Outer measure and set with full measure
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I've seen two different definitions for an outer measure $mu$ on $mathcal{P}(X)$, where $X$ is a set, obtained from a given probability measure on $X$.
D1 = a set of full outer measure is a subset $Asubseteq X$ such that $mu(A)=1$.
D2 = a set of full outer measure is a subset $A$ such that $mu(Xbackslash A)=0$.
Since outer measures are not additive, I'm having trouble seeing how those two definitions are equivalent. Clearly D2 implies that a set with full outer measure is measurable (for the sigma algebra generated by $mu$), however it's not clear to me that it's the case for D1.
So my question is : are those definitions equivalent ?
My follow up question is : when we say that $C([0,1])$ has full outer measure for the Wiener measure on $mathbb{R}^{[0,1]}$, are we using Definition D2, or D1 ?
measure-theory
asked Nov 28 at 0:03
Phil-W
28417
add a comment |
up vote
1
down vote
favorite
I've seen two different definitions for an outer measure $mu$ on $mathcal{P}(X)$, where $X$ is a set, obtained from a given probability measure on $X$.
D1 = a set of full outer measure is a subset $Asubseteq X$ such that $mu(A)=1$.
D2 = a set of full outer measure is a subset $A$ such that $mu(Xbackslash A)=0$.
Since outer measures are not additive, I'm having trouble seeing how those two definitions are equivalent. Clearly D2 implies that a set with full outer measure is measurable (for the sigma algebra generated by $mu$), however it's not clear to me that it's the case for D1.
So my question is : are those definitions equivalent ?
My follow up question is : when we say that $C([0,1])$ has full outer measure for the Wiener measure on $mathbb{R}^{[0,1]}$, are we using Definition D2, or D1 ?
measure-theory
asked Nov 28 at 0:03
Phil-W
28417
It appears to be a probability measure, so it will.
– dbx
Nov 28 at 0:10
didn't read it, thanks
– Robson
Nov 28 at 0:10
I suppose that $mu$ is constructed using the usual method (method I here en.wikipedia.org/wiki/Outer_measure) from a given probability measure. I don't know if that implies that $mu(X)=1$ however.
– Phil-W
Nov 28 at 0:19
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've seen two different definitions for an outer measure $mu$ on $mathcal{P}(X)$, where $X$ is a set, obtained from a given probability measure on $X$.
D1 = a set of full outer measure is a subset $Asubseteq X$ such that $mu(A)=1$.
D2 = a set of full outer measure is a subset $A$ such that $mu(Xbackslash A)=0$.
Since outer measures are not additive, I'm having trouble seeing how those two definitions are equivalent. Clearly D2 implies that a set with full outer measure is measurable (for the sigma algebra generated by $mu$), however it's not clear to me that it's the case for D1.
So my question is : are those definitions equivalent ?
My follow up question is : when we say that $C([0,1])$ has full outer measure for the Wiener measure on $mathbb{R}^{[0,1]}$, are we using Definition D2, or D1 ?
measure-theory
asked Nov 28 at 0:03
Phil-W
28417
I've seen two different definitions for an outer measure $mu$ on $mathcal{P}(X)$, where $X$ is a set, obtained from a given probability measure on $X$.
D1 = a set of full outer measure is a subset $Asubseteq X$ such that $mu(A)=1$.
D2 = a set of full outer measure is a subset $A$ such that $mu(Xbackslash A)=0$.
Since outer measures are not additive, I'm having trouble seeing how those two definitions are equivalent. Clearly D2 implies that a set with full outer measure is measurable (for the sigma algebra generated by $mu$), however it's not clear to me that it's the case for D1.
So my question is : are those definitions equivalent ?
My follow up question is : when we say that $C([0,1])$ has full outer measure for the Wiener measure on $mathbb{R}^{[0,1]}$, are we using Definition D2, or D1 ?
measure-theory
measure-theory
asked Nov 28 at 0:03
Phil-W
28417
asked Nov 28 at 0:03
Phil-W
28417
asked Nov 28 at 0:03
Phil-W
28417
asked Nov 28 at 0:03
Phil-W
28417
asked Nov 28 at 0:03
Phil-W
28417
28417
It appears to be a probability measure, so it will.
– dbx
Nov 28 at 0:10
didn't read it, thanks
– Robson
Nov 28 at 0:10
I suppose that $mu$ is constructed using the usual method (method I here en.wikipedia.org/wiki/Outer_measure) from a given probability measure. I don't know if that implies that $mu(X)=1$ however.
