Question on strictly positive measure of a level set
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Let $f in W^{2,p}(U)$ for $1 le p lt infty$ where $U$ is a compact subset of $mathbb R^2$ and $g in L^{infty}(U)$ with $g(f)=f$ almost everywhere on $U$. I have trouble understanding why the following holds:
Assume that the set ${f=0}$ is non empty and moreover that $g lt 1$ in ${f=0}$. Then it has a strictly
positive lebesgue measure.
I know that ${f=0}$ is a (relatively) closed set but that is not enough for the strict inequality. What do I miss? Is it because it contains an open set?
Any help/hints are much appreciated! Thanks in advance!
real-analysis analysis measure-theory lebesgue-measure
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
$endgroup$
add a comment |
$begingroup$
Let $f in W^{2,p}(U)$ for $1 le p lt infty$ where $U$ is a compact subset of $mathbb R^2$ and $g in L^{infty}(U)$ with $g(f)=f$ almost everywhere on $U$. I have trouble understanding why the following holds:
Assume that the set ${f=0}$ is non empty and moreover that $g lt 1$ in ${f=0}$. Then it has a strictly
positive lebesgue measure.
I know that ${f=0}$ is a (relatively) closed set but that is not enough for the strict inequality. What do I miss? Is it because it contains an open set?
Any help/hints are much appreciated! Thanks in advance!
real-analysis analysis measure-theory lebesgue-measure
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
$endgroup$
$begingroup$
Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
$endgroup$
– ktoi
Nov 28 '18 at 21:26
$begingroup$
@ktoi I just edited my question...
$endgroup$
– kaithkolesidou
Dec 3 '18 at 11:16
1
$begingroup$
What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 12:01
$begingroup$
@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
$endgroup$
– kaithkolesidou
Dec 3 '18 at 12:16
$begingroup$
@kaithkolesidou What do you mean by $g(f)$? If it's functional composition, then the assumption $g<1$ on ${f = 0}$ doesn't really tell you anything and taking $g(x) = x$ my example still remains valid. Also it seems like you need more regularity assumptions on $g$; if ${ f = 0 }$ is a null set then the condition $g < 1$ on ${f = 0}$ gives hardly any restriction on $g.$
$endgroup$
– ktoi
Dec 4 '18 at 19:07
add a comment |
$begingroup$
Let $f in W^{2,p}(U)$ for $1 le p lt infty$ where $U$ is a compact subset of $mathbb R^2$ and $g in L^{infty}(U)$ with $g(f)=f$ almost everywhere on $U$. I have trouble understanding why the following holds:
Assume that the set ${f=0}$ is non empty and moreover that $g lt 1$ in ${f=0}$. Then it has a strictly
positive lebesgue measure.
I know that ${f=0}$ is a (relatively) closed set but that is not enough for the strict inequality. What do I miss? Is it because it contains an open set?
Any help/hints are much appreciated! Thanks in advance!
real-analysis analysis measure-theory lebesgue-measure
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
$endgroup$
Let $f in W^{2,p}(U)$ for $1 le p lt infty$ where $U$ is a compact subset of $mathbb R^2$ and $g in L^{infty}(U)$ with $g(f)=f$ almost everywhere on $U$. I have trouble understanding why the following holds:
Assume that the set ${f=0}$ is non empty and moreover that $g lt 1$ in ${f=0}$. Then it has a strictly
positive lebesgue measure.
I know that ${f=0}$ is a (relatively) closed set but that is not enough for the strict inequality. What do I miss? Is it because it contains an open set?
Any help/hints are much appreciated! Thanks in advance!
real-analysis analysis measure-theory lebesgue-measure
real-analysis analysis measure-theory lebesgue-measure
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
edited Dec 3 '18 at 12:15
kaithkolesidou
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
asked Nov 28 '18 at 12:55
kaithkolesidoukaithkolesidou
959511
959511
$begingroup$
Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
$endgroup$
– ktoi
Nov 28 '18 at 21:26
$begingroup$
@ktoi I just edited my question...
