Question on strictly positive measure of a level set












2














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!










share|cite|improve this question
























  • Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
    – ktoi
    Nov 28 '18 at 21:26












  • @ktoi I just edited my question...
    – kaithkolesidou
    Dec 3 '18 at 11:16






  • 1




    What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
    – Kavi Rama Murthy
    Dec 3 '18 at 12:01










  • @KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
    – kaithkolesidou
    Dec 3 '18 at 12:16










  • @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 $g.$
    – ktoi
    Dec 4 '18 at 19:07
















2














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!










share|cite|improve this question
























  • Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
    – ktoi
    Nov 28 '18 at 21:26












  • @ktoi I just edited my question...
    – kaithkolesidou
    Dec 3 '18 at 11:16






  • 1




    What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
    – Kavi Rama Murthy
    Dec 3 '18 at 12:01










  • @KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
    – kaithkolesidou
    Dec 3 '18 at 12:16










  • @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 $g.$
    – ktoi
    Dec 4 '18 at 19:07














2












2








2


2





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!










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 12:15

























asked Nov 28 '18 at 12:55









kaithkolesidou

959511




959511












  • Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
    – ktoi
    Nov 28 '18 at 21:26












  • @ktoi I just edited my question...
    – kaithkolesidou
    Dec 3 '18 at 11:16






  • 1




    What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
    – Kavi Rama Murthy
    Dec 3 '18 at 12:01










  • @KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
    – kaithkolesidou
    Dec 3 '18 at 12:16










  • @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 $g.$
    – ktoi
    Dec 4 '18 at 19:07


















  • Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
    – ktoi
    Nov 28 '18 at 21:26












  • @ktoi I just edited my question...
    – kaithkolesidou
    Dec 3 '18 at 11:16






  • 1




    What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
    – Kavi Rama Murthy
    Dec 3 '18 at 12:01










  • @KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
    – kaithkolesidou
    Dec 3 '18 at 12:16










  • @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 $g.$
    – ktoi
    Dec 4 '18 at 19:07
















Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
– ktoi
Nov 28 '18 at 21:26






Do you not have any additional assumptions? Unless I'm missing something, $f(x) = x_1^2+x_2^2$ is an easy counterexample.
– ktoi
Nov 28 '18 at 21:26














@ktoi I just edited my question...
– kaithkolesidou
Dec 3 '18 at 11:16




@ktoi I just edited my question...
– kaithkolesidou
Dec 3 '18 at 11:16




1




1




What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
– Kavi Rama Murthy
Dec 3 '18 at 12:01




What is the connection between $f$ and $g$. The example given by ktoi is still valid with $g=0$.
– Kavi Rama Murthy
Dec 3 '18 at 12:01












@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
– kaithkolesidou
Dec 3 '18 at 12:16




@KaviRamaMurthy I'm sorry I didn't mention it from the beginning. I edited
– kaithkolesidou
Dec 3 '18 at 12:16












@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 $g.$
– ktoi
Dec 4 '18 at 19:07




@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 $g.$
– ktoi
Dec 4 '18 at 19:07










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