroup$
+1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:07
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.
To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.
To see that this set is simply connected is best to just draw it.
edited Jan 16 at 23:10
DanielWainfleet
35k31648
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
$endgroup$
2
$begingroup$
Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
$endgroup$
– stressed out
Jan 16 at 8:24
3
$begingroup$
the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
$endgroup$
– postmortes
Jan 16 at 9:33
2
$begingroup$
My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
$endgroup$
– DanielWainfleet
Jan 16 at 23:11
1
$begingroup$
@stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:02
$begingroup$
+1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:07
|
show 1 more comment
$begingroup$
Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.
To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.
To see that this set is simply connected is best to just draw it.
edited Jan 16 at 23:10
DanielWainfleet
35k31648
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
$endgroup$
2
$begingroup$
Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
$endgroup$
– stressed out
Jan 16 at 8:24
3
$begingroup$
the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
$endgroup$
– postmortes
Jan 16 at 9:33
2
$begingroup$
My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
$endgroup$
– DanielWainfleet
Jan 16 at 23:11
1
$begingroup$
@stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:02
$begingroup$
+1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:07
|
show 1 more comment
$begingroup$
Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.
To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.
To see that this set is simply connected is best to just draw it.
edited Jan 16 at 23:10
DanielWainfleet
35k31648
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
$endgroup$
Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.
To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.
To see that this set is simply connected is best to just draw it.
edited Jan 16 at 23:10
DanielWainfleet
35k31648
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
edited Jan 16 at 23:10
DanielWainfleet
35k31648
edited Jan 16 at 23:10
DanielWainfleet
35k31648
edited Jan 16 at 23:10
DanielWainfleet
35k31648
35k31648
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
answered Jan 16 at 8:18
Alessandro CodenottiAlessandro Codenotti
3,81311539
3,81311539
2
$begingroup$
Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
$endgroup$
– stressed out
Jan 16 at 8:24
3
$begingroup$
the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
$endgroup$
– postmortes
Jan 16 at 9:33
2
$begingroup$
My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
$endgroup$
– DanielWainfleet
Jan 16 at 23:11
1
$begingroup$
@stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:02
$begingroup$
+1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:07
|
show 1 more comment
2
$begingroup$
Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
$endgroup$
– stressed out
Jan 16 at 8:24
3
$begingroup$
the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
$endgroup$
– postmortes
Jan 16 at 9:33
2
$begingroup$
My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
$endgroup$
– DanielWainfleet
Jan 16 at 23:11
1
$begingroup$
@stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:02
$begingroup$
+1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:07
2
2
$begingroup$
Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
$endgroup$
– stressed out
Jan 16 at 8:24
$begingroup$
Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
$endgroup$
– stressed out
Jan 16 at 8:24
3
3
$begingroup$
the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
$endgroup$
– postmortes
Jan 16 at 9:33
$begingroup$
the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
$endgroup$
– postmortes
Jan 16 at 9:33
2
2
$begingroup$
My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
$endgroup$
– DanielWainfleet
Jan 16 at 23:11
$begingroup$
My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
$endgroup$
– DanielWainfleet
Jan 16 at 23:11
1
1
$begingroup$
@stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:02
$begingroup$
@stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:02
$begingroup$
+1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:07
$begingroup$
+1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
$endgroup$
– Ben Blum-Smith
Jan 17 at 5:07
|
show 1 more comment
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