Find the Lebesgue set w.r.t. to the functions












2














I'm working on a problem in measure theory:



Find the Lebesgue set $L_f={x:lim_{rto 0} frac{1}{m(B(r,x))} int_{B(r,x)}lvert f(y)-f(x)rvert dy=0}$ if:





  1. $f:mathbb R^n to mathbb C$ is continuous


  2. $f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.


  3. $f(x)=lfloor xrfloor$ on $mathbb R$


I'm having a bit of a tough time wrapping my head around what the Lebesgue sets really are, and consequently is struggling to characterize the Lebesgue sets w.r.t to these functions. I could really use some help on these!










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  • For 1) all points are Lebesgue points. For 2) $x$ is a Lebesgue point iff $xin E$. For 3) Lebesgue points are all points except integers. All these follow easily from definition.
    – Kavi Rama Murthy
    Nov 30 at 0:37










  • @KaviRamaMurthy maybe I have an ill understanding of the definition... could you elaborate a bit more?
    – Sank
    Nov 30 at 0:52










  • What do you have as the definition?
    – Robert Israel
    Nov 30 at 1:15










  • @RobertIsrael I edited the post with the definition in the question.
    – Sank
    Nov 30 at 1:22
















2














I'm working on a problem in measure theory:



Find the Lebesgue set $L_f={x:lim_{rto 0} frac{1}{m(B(r,x))} int_{B(r,x)}lvert f(y)-f(x)rvert dy=0}$ if:





  1. $f:mathbb R^n to mathbb C$ is continuous


  2. $f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.


  3. $f(x)=lfloor xrfloor$ on $mathbb R$


I'm having a bit of a tough time wrapping my head around what the Lebesgue sets really are, and consequently is struggling to characterize the Lebesgue sets w.r.t to these functions. I could really use some help on these!










share|cite|improve this question
























  • For 1) all points are Lebesgue points. For 2) $x$ is a Lebesgue point iff $xin E$. For 3) Lebesgue points are all points except integers. All these follow easily from definition.
    – Kavi Rama Murthy
    Nov 30 at 0:37










  • @KaviRamaMurthy maybe I have an ill understanding of the definition... could you elaborate a bit more?
    – Sank
    Nov 30 at 0:52










  • What do you have as the definition?
    – Robert Israel
    Nov 30 at 1:15










  • @RobertIsrael I edited the post with the definition in the question.
    – Sank
    Nov 30 at 1:22














2












2








2







I'm working on a problem in measure theory:



Find the Lebesgue set $L_f={x:lim_{rto 0} frac{1}{m(B(r,x))} int_{B(r,x)}lvert f(y)-f(x)rvert dy=0}$ if:





  1. $f:mathbb R^n to mathbb C$ is continuous


  2. $f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.


  3. $f(x)=lfloor xrfloor$ on $mathbb R$


I'm having a bit of a tough time wrapping my head around what the Lebesgue sets really are, and consequently is struggling to characterize the Lebesgue sets w.r.t to these functions. I could really use some help on these!










share|cite|improve this question















I'm working on a problem in measure theory:



Find the Lebesgue set $L_f={x:lim_{rto 0} frac{1}{m(B(r,x))} int_{B(r,x)}lvert f(y)-f(x)rvert dy=0}$ if:





  1. $f:mathbb R^n to mathbb C$ is continuous


  2. $f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.


  3. $f(x)=lfloor xrfloor$ on $mathbb R$


I'm having a bit of a tough time wrapping my head around what the Lebesgue sets really are, and consequently is struggling to characterize the Lebesgue sets w.r.t to these functions. I could really use some help on these!







real-analysis measure-theory






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edited Nov 30 at 1:21

























asked Nov 30 at 0:21









Sank

16511




16511












  • For 1) all points are Lebesgue points. For 2) $x$ is a Lebesgue point iff $xin E$. For 3) Lebesgue points are all points except integers. All these follow easily from definition.
    – Kavi Rama Murthy
    Nov 30 at 0:37










  • @KaviRamaMurthy maybe I have an ill understanding of the definition... could you elaborate a bit more?
    – Sank
    Nov 30 at 0:52










  • What do you have as the definition?
    – Robert Israel
    Nov 30 at 1:15










  • @RobertIsrael I edited the post with the definition in the question.
    – Sank
    Nov 30 at 1:22


















  • For 1) all points are Lebesgue points. For 2) $x$ is a Lebesgue point iff $xin E$. For 3) Lebesgue points are all points except integers. All these follow easily from definition.
    – Kavi Rama Murthy
    Nov 30 at 0:37










