If $X$ and $Y$ are measurable, is $left{X=Yright}$ measurable?
Let $X,Y$ be measurable functions between measurable spaces $(Omega,mathcal A)$ and $(E,mathcal E)$. Is $left{X=Yright}inmathcal A$?
Clearly, if $E$ is a topological vector space and $mathcal E$ is the Borel $sigma$-algebra $mathcal B(E)$ on $E$ for which $left{0right}inmathcal E$ (e.g. $E$ being a $T_1$-space), then $left{X=Yright}=(X-Y)^{-1}(left{0right})inmathcal A$.
measure-theory
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
add a comment |
Let $X,Y$ be measurable functions between measurable spaces $(Omega,mathcal A)$ and $(E,mathcal E)$. Is $left{X=Yright}inmathcal A$?
Clearly, if $E$ is a topological vector space and $mathcal E$ is the Borel $sigma$-algebra $mathcal B(E)$ on $E$ for which $left{0right}inmathcal E$ (e.g. $E$ being a $T_1$-space), then $left{X=Yright}=(X-Y)^{-1}(left{0right})inmathcal A$.
measure-theory
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
What is $X-Y$ for $E$ topological space?
– Ennar
2 days ago
@Ennar Sorry, intended to write topological vector space.
– 0xbadf00d
2 days ago
add a comment |
Let $X,Y$ be measurable functions between measurable spaces $(Omega,mathcal A)$ and $(E,mathcal E)$. Is $left{X=Yright}inmathcal A$?
Clearly, if $E$ is a topological vector space and $mathcal E$ is the Borel $sigma$-algebra $mathcal B(E)$ on $E$ for which $left{0right}inmathcal E$ (e.g. $E$ being a $T_1$-space), then $left{X=Yright}=(X-Y)^{-1}(left{0right})inmathcal A$.
measure-theory
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
Let $X,Y$ be measurable functions between measurable spaces $(Omega,mathcal A)$ and $(E,mathcal E)$. Is $left{X=Yright}inmathcal A$?
Clearly, if $E$ is a topological vector space and $mathcal E$ is the Borel $sigma$-algebra $mathcal B(E)$ on $E$ for which $left{0right}inmathcal E$ (e.g. $E$ being a $T_1$-space), then $left{X=Yright}=(X-Y)^{-1}(left{0right})inmathcal A$.
measure-theory
measure-theory
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
edited 2 days ago
0xbadf00d
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
asked 2 days ago
0xbadf00d0xbadf00d
1,77441430
1,77441430
What is $X-Y$ for $E$ topological space?
– Ennar
2 days ago
@Ennar Sorry, intended to write topological vector space.
– 0xbadf00d
2 days ago
add a comment |
What is $X-Y$ for $E$ topological space?
– Ennar
2 days ago
@Ennar Sorry, intended to write topological vector space.
– 0xbadf00d
2 days ago
What is $X-Y$ for $E$ topological space?
– Ennar
2 days ago
What is $X-Y$ for $E$ topological space?
– Ennar
2 days ago
@Ennar Sorry, intended to write topological vector space.
– 0xbadf00d
2 days ago
@Ennar Sorry, intended to write topological vector space.
– 0xbadf00d
2 days ago
add a comment |
1 Answer
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In your example, you took $E$ a topological vector space.
Take $Omega = E times E$, with the corresponding identity on $sigma$-algebras, $X$ and $Y$ be the coordinate projections, the question proves to be equivalent to « is the diagonal of $E times E$ measurable? »
Take $E={0,1,2}$, and $mathcal{E}=sigma({0},{1,2})$, you can check that the answer is no.
answered 2 days ago
MindlackMindlack
2,06217
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
In your example, you took $E$ a topological vector space.
Take $Omega = E times E$, with the corresponding identity on $sigma$-algebras, $X$ and $Y$ be the coordinate projections, the question proves to be equivalent to « is the diagonal of $E times E$ measurable? »
Take $E={0,1,2}$, and $mathcal{E}=sigma({0},{1,2})$, you can check that the answer is no.
answered 2 days ago
MindlackMindlack
2,06217
add a comment |
In your example, you took $E$ a topological vector space.
Take $Omega = E times E$, with the corresponding identity on $sigma$-algebras, $X$ and $Y$ be the coordinate projections, the question proves to be equivalent to « is the diagonal of $E times E$ measurable? »
Take $E={0,1,2}$, and $mathcal{E}=sigma({0},{1,2})$, you can check that the answer is no.
answered 2 days ago
MindlackMindlack
2,06217
add a comment |
In your example, you took $E$ a topological vector space.
Take $Omega = E times E$, with the corresponding identity on $sigma$-algebras, $X$ and $Y$ be the coordinate projections, the question proves to be equivalent to « is the diagonal of $E times E$ measurable? »
Take $E={0,1,2}$, and $mathcal{E}=sigma({0},{1,2})$, you can check that the answer is no.
answered 2 days ago
MindlackMindlack
2,06217
In your example, you took $E$ a topological vector space.
Take $Omega = E times E$, with the corresponding identity on $sigma$-algebras, $X$ and $Y$ be the coordinate projections, the question proves to be equivalent to « is the diagonal of $E times E$ measurable? »
Take $E={0,1,2}$, and $mathcal{E}=sigma({0},{1,2})$, you can check that the answer is no.
answered 2 days ago
MindlackMindlack
2,06217
edited 2 days ago
answered 2 days ago
MindlackMindlack
2,06217
answered 2 days ago
MindlackMindlack
2,06217
answered 2 days ago
MindlackMindlack
2,06217
2,06217
add a comment |
add a comment |
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What is $X-Y$ for $E$ topological space?
– Ennar
2 days ago
@Ennar Sorry, intended to write topological vector space.
– 0xbadf00d
2 days ago