Must every real function be bounded on a non-negligble set?
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Let $f:mathbb{R} to mathbb{R}$ be any real function. Must there be a "non-negligible" set $S subset mathbb{R}$ such that $f(S)$ is bounded? "Non-negligible" is open to interpretation. We can take it to mean that the closure of $S$ has nonempty interior. Alternatively, we can take it to mean that $S$ has positive measure. A̶l̶t̶e̶r̶n̶a̶t̶i̶v̶e̶l̶y̶,̶ ̶w̶e̶ ̶c̶a̶n̶ ̶s̶i̶m̶p̶l̶y̶ ̶t̶a̶k̶e̶ ̶i̶t̶ ̶t̶o̶ ̶m̶e̶a̶n̶ ̶t̶h̶a̶t̶ ̶$̶S̶$̶ ̶i̶s̶ ̶u̶n̶c̶o̶u̶n̶t̶a̶b̶l̶e̶ (edit: solution to this below). The answerer may choose any of these interpretations (or an alternative one, if he feels it is relevant).
I'm aware of the existence of some pathological functions like the Conway Base $13$ function. This is a function $f:mathbb{R} to mathbb{R}$ with the property that any nonempty open set is mapped to $mathbb{R}$. However, I'm not sure whether or not this is a counterexample to any of the claims above.
We can indeed find an uncountable $S subset mathbb{R}$ with $f(S)$ bounded. Consider that $$mathbb{R} = bigcup_{n in mathbb{Z}} f^{-1}([n, n+1])$$ hence there must exist an $n$ such that $f^{-1}([n, n+1])$ is uncountable. Hoping for a stronger result.
real-analysis
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
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add a comment |
$begingroup$
Let $f:mathbb{R} to mathbb{R}$ be any real function. Must there be a "non-negligible" set $S subset mathbb{R}$ such that $f(S)$ is bounded? "Non-negligible" is open to interpretation. We can take it to mean that the closure of $S$ has nonempty interior. Alternatively, we can take it to mean that $S$ has positive measure. A̶l̶t̶e̶r̶n̶a̶t̶i̶v̶e̶l̶y̶,̶ ̶w̶e̶ ̶c̶a̶n̶ ̶s̶i̶m̶p̶l̶y̶ ̶t̶a̶k̶e̶ ̶i̶t̶ ̶t̶o̶ ̶m̶e̶a̶n̶ ̶t̶h̶a̶t̶ ̶$̶S̶$̶ ̶i̶s̶ ̶u̶n̶c̶o̶u̶n̶t̶a̶b̶l̶e̶ (edit: solution to this below). The answerer may choose any of these interpretations (or an alternative one, if he feels it is relevant).
I'm aware of the existence of some pathological functions like the Conway Base $13$ function. This is a function $f:mathbb{R} to mathbb{R}$ with the property that any nonempty open set is mapped to $mathbb{R}$. However, I'm not sure whether or not this is a counterexample to any of the claims above.
We can indeed find an uncountable $S subset mathbb{R}$ with $f(S)$ bounded. Consider that $$mathbb{R} = bigcup_{n in mathbb{Z}} f^{-1}([n, n+1])$$ hence there must exist an $n$ such that $f^{-1}([n, n+1])$ is uncountable. Hoping for a stronger result.
real-analysis
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
$endgroup$
$begingroup$
I suggest you make up your mind and ask a concrete question.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:10
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I realized that this question is a partial duplicate of this. In that answer, a non-measurable $f$ is constructed which is unbounded on all sets of positive measure.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:03
$begingroup$
It is true, however, that there is an $S$ whose closure has nonempty interior. This follows immediately from this result, which I was actually aware of beforehand, but didn't realize the connection.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:04
add a comment |
$begingroup$
Let $f:mathbb{R} to mathbb{R}$ be any real function. Must there be a "non-negligible" set $S subset mathbb{R}$ such that $f(S)$ is bounded? "Non-negligible" is open to interpretation. We can take it to mean that the closure of $S$ has nonempty interior. Alternatively, we can take it to mean that $S$ has positive measure. A̶l̶t̶e̶r̶n̶a̶t̶i̶v̶e̶l̶y̶,̶ ̶w̶e̶ ̶c̶a̶n̶ ̶s̶i̶m̶p̶l̶y̶ ̶t̶a̶k̶e̶ ̶i̶t̶ ̶t̶o̶ ̶m̶e̶a̶n̶ ̶t̶h̶a̶t̶ ̶$̶S̶$̶ ̶i̶s̶ ̶u̶n̶c̶o̶u̶n̶t̶a̶b̶l̶e̶ (edit: solution to this below). The answerer may choose any of these interpretations (or an alternative one, if he feels it is relevant).
I'm aware of the existence of some pathological functions like the Conway Base $13$ function. This is a function $f:mathbb{R} to mathbb{R}$ with the property that any nonempty open set is mapped to $mathbb{R}$. However, I'm not sure whether or not this is a counterexample to any of the claims above.
