Doubt about the topological space of Measurable Functions
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My teacher wrote this on the blackboard but I can't see why it's true or not.
The vector space of (Lebesgue) measurable functions with the point-wise topology is a topological vector space which is closed.
I don't really get it, because in every topology, the whole set, is open and closed, so I thought he might have wanted to say complete. I don't know, I'm quite confused.
EDIT: I asked him what he meant, by email, and stated that he might've wanted to say sequentially closed but this was his answer:
Consider the vector space of functions with domain on a measurable set, endowed with the point-wise convergence topology, then the subset of measurable ones is closed.
general-topology functional-analysis measure-theory
asked Nov 23 at 13:51
M. Navarro
516
add a comment |
up vote
2
down vote
favorite
My teacher wrote this on the blackboard but I can't see why it's true or not.
The vector space of (Lebesgue) measurable functions with the point-wise topology is a topological vector space which is closed.
I don't really get it, because in every topology, the whole set, is open and closed, so I thought he might have wanted to say complete. I don't know, I'm quite confused.
EDIT: I asked him what he meant, by email, and stated that he might've wanted to say sequentially closed but this was his answer:
Consider the vector space of functions with domain on a measurable set, endowed with the point-wise convergence topology, then the subset of measurable ones is closed.
general-topology functional-analysis measure-theory
asked Nov 23 at 13:51
M. Navarro
516
4
Maybe he meant that if a sequence of Lebesgue measurable function converges pointwise to some function, then the limiting function is also Lebesgue measurable, i,e; the space is closed under talking limit.
– Tomath
Nov 23 at 13:56
I've edited the question with the answer my teacher gave me.
– M. Navarro
Nov 24 at 9:53
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My teacher wrote this on the blackboard but I can't see why it's true or not.
The vector space of (Lebesgue) measurable functions with the point-wise topology is a topological vector space which is closed.
I don't really get it, because in every topology, the whole set, is open and closed, so I thought he might have wanted to say complete. I don't know, I'm quite confused.
EDIT: I asked him what he meant, by email, and stated that he might've wanted to say sequentially closed but this was his answer:
Consider the vector space of functions with domain on a measurable set, endowed with the point-wise convergence topology, then the subset of measurable ones is closed.
general-topology functional-analysis measure-theory
asked Nov 23 at 13:51
M. Navarro
516
My teacher wrote this on the blackboard but I can't see why it's true or not.
The vector space of (Lebesgue) measurable functions with the point-wise topology is a topological vector space which is closed.
I don't really get it, because in every topology, the whole set, is open and closed, so I thought he might have wanted to say complete. I don't know, I'm quite confused.
EDIT: I asked him what he meant, by email, and stated that he might've wanted to say sequentially closed but this was his answer:
Consider the vector space of functions with domain on a measurable set, endowed with the point-wise convergence topology, then the subset of measurable ones is closed.
general-topology functional-analysis measure-theory
general-topology functional-analysis measure-theory
asked Nov 23 at 13:51
M. Navarro
516
asked Nov 23 at 13:51
M. Navarro
516
edited Nov 24 at 9:52
asked Nov 23 at 13:51
M. Navarro
516
asked Nov 23 at 13:51
M. Navarro
516
asked Nov 23 at 13:51
M. Navarro
516
516
4
Maybe he meant that if a sequence of Lebesgue measurable function converges pointwise to some function, then the limiting function is also Lebesgue measurable, i,e; the space is closed under talking limit.
– Tomath
Nov 23 at 13:56
I've edited the question with the answer my teacher gave me.
– M. Navarro
Nov 24 at 9:53
add a comment |
4
Maybe he meant that if a sequence of Lebesgue measurable function converges pointwise to some function, then the limiting function is also Lebesgue measurable, i,e; the space is closed under talking limit.
– Tomath
Nov 23 at 13:56
I've edited the question with the answer my teacher gave me.
– M. Navarro
Nov 24 at 9:53
4
4
Maybe he meant that if a sequence of Lebesgue measurable function converges pointwise to some function, then the limiting function is also Lebesgue measurable, i,e; the space is closed under talking limit.
– Tomath
Nov 23 at 13:56
Maybe he meant that if a sequence of Lebesgue measurable function converges pointwise to some function, then the limiting function is also Lebesgue measurable, i,e; the space is closed under talking limit.
– Tomath
Nov 23 at 13:56
I've edited the question with the answer my teacher gave me.
– M. Navarro
Nov 24 at 9:53
I've edited the question with the answer my teacher gave me.
