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Is the maximal set of injectivity of a measurable map a measurable set?
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Consider a measurable function $f colon X to Y$, where $X$ and $Y$ are measurable spaces. The following definition makes sense even with no measurable structure, but the main question does not.
Consider the sets $A$ that satisfy the following condition. The sets those where the function $f$ is injective with respect to the entire domain $X$:
$$
forall x_a in A ; forall x in X text we have f(x_a) = f(x) implies x= x_a.
$$
In particular, this is a stronger condition than only requiring $f|_A$ to be injective, as $f$ may not take the values it takes in $A$ outside of $A$, either.
Call such a set a good set. Based on blackboard sketches it seems that an equivalent definition for a set to be good is that, for all subsets $S subseteq A$, we have $f^-1(f(S)) = S$.
Unions of good sets are still good sets, so there is a maximal good set.
Question: Is the maximal good set measurable? What about if $X = Y = mathbbR$ (with Borel or Lebesgue measure, for example)? Blackboard sketching suggests that there is some hope for continuous real-valued functions.
Also, if this kind of set of injectivity has a standard name or work done on it, then I would be happy to know of it.
measure-theory functions measurable-functions
asked Jul 31 at 14:59
Tommi Brander
900822
add a comment |Â
up vote
2
down vote
favorite
Consider a measurable function $f colon X to Y$, where $X$ and $Y$ are measurable spaces. The following definition makes sense even with no measurable structure, but the main question does not.
Consider the sets $A$ that satisfy the following condition. The sets those where the function $f$ is injective with respect to the entire domain $X$:
$$
forall x_a in A ; forall x in X text we have f(x_a) = f(x) implies x= x_a.
$$
In particular, this is a stronger condition than only requiring $f|_A$ to be injective, as $f$ may not take the values it takes in $A$ outside of $A$, either.
Call such a set a good set. Based on blackboard sketches it seems that an equivalent definition for a set to be good is that, for all subsets $S subseteq A$, we have $f^-1(f(S)) = S$.
Unions of good sets are still good sets, so there is a maximal good set.
Question: Is the maximal good set measurable? What about if $X = Y = mathbbR$ (with Borel or Lebesgue measure, for example)? Blackboard sketching suggests that there is some hope for continuous real-valued functions.
Also, if this kind of set of injectivity has a standard name or work done on it, then I would be happy to know of it.
measure-theory functions measurable-functions
asked Jul 31 at 14:59
Tommi Brander
900822
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider a measurable function $f colon X to Y$, where $X$ and $Y$ are measurable spaces. The following definition makes sense even with no measurable structure, but the main question does not.
Consider the sets $A$ that satisfy the following condition. The sets those where the function $f$ is injective with respect to the entire domain $X$:
$$
forall x_a in A ; forall x in X text we have f(x_a) = f(x) implies x= x_a.
$$
In particular, this is a stronger condition than only requiring $f|_A$ to be injective, as $f$ may not take the values it takes in $A$ outside of $A$, either.
Call such a set a good set. Based on blackboard sketches it seems that an equivalent definition for a set to be good is that, for all subsets $S subseteq A$, we have $f^-1(f(S)) = S$.
Unions of good sets are still good sets, so there is a maximal good set.
Question: Is the maximal good set measurable? What about if $X = Y = mathbbR$ (with Borel or Lebesgue measure, for example)? Blackboard sketching suggests that there is some hope for continuous real-valued functions.
Also, if this kind of set of injectivity has a standard name or work done on it, then I would be happy to know of it.
measure-theory functions measurable-functions
asked Jul 31 at 14:59
Tommi Brander
900822
Consider a measurable function $f colon X to Y$, where $X$ and $Y$ are measurable spaces. The following definition makes sense even with no measurable structure, but the main question does not.
Consider the sets $A$ that satisfy the following condition. The sets those where the function $f$ is injective with respect to the entire domain $X$:
$$
forall x_a in A ; forall x in X text we have f(x_a) = f(x) implies x= x_a.
$$
In particular, this is a stronger condition than only requiring $f|_A$ to be injective, as $f$ may not take the values it takes in $A$ outside of $A$, either.
Call such a set a good set. Based on blackboard sketches it seems that an equivalent definition for a set to be good is that, for all subsets $S subseteq A$, we have $f^-1(f(S)) = S$.
Unions of good sets are still good sets, so there is a maximal good set.
Question: Is the maximal good set measurable? What about if $X = Y = mathbbR$ (with Borel or Lebesgue measure, for example)? Blackboard sketching suggests that there is some hope for continuous real-valued functions.
Also, if this kind of set of injectivity has a standard name or work done on it, then I would be happy to know of it.
measure-theory functions measurable-functions
asked Jul 31 at 14:59
Tommi Brander
900822
asked Jul 31 at 14:59
Tommi Brander
900822
asked Jul 31 at 14:59
Tommi Brander
900822
asked Jul 31 at 14:59
Tommi Brander
900822
900822
add a comment |Â
add a comment |Â
1 Answer
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The maximal good set certainly isn't measurable in general. For instance, if the $sigma$-algebra on $Y$ is $emptyset,Y$, then any function is measurable, regardless of the $sigma$-algebra on $X$.
I don't know what happens for the nice case of Borel functions on $mathbbR$. The maximal good set is always coanalytic, since it is the complement of the projection of the Borel set $(x,y):f(x)=f(y),xneq ysubsetmathbbR^2$ (onto either coordinate).
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The maximal good set certainly isn't measurable in general. For instance, if the $sigma$-algebra on $Y$ is $emptyset,Y$, then any function is measurable, regardless of the $sigma$-algebra on $X$.
I don't know what happens for the nice case of Borel functions on $mathbbR$. The maximal good set is always coanalytic, since it is the complement of the projection of the Borel set $(x,y):f(x)=f(y),xneq ysubsetmathbbR^2$ (onto either coordinate).
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
add a comment |Â
up vote
1
down vote
The maximal good set certainly isn't measurable in general. For instance, if the $sigma$-algebra on $Y$ is $emptyset,Y$, then any function is measurable, regardless of the $sigma$-algebra on $X$.
I don't know what happens for the nice case of Borel functions on $mathbbR$. The maximal good set is always coanalytic, since it is the complement of the projection of the Borel set $(x,y):f(x)=f(y),xneq ysubsetmathbbR^2$ (onto either coordinate).
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The maximal good set certainly isn't measurable in general. For instance, if the $sigma$-algebra on $Y$ is $emptyset,Y$, then any function is measurable, regardless of the $sigma$-algebra on $X$.
I don't know what happens for the nice case of Borel functions on $mathbbR$. The maximal good set is always coanalytic, since it is the complement of the projection of the Borel set $(x,y):f(x)=f(y),xneq ysubsetmathbbR^2$ (onto either coordinate).
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
The maximal good set certainly isn't measurable in general. For instance, if the $sigma$-algebra on $Y$ is $emptyset,Y$, then any function is measurable, regardless of the $sigma$-algebra on $X$.
I don't know what happens for the nice case of Borel functions on $mathbbR$. The maximal good set is always coanalytic, since it is the complement of the projection of the Borel set $(x,y):f(x)=f(y),xneq ysubsetmathbbR^2$ (onto either coordinate).
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
answered Jul 31 at 18:21
Eric Wofsey
161k12188297
161k12188297
add a comment |Â
add a comment |Â
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