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Functions with non-zero derivative sending sets of positive measure to sets of positive measure
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For the purposes of this question, assume that when "measure" is mentioned, we refer to the Lebesgue measure.
Suppose I have a function $f : mathbbR to mathbbR$ such that $fracdfdx$ is non-zero at every point. Is it then the case that for any set of positive measure $X$, $f(X)$ also has positive measure? If not, what precondition is required for this to be the case.
I'm not sure if it makes a difference, but I would be satisfied with an answer for the case: $f : [a,b] rightarrow mathbbR$
real-analysis measure-theory lebesgue-measure
edited Jul 30 at 3:04
Xander Henderson
13.1k83150
asked Jul 30 at 2:47
Drathora
62
add a comment |Â
up vote
1
down vote
favorite
For the purposes of this question, assume that when "measure" is mentioned, we refer to the Lebesgue measure.
Suppose I have a function $f : mathbbR to mathbbR$ such that $fracdfdx$ is non-zero at every point. Is it then the case that for any set of positive measure $X$, $f(X)$ also has positive measure? If not, what precondition is required for this to be the case.
I'm not sure if it makes a difference, but I would be satisfied with an answer for the case: $f : [a,b] rightarrow mathbbR$
real-analysis measure-theory lebesgue-measure
edited Jul 30 at 3:04
Xander Henderson
13.1k83150
asked Jul 30 at 2:47
Drathora
62
There are a couple of little technicalities that I am not sure of, but I think that the basic argument should follow from Darboux. It follows from Darboux's theorem that we may assume that $f'$ is positive (or negative, resp.) on $Bbb R$, from which it follows that $f$ is increasing (or decreasing, resp.). But then $f(a) < f(b)$ (or $f(a) > f(b)$, resp.) for any $a<b$, from which we can conclude that $mu(f([a,b])) = |f(b)-f(a)|$. So non-degenerate intervals are sent to non-degenerate intervals. I think that you ought to be able to finish the argument via some kind of appeal to regularity.
– Xander Henderson
Jul 30 at 3:17
This argument does assumes, however, that $f'(x)$ exists for all $x$ (not just almost all $x$), but this seems to be implied by the hypothesis that $f'(x) ne 0$ for all $x$.
– Xander Henderson
Jul 30 at 3:18
I'm with you up until the end. The case that is unclear to me are when the positive-measure set is something that contains no positive length intervals. For example, the intersection of [a,b] with the irrationals. I assume that this is where the regularity comes into play, but I'm not familiar enough with the notion to know for sure. Is there any chance you could shed some light on this? Thanks
– Drathora
Jul 30 at 3:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For the purposes of this question, assume that when "measure" is mentioned, we refer to the Lebesgue measure.
Suppose I have a function $f : mathbbR to mathbbR$ such that $fracdfdx$ is non-zero at every point. Is it then the case that for any set of positive measure $X$, $f(X)$ also has positive measure? If not, what precondition is required for this to be the case.
I'm not sure if it makes a difference, but I would be satisfied with an answer for the case: $f : [a,b] rightarrow mathbbR$
real-analysis measure-theory lebesgue-measure
edited Jul 30 at 3:04
Xander Henderson
13.1k83150
asked Jul 30 at 2:47
Drathora
62
For the purposes of this question, assume that when "measure" is mentioned, we refer to the Lebesgue measure.
Suppose I have a function $f : mathbbR to mathbbR$ such that $fracdfdx$ is non-zero at every point. Is it then the case that for any set of positive measure $X$, $f(X)$ also has positive measure? If not, what precondition is required for this to be the case.
I'm not sure if it makes a difference, but I would be satisfied with an answer for the case: $f : [a,b] rightarrow mathbbR$
real-analysis measure-theory lebesgue-measure
edited Jul 30 at 3:04
Xander Henderson
13.1k83150
asked Jul 30 at 2:47
Drathora
62
edited Jul 30 at 3:04
Xander Henderson
13.1k83150
edited Jul 30 at 3:04
Xander Henderson
13.1k83150
edited Jul 30 at 3:04
Xander Henderson
13.1k83150
13.1k83150
asked Jul 30 at 2:47
Drathora
62
asked Jul 30 at 2:47
Drathora
62
asked Jul 30 at 2:47
Drathora
62
62
There are a couple of little technicalities that I am not sure of, but I think that the basic argument should follow from Darboux. It follows from Darboux's theorem that we may assume that $f'$ is positive (or negative, resp.) on $Bbb R$, from which it follows that $f$ is increasing (or decreasing, resp.). But then $f(a) < f(b)$ (or $f(a) > f(b)$, resp.) for any $a<b$, from which we can conclude that $mu(f([a,b])) = |f(b)-f(a)|$. So non-degenerate intervals are sent to non-degenerate intervals. I think that you ought to be able to finish the argument via some kind of appeal to regularity.