$endgroup$
– kaithkolesidou
Dec 3 '18 at 11:16
1
$begingroup$
What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 12:01
$begingroup$
@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
$endgroup$
– kaithkolesidou
Dec 3 '18 at 12:16
$begingroup$
@kaithkolesidou What do you mean by $g(f)$? If it's functional composition, then the assumption $g<1$ on ${f = 0}$ doesn't really tell you anything and taking $g(x) = x$ my example still remains valid. Also it seems like you need more regularity assumptions on $g$; if ${ f = 0 }$ is a null set then the condition $g < 1$ on ${f = 0}$ gives hardly any restriction on $g.$
$endgroup$
– ktoi
Dec 4 '18 at 19:07
add a comment |
$begingroup$
Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
$endgroup$
– ktoi
Nov 28 '18 at 21:26
$begingroup$
@ktoi I just edited my question...
$endgroup$
– kaithkolesidou
Dec 3 '18 at 11:16
1
$begingroup$
What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 12:01
$begingroup$
@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
$endgroup$
– kaithkolesidou
Dec 3 '18 at 12:16
$begingroup$
@kaithkolesidou What do you mean by $g(f)$? If it's functional composition, then the assumption $g<1$ on ${f = 0}$ doesn't really tell you anything and taking $g(x) = x$ my example still remains valid. Also it seems like you need more regularity assumptions on $g$; if ${ f = 0 }$ is a null set then the condition $g < 1$ on ${f = 0}$ gives hardly any restriction on $g.$
$endgroup$
– ktoi
Dec 4 '18 at 19:07
$begingroup$
Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
$endgroup$
– ktoi
Nov 28 '18 at 21:26
$begingroup$
Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
$endgroup$
– ktoi
Nov 28 '18 at 21:26
$begingroup$
@ktoi I just edited my question...
$endgroup$
– kaithkolesidou
Dec 3 '18 at 11:16
$begingroup$
@ktoi I just edited my question...
$endgroup$
– kaithkolesidou
Dec 3 '18 at 11:16
1
1
$begingroup$
What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 12:01
$begingroup$
What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 12:01
$begingroup$
@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
$endgroup$
– kaithkolesidou
Dec 3 '18 at 12:16
$begingroup$
@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
$endgroup$
– kaithkolesidou
Dec 3 '18 at 12:16
$begingroup$
@kaithkolesidou What do you mean by $g(f)$? If it's functional composition, then the assumption $g<1$ on ${f = 0}$ doesn't really tell you anything and taking $g(x) = x$ my example still remains valid. Also it seems like you need more regularity assumptions on $g$; if ${ f = 0 }$ is a null set then the condition $g < 1$ on ${f = 0}$ gives hardly any restriction on $g.$
$endgroup$
– ktoi
Dec 4 '18 at 19:07
$begingroup$
@kaithkolesidou What do you mean by $g(f)$? If it's functional composition, then the assumption $g<1$ on ${f = 0}$ doesn't really tell you anything and taking $g(x) = x$ my example still remains valid. Also it seems like you need more regularity assumptions on $g$; if ${ f = 0 }$ is a null set then the condition $g < 1$ on ${f = 0}$ gives hardly any restriction on $g.$
$endgroup$
– ktoi
Dec 4 '18 at 19:07
add a comment |
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$begingroup$
Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
$endgroup$
– ktoi
Nov 28 '18 at 21:26
$begingroup$
@ktoi I just edited my question...
$endgroup$
– kaithkolesidou
Dec 3 '18 at 11:16
1
$begingroup$
What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 12:01
$begingroup$
@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
$endgroup$
– kaithkolesidou
Dec 3 '18 at 12:16
$begingroup$
@kaithkolesidou What do you mean by $g(f)$? If it's functional composition, then the assumption $g<1$ on ${f = 0}$ doesn't really tell you anything and taking $g(x) = x$ my example still remains valid. Also it seems like you need more regularity assumptions on $g$; if ${ f = 0 }$ is a null set then the condition $g < 1$ on ${f = 0}$ gives hardly any restriction on $g.$
$endgroup$
– ktoi
Dec 4 '18 at 19:07