– M. Navarro
Nov 24 at 9:53
add a comment |
1 Answer
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2
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The statement is rather controversial, because its most natural interpretation
The space of Lebesgue measurable functions is a closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
is false (provided we believe in non-measurable functions). In point of fact, the subset $S$ of all the functions $f:Bbb RtoBbb R$ such that ${xinBbb R,:, f(x)ne 0}$ has finite cardinality is already dense in that topology.
The second most natural guess would be
The space of Lebesgue measurable functions is a sequentially closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
Which is, of course, true: the pointwise limit of a sequence ${f_n}_{ninBbb N}$ of Lebesgue measurable functions is Lebesgue measurable.
In my opinion, you may want to ask your teacher for clarifications.
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
I'll ask him, and if this is what he really means, I'll accept your answer. Thanks in advance!!
– M. Navarro
Nov 23 at 16:06
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
2
down vote
The statement is rather controversial, because its most natural interpretation
The space of Lebesgue measurable functions is a closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
is false (provided we believe in non-measurable functions). In point of fact, the subset $S$ of all the functions $f:Bbb RtoBbb R$ such that ${xinBbb R,:, f(x)ne 0}$ has finite cardinality is already dense in that topology.
The second most natural guess would be
The space of Lebesgue measurable functions is a sequentially closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
Which is, of course, true: the pointwise limit of a sequence ${f_n}_{ninBbb N}$ of Lebesgue measurable functions is Lebesgue measurable.
In my opinion, you may want to ask your teacher for clarifications.
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
I'll ask him, and if this is what he really means, I'll accept your answer. Thanks in advance!!
– M. Navarro
Nov 23 at 16:06
add a comment |
up vote
2
down vote
The statement is rather controversial, because its most natural interpretation
The space of Lebesgue measurable functions is a closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
is false (provided we believe in non-measurable functions). In point of fact, the subset $S$ of all the functions $f:Bbb RtoBbb R$ such that ${xinBbb R,:, f(x)ne 0}$ has finite cardinality is already dense in that topology.
The second most natural guess would be
The space of Lebesgue measurable functions is a sequentially closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
Which is, of course, true: the pointwise limit of a sequence ${f_n}_{ninBbb N}$ of Lebesgue measurable functions is Lebesgue measurable.
In my opinion, you may want to ask your teacher for clarifications.
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
I'll ask him, and if this is what he really means, I'll accept your answer. Thanks in advance!!
– M. Navarro
Nov 23 at 16:06
add a comment |
up vote
2
down vote
up vote
2
down vote
The statement is rather controversial, because its most natural interpretation
The space of Lebesgue measurable functions is a closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
is false (provided we believe in non-measurable functions). In point of fact, the subset $S$ of all the functions $f:Bbb RtoBbb R$ such that ${xinBbb R,:, f(x)ne 0}$ has finite cardinality is already dense in that topology.
The second most natural guess would be
The space of Lebesgue measurable functions is a sequentially closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
Which is, of course, true: the pointwise limit of a sequence ${f_n}_{ninBbb N}$ of Lebesgue measurable functions is Lebesgue measurable.
In my opinion, you may want to ask your teacher for clarifications.
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
The statement is rather controversial, because its most natural interpretation
The space of Lebesgue measurable functions is a closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
is false (provided we believe in non-measurable functions). In point of fact, the subset $S$ of all the functions $f:Bbb RtoBbb R$ such that ${xinBbb R,:, f(x)ne 0}$ has finite cardinality is already dense in that topology.
The second most natural guess would be
The space of Lebesgue measurable functions is a sequentially closed subspace of $Bbb R^{Bbb R}$ endowed with the product topology.
Which is, of course, true: the pointwise limit of a sequence ${f_n}_{ninBbb N}$ of Lebesgue measurable functions is Lebesgue measurable.
In my opinion, you may want to ask your teacher for clarifications.
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
edited Nov 23 at 16:25
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
answered Nov 23 at 16:02
Saucy O'Path
5,8171626
5,8171626
I'll ask him, and if this is what he really means, I'll accept your answer. Thanks in advance!!
– M. Navarro
Nov 23 at 16:06
add a comment |
I'll ask him, and if this is what he really means, I'll accept your answer. Thanks in advance!!
– M. Navarro
Nov 23 at 16:06
I'll ask him, and if this is what he really means, I'll accept your answer. Thanks in advance!!
– M. Navarro
Nov 23 at 16:06
I'll ask him, and if this is what he really means, I'll accept your answer. Thanks in advance!!
– M. Navarro
Nov 23 at 16:06
add a comment |
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Maybe he meant that if a sequence of Lebesgue measurable function converges pointwise to some function, then the limiting function is also Lebesgue measurable, i,e; the space is closed under talking limit.
– Tomath
Nov 23 at 13:56
I've edited the question with the answer my teacher gave me.
– M. Navarro
Nov 24 at 9:53