– Xander Henderson
Jul 30 at 3:17
This argument does assumes, however, that $f'(x)$ exists for all $x$ (not just almost all $x$), but this seems to be implied by the hypothesis that $f'(x) ne 0$ for all $x$.
– Xander Henderson
Jul 30 at 3:18
I'm with you up until the end. The case that is unclear to me are when the positive-measure set is something that contains no positive length intervals. For example, the intersection of [a,b] with the irrationals. I assume that this is where the regularity comes into play, but I'm not familiar enough with the notion to know for sure. Is there any chance you could shed some light on this? Thanks
– Drathora
Jul 30 at 3:28
add a comment |Â
There are a couple of little technicalities that I am not sure of, but I think that the basic argument should follow from Darboux. It follows from Darboux's theorem that we may assume that $f'$ is positive (or negative, resp.) on $Bbb R$, from which it follows that $f$ is increasing (or decreasing, resp.). But then $f(a) < f(b)$ (or $f(a) > f(b)$, resp.) for any $a<b$, from which we can conclude that $mu(f([a,b])) = |f(b)-f(a)|$. So non-degenerate intervals are sent to non-degenerate intervals. I think that you ought to be able to finish the argument via some kind of appeal to regularity.
– Xander Henderson
Jul 30 at 3:17
This argument does assumes, however, that $f'(x)$ exists for all $x$ (not just almost all $x$), but this seems to be implied by the hypothesis that $f'(x) ne 0$ for all $x$.
– Xander Henderson
Jul 30 at 3:18
I'm with you up until the end. The case that is unclear to me are when the positive-measure set is something that contains no positive length intervals. For example, the intersection of [a,b] with the irrationals. I assume that this is where the regularity comes into play, but I'm not familiar enough with the notion to know for sure. Is there any chance you could shed some light on this? Thanks
– Drathora
Jul 30 at 3:28
There are a couple of little technicalities that I am not sure of, but I think that the basic argument should follow from Darboux. It follows from Darboux's theorem that we may assume that $f'$ is positive (or negative, resp.) on $Bbb R$, from which it follows that $f$ is increasing (or decreasing, resp.). But then $f(a) < f(b)$ (or $f(a) > f(b)$, resp.) for any $a<b$, from which we can conclude that $mu(f([a,b])) = |f(b)-f(a)|$. So non-degenerate intervals are sent to non-degenerate intervals. I think that you ought to be able to finish the argument via some kind of appeal to regularity.
– Xander Henderson
Jul 30 at 3:17
There are a couple of little technicalities that I am not sure of, but I think that the basic argument should follow from Darboux. It follows from Darboux's theorem that we may assume that $f'$ is positive (or negative, resp.) on $Bbb R$, from which it follows that $f$ is increasing (or decreasing, resp.). But then $f(a) < f(b)$ (or $f(a) > f(b)$, resp.) for any $a<b$, from which we can conclude that $mu(f([a,b])) = |f(b)-f(a)|$. So non-degenerate intervals are sent to non-degenerate intervals. I think that you ought to be able to finish the argument via some kind of appeal to regularity.
– Xander Henderson
Jul 30 at 3:17
This argument does assumes, however, that $f'(x)$ exists for all $x$ (not just almost all $x$), but this seems to be implied by the hypothesis that $f'(x) ne 0$ for all $x$.
– Xander Henderson
Jul 30 at 3:18
This argument does assumes, however, that $f'(x)$ exists for all $x$ (not just almost all $x$), but this seems to be implied by the hypothesis that $f'(x) ne 0$ for all $x$.