We can indeed find an uncountable $S subset mathbb{R}$ with $f(S)$ bounded. Consider that $$mathbb{R} = bigcup_{n in mathbb{Z}} f^{-1}([n, n+1])$$ hence there must exist an $n$ such that $f^{-1}([n, n+1])$ is uncountable. Hoping for a stronger result.
real-analysis
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
$endgroup$
Let $f:mathbb{R} to mathbb{R}$ be any real function. Must there be a "non-negligible" set $S subset mathbb{R}$ such that $f(S)$ is bounded? "Non-negligible" is open to interpretation. We can take it to mean that the closure of $S$ has nonempty interior. Alternatively, we can take it to mean that $S$ has positive measure. A̶l̶t̶e̶r̶n̶a̶t̶i̶v̶e̶l̶y̶,̶ ̶w̶e̶ ̶c̶a̶n̶ ̶s̶i̶m̶p̶l̶y̶ ̶t̶a̶k̶e̶ ̶i̶t̶ ̶t̶o̶ ̶m̶e̶a̶n̶ ̶t̶h̶a̶t̶ ̶$̶S̶$̶ ̶i̶s̶ ̶u̶n̶c̶o̶u̶n̶t̶a̶b̶l̶e̶ (edit: solution to this below). The answerer may choose any of these interpretations (or an alternative one, if he feels it is relevant).
I'm aware of the existence of some pathological functions like the Conway Base $13$ function. This is a function $f:mathbb{R} to mathbb{R}$ with the property that any nonempty open set is mapped to $mathbb{R}$. However, I'm not sure whether or not this is a counterexample to any of the claims above.
We can indeed find an uncountable $S subset mathbb{R}$ with $f(S)$ bounded. Consider that $$mathbb{R} = bigcup_{n in mathbb{Z}} f^{-1}([n, n+1])$$ hence there must exist an $n$ such that $f^{-1}([n, n+1])$ is uncountable. Hoping for a stronger result.
real-analysis
real-analysis
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
edited Dec 29 '18 at 9:09
MathematicsStudent1122
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
asked Dec 29 '18 at 8:44
MathematicsStudent1122MathematicsStudent1122
9,03932669
9,03932669
$begingroup$
I suggest you make up your mind and ask a concrete question.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:10
$begingroup$
I realized that this question is a partial duplicate of this. In that answer, a non-measurable $f$ is constructed which is unbounded on all sets of positive measure.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:03
$begingroup$
It is true, however, that there is an $S$ whose closure has nonempty interior. This follows immediately from this result, which I was actually aware of beforehand, but didn't realize the connection.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:04
add a comment |
$begingroup$
I suggest you make up your mind and ask a concrete question.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:10
$begingroup$
I realized that this question is a partial duplicate of this. In that answer, a non-measurable $f$ is constructed which is unbounded on all sets of positive measure.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:03
$begingroup$
It is true, however, that there is an $S$ whose closure has nonempty interior. This follows immediately from this result, which I was actually aware of beforehand, but didn't realize the connection.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:04
$begingroup$
I suggest you make up your mind and ask a concrete question.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:10
$begingroup$
I suggest you make up your mind and ask a concrete question.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:10
$begingroup$
I realized that this question is a partial duplicate of this. In that answer, a non-measurable $f$ is constructed which is unbounded on all sets of positive measure.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:03
$begingroup$
I realized that this question is a partial duplicate of this. In that answer, a non-measurable $f$ is constructed which is unbounded on all sets of positive measure.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:03
$begingroup$
It is true, however, that there is an $S$ whose closure has nonempty interior. This follows immediately from this result, which I was actually aware of beforehand, but didn't realize the connection.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:04
$begingroup$
It is true, however, that there is an $S$ whose closure has nonempty interior. This follows immediately from this result, which I was actually aware of beforehand, but didn't realize the connection.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:04
add a comment |
1 Answer
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A similar argument to the one you provide shows that if $f$ is measurable, then there must be a positive-measure set on which $f$ is bounded: since $infty=m(mathbb R)=sum m(f^{-1}([n,n+1]))$, $f^{-1}([n,n+1])$ must have positive measure for some $n$.
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
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$begingroup$
A similar argument to the one you provide shows that if $f$ is measurable, then there must be a positive-measure set on which $f$ is bounded: since $infty=m(mathbb R)=sum m(f^{-1}([n,n+1]))$, $f^{-1}([n,n+1])$ must have positive measure for some $n$.
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
$endgroup$
add a comment |
$begingroup$
A similar argument to the one you provide shows that if $f$ is measurable, then there must be a positive-measure set on which $f$ is bounded: since $infty=m(mathbb R)=sum m(f^{-1}([n,n+1]))$, $f^{-1}([n,n+1])$ must have positive measure for some $n$.
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
$endgroup$
add a comment |
$begingroup$
A similar argument to the one you provide shows that if $f$ is measurable, then there must be a positive-measure set on which $f$ is bounded: since $infty=m(mathbb R)=sum m(f^{-1}([n,n+1]))$, $f^{-1}([n,n+1])$ must have positive measure for some $n$.
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
$endgroup$
A similar argument to the one you provide shows that if $f$ is measurable, then there must be a positive-measure set on which $f$ is bounded: since $infty=m(mathbb R)=sum m(f^{-1}([n,n+1]))$, $f^{-1}([n,n+1])$ must have positive measure for some $n$.
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
answered Dec 29 '18 at 9:16
CarmeisterCarmeister
2,8692924
2,8692924
add a comment |
add a comment |
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$begingroup$
I suggest you make up your mind and ask a concrete question.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:10
$begingroup$
I realized that this question is a partial duplicate of this. In that answer, a non-measurable $f$ is constructed which is unbounded on all sets of positive measure.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:03
$begingroup$
It is true, however, that there is an $S$ whose closure has nonempty interior. This follows immediately from this result, which I was actually aware of beforehand, but didn't realize the connection.
$endgroup$
– MathematicsStudent1122
Dec 29 '18 at 10:04