– Phil-W
Nov 28 at 0:19
add a comment |
It appears to be a probability measure, so it will.
– dbx
Nov 28 at 0:10
didn't read it, thanks
– Robson
Nov 28 at 0:10
I suppose that $mu$ is constructed using the usual method (method I here en.wikipedia.org/wiki/Outer_measure) from a given probability measure. I don't know if that implies that $mu(X)=1$ however.
– Phil-W
Nov 28 at 0:19
It appears to be a probability measure, so it will.
– dbx
Nov 28 at 0:10
It appears to be a probability measure, so it will.
– dbx
Nov 28 at 0:10
didn't read it, thanks
– Robson
Nov 28 at 0:10
didn't read it, thanks
– Robson
Nov 28 at 0:10
I suppose that $mu$ is constructed using the usual method (method I here en.wikipedia.org/wiki/Outer_measure) from a given probability measure. I don't know if that implies that $mu(X)=1$ however.
– Phil-W
Nov 28 at 0:19
I suppose that $mu$ is constructed using the usual method (method I here en.wikipedia.org/wiki/Outer_measure) from a given probability measure. I don't know if that implies that $mu(X)=1$ however.
– Phil-W
Nov 28 at 0:19
add a comment |
1 Answer
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votes
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0
down vote
accepted
D1 and D2 are not equivalent in general.
Consider $X=[0,1]$ and $mu=$ Lebesgue outer measure
By this thread, there is a non-measurable set $V$ with $mu(V)=1$, so $V$ is a set of full measure in the sense of D1. However, as you pointed out, any set of full measure in the sense of D2 must be measurable, so $V$ is not such a set.
I do not have knowledge on Wiener measure so I leave it to someone who know it well.
answered Nov 28 at 13:31
RunningMeatball
435
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
D1 and D2 are not equivalent in general.
Consider $X=[0,1]$ and $mu=$ Lebesgue outer measure
By this thread, there is a non-measurable set $V$ with $mu(V)=1$, so $V$ is a set of full measure in the sense of D1. However, as you pointed out, any set of full measure in the sense of D2 must be measurable, so $V$ is not such a set.
I do not have knowledge on Wiener measure so I leave it to someone who know it well.
answered Nov 28 at 13:31
RunningMeatball
435
add a comment |
up vote
0
down vote
accepted
D1 and D2 are not equivalent in general.
Consider $X=[0,1]$ and $mu=$ Lebesgue outer measure
By this thread, there is a non-measurable set $V$ with $mu(V)=1$, so $V$ is a set of full measure in the sense of D1. However, as you pointed out, any set of full measure in the sense of D2 must be measurable, so $V$ is not such a set.
I do not have knowledge on Wiener measure so I leave it to someone who know it well.
answered Nov 28 at 13:31
RunningMeatball
435
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
D1 and D2 are not equivalent in general.
Consider $X=[0,1]$ and $mu=$ Lebesgue outer measure
By this thread, there is a non-measurable set $V$ with $mu(V)=1$, so $V$ is a set of full measure in the sense of D1. However, as you pointed out, any set of full measure in the sense of D2 must be measurable, so $V$ is not such a set.
I do not have knowledge on Wiener measure so I leave it to someone who know it well.
answered Nov 28 at 13:31
RunningMeatball
435
D1 and D2 are not equivalent in general.
Consider $X=[0,1]$ and $mu=$ Lebesgue outer measure
By this thread, there is a non-measurable set $V$ with $mu(V)=1$, so $V$ is a set of full measure in the sense of D1. However, as you pointed out, any set of full measure in the sense of D2 must be measurable, so $V$ is not such a set.
I do not have knowledge on Wiener measure so I leave it to someone who know it well.
answered Nov 28 at 13:31
RunningMeatball
435
answered Nov 28 at 13:31
RunningMeatball
435
answered Nov 28 at 13:31
RunningMeatball
435
answered Nov 28 at 13:31
RunningMeatball
435
435
add a comment |
add a comment |
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It appears to be a probability measure, so it will.
– dbx
Nov 28 at 0:10
didn't read it, thanks
– Robson
Nov 28 at 0:10
I suppose that $mu$ is constructed using the usual method (method I here en.wikipedia.org/wiki/Outer_measure) from a given probability measure. I don't know if that implies that $mu(X)=1$ however.
– Phil-W
Nov 28 at 0:19