– Xander Henderson
Jul 30 at 3:18
I'm with you up until the end. The case that is unclear to me are when the positive-measure set is something that contains no positive length intervals. For example, the intersection of [a,b] with the irrationals. I assume that this is where the regularity comes into play, but I'm not familiar enough with the notion to know for sure. Is there any chance you could shed some light on this? Thanks
– Drathora
Jul 30 at 3:28
I'm with you up until the end. The case that is unclear to me are when the positive-measure set is something that contains no positive length intervals. For example, the intersection of [a,b] with the irrationals. I assume that this is where the regularity comes into play, but I'm not familiar enough with the notion to know for sure. Is there any chance you could shed some light on this? Thanks
– Drathora
Jul 30 at 3:28
add a comment |Â
1 Answer
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Yes. By Darboux's theorem, $f'$ must have constant sign, so $f$ is monotone. It follows that $f$ is locally absolutely continuous (see Non-decreasing and everywhere differentiable on $[0,1]$ implies absolutelly continuous?), and so for any interval $I$, $$mu(f(I))=int_I |f'|.$$ Both sides of this equation are measures when considered as functions of $I$ (here we use the fact that $f$ is injective so $f(cdot)$ preserves disjointness of sets), and it follows that the two sides are equal for all Borel sets as well: for any Borel set $X$, $$mu(f(X))=int_X|f'|.$$ If $X$ has positive measure, then the integral on the right is positive since $f'$ is always nonzero, so $mu(f(X))>0$.
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
Thank you. I suppose this also holds for open intervals by simply taking a suitably large sub-interval that is closed?
– Drathora
Jul 30 at 3:53
Not sure what you mean by "this". I'm not assuming any interval is closed here. In any case, everything in this question is local, so it doesn't matter.
– Eric Wofsey
Jul 30 at 3:55
Apologies, I should have been more verbose. The preconditions for Darboux's theorem stated that $I$ should be a closed interval, which was the motivation behind the followup question. The question you linked also used a closed interval as an example, so I just wanted to make sure that everything still held when talking about open intervals.
– Drathora
Jul 30 at 3:58
I guess maybe I should have said "locally absolutely continuous", since it seems that definitions of absolute continuity on non-compact intervals vary and may be stronger than what I am asserting here.
– Eric Wofsey
Jul 30 at 4:00
Right, everything here is local (we only need all these statements to hold in some neighborhood of every point), so we can freely use theorems that are stated only for closed intervals.
– Eric Wofsey
Jul 30 at 4:01
 |Â
show 1 more comment
1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes. By Darboux's theorem, $f'$ must have constant sign, so $f$ is monotone. It follows that $f$ is locally absolutely continuous (see Non-decreasing and everywhere differentiable on $[0,1]$ implies absolutelly continuous?), and so for any interval $I$, $$mu(f(I))=int_I |f'|.$$ Both sides of this equation are measures when considered as functions of $I$ (here we use the fact that $f$ is injective so $f(cdot)$ preserves disjointness of sets), and it follows that the two sides are equal for all Borel sets as well: for any Borel set $X$, $$mu(f(X))=int_X|f'|.$$ If $X$ has positive measure, then the integral on the right is positive since $f'$ is always nonzero, so $mu(f(X))>0$.
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
Thank you. I suppose this also holds for open intervals by simply taking a suitably large sub-interval that is closed?
– Drathora
Jul 30 at 3:53
Not sure what you mean by "this". I'm not assuming any interval is closed here. In any case, everything in this question is local, so it doesn't matter.
– Eric Wofsey
Jul 30 at 3:55
Apologies, I should have been more verbose. The preconditions for Darboux's theorem stated that $I$ should be a closed interval, which was the motivation behind the followup question. The question you linked also used a closed interval as an example, so I just wanted to make sure that everything still held when talking about open intervals.
– Drathora
Jul 30 at 3:58
I guess maybe I should have said "locally absolutely continuous", since it seems that definitions of absolute continuity on non-compact intervals vary and may be stronger than what I am asserting here.
– Eric Wofsey
Jul 30 at 4:00
Right, everything here is local (we only need all these statements to hold in some neighborhood of every point), so we can freely use theorems that are stated only for closed intervals.
– Eric Wofsey
Jul 30 at 4:01
 |Â
show 1 more comment
up vote
2
down vote
Yes. By Darboux's theorem, $f'$ must have constant sign, so $f$ is monotone. It follows that $f$ is locally absolutely continuous (see Non-decreasing and everywhere differentiable on $[0,1]$ implies absolutelly continuous?), and so for any interval $I$, $$mu(f(I))=int_I |f'|.$$ Both sides of this equation are measures when considered as functions of $I$ (here we use the fact that $f$ is injective so $f(cdot)$ preserves disjointness of sets), and it follows that the two sides are equal for all Borel sets as well: for any Borel set $X$, $$mu(f(X))=int_X|f'|.$$ If $X$ has positive measure, then the integral on the right is positive since $f'$ is always nonzero, so $mu(f(X))>0$.
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
Thank you. I suppose this also holds for open intervals by simply taking a suitably large sub-interval that is closed?
– Drathora
Jul 30 at 3:53
Not sure what you mean by "this". I'm not assuming any interval is closed here. In any case, everything in this question is local, so it doesn't matter.
– Eric Wofsey
Jul 30 at 3:55
Apologies, I should have been more verbose. The preconditions for Darboux's theorem stated that $I$ should be a closed interval, which was the motivation behind the followup question. The question you linked also used a closed interval as an example, so I just wanted to make sure that everything still held when talking about open intervals.
– Drathora
Jul 30 at 3:58
I guess maybe I should have said "locally absolutely continuous", since it seems that definitions of absolute continuity on non-compact intervals vary and may be stronger than what I am asserting here.
– Eric Wofsey
Jul 30 at 4:00
Right, everything here is local (we only need all these statements to hold in some neighborhood of every point), so we can freely use theorems that are stated only for closed intervals.
– Eric Wofsey
Jul 30 at 4:01
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
Yes. By Darboux's theorem, $f'$ must have constant sign, so $f$ is monotone. It follows that $f$ is locally absolutely continuous (see Non-decreasing and everywhere differentiable on $[0,1]$ implies absolutelly continuous?), and so for any interval $I$, $$mu(f(I))=int_I |f'|.$$ Both sides of this equation are measures when considered as functions of $I$ (here we use the fact that $f$ is injective so $f(cdot)$ preserves disjointness of sets), and it follows that the two sides are equal for all Borel sets as well: for any Borel set $X$, $$mu(f(X))=int_X|f'|.$$ If $X$ has positive measure, then the integral on the right is positive since $f'$ is always nonzero, so $mu(f(X))>0$.
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
Yes. By Darboux's theorem, $f'$ must have constant sign, so $f$ is monotone. It follows that $f$ is locally absolutely continuous (see Non-decreasing and everywhere differentiable on $[0,1]$ implies absolutelly continuous?), and so for any interval $I$, $$mu(f(I))=int_I |f'|.$$ Both sides of this equation are measures when considered as functions of $I$ (here we use the fact that $f$ is injective so $f(cdot)$ preserves disjointness of sets), and it follows that the two sides are equal for all Borel sets as well: for any Borel set $X$, $$mu(f(X))=int_X|f'|.$$ If $X$ has positive measure, then the integral on the right is positive since $f'$ is always nonzero, so $mu(f(X))>0$.
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
edited Jul 30 at 4:00
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
answered Jul 30 at 3:41
Eric Wofsey
162k12188298
162k12188298
Thank you. I suppose this also holds for open intervals by simply taking a suitably large sub-interval that is closed?
– Drathora
Jul 30 at 3:53
Not sure what you mean by "this". I'm not assuming any interval is closed here. In any case, everything in this question is local, so it doesn't matter.
– Eric Wofsey
Jul 30 at 3:55
Apologies, I should have been more verbose. The preconditions for Darboux's theorem stated that $I$ should be a closed interval, which was the motivation behind the followup question. The question you linked also used a closed interval as an example, so I just wanted to make sure that everything still held when talking about open intervals.
– Drathora
Jul 30 at 3:58
I guess maybe I should have said "locally absolutely continuous", since it seems that definitions of absolute continuity on non-compact intervals vary and may be stronger than what I am asserting here.
– Eric Wofsey
Jul 30 at 4:00
Right, everything here is local (we only need all these statements to hold in some neighborhood of every point), so we can freely use theorems that are stated only for closed intervals.
– Eric Wofsey
Jul 30 at 4:01
 |Â
show 1 more comment
Thank you. I suppose this also holds for open intervals by simply taking a suitably large sub-interval that is closed?
– Drathora
Jul 30 at 3:53
Not sure what you mean by "this". I'm not assuming any interval is closed here. In any case, everything in this question is local, so it doesn't matter.
– Eric Wofsey
Jul 30 at 3:55
Apologies, I should have been more verbose. The preconditions for Darboux's theorem stated that $I$ should be a closed interval, which was the motivation behind the followup question. The question you linked also used a closed interval as an example, so I just wanted to make sure that everything still held when talking about open intervals.
– Drathora
Jul 30 at 3:58
I guess maybe I should have said "locally absolutely continuous", since it seems that definitions of absolute continuity on non-compact intervals vary and may be stronger than what I am asserting here.
– Eric Wofsey
Jul 30 at 4:00
Right, everything here is local (we only need all these statements to hold in some neighborhood of every point), so we can freely use theorems that are stated only for closed intervals.
– Eric Wofsey
Jul 30 at 4:01
Thank you. I suppose this also holds for open intervals by simply taking a suitably large sub-interval that is closed?
– Drathora
Jul 30 at 3:53
Thank you. I suppose this also holds for open intervals by simply taking a suitably large sub-interval that is closed?
– Drathora
Jul 30 at 3:53
Not sure what you mean by "this". I'm not assuming any interval is closed here. In any case, everything in this question is local, so it doesn't matter.
– Eric Wofsey
Jul 30 at 3:55
Not sure what you mean by "this". I'm not assuming any interval is closed here. In any case, everything in this question is local, so it doesn't matter.
– Eric Wofsey
Jul 30 at 3:55
Apologies, I should have been more verbose. The preconditions for Darboux's theorem stated that $I$ should be a closed interval, which was the motivation behind the followup question. The question you linked also used a closed interval as an example, so I just wanted to make sure that everything still held when talking about open intervals.
– Drathora
Jul 30 at 3:58
Apologies, I should have been more verbose. The preconditions for Darboux's theorem stated that $I$ should be a closed interval, which was the motivation behind the followup question. The question you linked also used a closed interval as an example, so I just wanted to make sure that everything still held when talking about open intervals.
– Drathora
Jul 30 at 3:58
I guess maybe I should have said "locally absolutely continuous", since it seems that definitions of absolute continuity on non-compact intervals vary and may be stronger than what I am asserting here.
– Eric Wofsey
Jul 30 at 4:00
I guess maybe I should have said "locally absolutely continuous", since it seems that definitions of absolute continuity on non-compact intervals vary and may be stronger than what I am asserting here.
– Eric Wofsey
Jul 30 at 4:00
Right, everything here is local (we only need all these statements to hold in some neighborhood of every point), so we can freely use theorems that are stated only for closed intervals.
– Eric Wofsey
Jul 30 at 4:01
Right, everything here is local (we only need all these statements to hold in some neighborhood of every point), so we can freely use theorems that are stated only for closed intervals.
– Eric Wofsey
Jul 30 at 4:01
 |Â
show 1 more comment
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There are a couple of little technicalities that I am not sure of, but I think that the basic argument should follow from Darboux. It follows from Darboux's theorem that we may assume that $f'$ is positive (or negative, resp.) on $Bbb R$, from which it follows that $f$ is increasing (or decreasing, resp.). But then $f(a) < f(b)$ (or $f(a) > f(b)$, resp.) for any $a<b$, from which we can conclude that $mu(f([a,b])) = |f(b)-f(a)|$. So non-degenerate intervals are sent to non-degenerate intervals. I think that you ought to be able to finish the argument via some kind of appeal to regularity.
– Xander Henderson
Jul 30 at 3:17
This argument does assumes, however, that $f'(x)$ exists for all $x$ (not just almost all $x$), but this seems to be implied by the hypothesis that $f'(x) ne 0$ for all $x$.
– Xander Henderson
Jul 30 at 3:18
I'm with you up until the end. The case that is unclear to me are when the positive-measure set is something that contains no positive length intervals. For example, the intersection of [a,b] with the irrationals. I assume that this is where the regularity comes into play, but I'm not familiar enough with the notion to know for sure. Is there any chance you could shed some light on this? Thanks
– Drathora
Jul 30 at 3:28