  • @KaviRamaMurthy maybe I have an ill understanding of the definition... could you elaborate a bit more?
    – Sank
    Nov 30 at 0:52










  • What do you have as the definition?
    – Robert Israel
    Nov 30 at 1:15










  • @RobertIsrael I edited the post with the definition in the question.
    – Sank
    Nov 30 at 1:22
















For 1) all points are Lebesgue points. For 2) $x$ is a Lebesgue point iff $xin E$. For 3) Lebesgue points are all points except integers. All these follow easily from definition.
– Kavi Rama Murthy
Nov 30 at 0:37




For 1) all points are Lebesgue points. For 2) $x$ is a Lebesgue point iff $xin E$. For 3) Lebesgue points are all points except integers. All these follow easily from definition.
– Kavi Rama Murthy
Nov 30 at 0:37












@KaviRamaMurthy maybe I have an ill understanding of the definition... could you elaborate a bit more?
– Sank
Nov 30 at 0:52




@KaviRamaMurthy maybe I have an ill understanding of the definition... could you elaborate a bit more?
– Sank
Nov 30 at 0:52












What do you have as the definition?
– Robert Israel
Nov 30 at 1:15




What do you have as the definition?
– Robert Israel
Nov 30 at 1:15












@RobertIsrael I edited the post with the definition in the question.
– Sank
Nov 30 at 1:22




@RobertIsrael I edited the post with the definition in the question.
– Sank
Nov 30 at 1:22










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If $f$ is continuous at $x$ then $|f(y)-f(x)| <epsilon$ foe all $y in B(x,r)$ provided $r$ is small enough. Hence $x$ is a Lebesgue point. In 2) $f=0$ a.e. so you can simply repalce $f(y)$ by $0$ in the definition without changing the integral. Hence $x$ is a Lebesgue point iff $f(x)=0$ which means $x in E^{c}$. (There was mistake in my comment. I meant $x in E^{c}$ not $x in E$). In 3) all points except integers are points of continuity and hence they are Lebesgue points. At integer points it is easy to see that the limit in the definition equals $frac 1 2$, not $0$.






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    If $f$ is continuous at $x$ then $|f(y)-f(x)| <epsilon$ foe all $y in B(x,r)$ provided $r$ is small enough. Hence $x$ is a Lebesgue point. In 2) $f=0$ a.e. so you can simply repalce $f(y)$ by $0$ in the definition without changing the integral. Hence $x$ is a Lebesgue point iff $f(x)=0$ which means $x in E^{c}$. (There was mistake in my comment. I meant $x in E^{c}$ not $x in E$). In 3) all points except integers are points of continuity and hence they are Lebesgue points. At integer points it is easy to see that the limit in the definition equals $frac 1 2$, not $0$.






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      If $f$ is continuous at $x$ then $|f(y)-f(x)| <epsilon$ foe all $y in B(x,r)$ provided $r$ is small enough. Hence $x$ is a Lebesgue point. In 2) $f=0$ a.e. so you can simply repalce $f(y)$ by $0$ in the definition without changing the integral. Hence $x$ is a Lebesgue point iff $f(x)=0$ which means $x in E^{c}$. (There was mistake in my comment. I meant $x in E^{c}$ not $x in E$). In 3) all points except integers are points of continuity and hence they are Lebesgue points. At integer points it is easy to see that the limit in the definition equals $frac 1 2$, not $0$.






      share|cite|improve this answer
























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        If $f$ is continuous at $x$ then $|f(y)-f(x)| <epsilon$ foe all $y in B(x,r)$ provided $r$ is small enough. Hence $x$ is a Lebesgue point. In 2) $f=0$ a.e. so you can simply repalce $f(y)$ by $0$ in the definition without changing the integral. Hence $x$ is a Lebesgue point iff $f(x)=0$ which means $x in E^{c}$. (There was mistake in my comment. I meant $x in E^{c}$ not $x in E$). In 3) all points except integers are points of continuity and hence they are Lebesgue points. At integer points it is easy to see that the limit in the definition equals $frac 1 2$, not $0$.






        share|cite|improve this answer












        If $f$ is continuous at $x$ then $|f(y)-f(x)| <epsilon$ foe all $y in B(x,r)$ provided $r$ is small enough. Hence $x$ is a Lebesgue point. In 2) $f=0$ a.e. so you can simply repalce $f(y)$ by $0$ in the definition without changing the integral. Hence $x$ is a Lebesgue point iff $f(x)=0$ which means $x in E^{c}$. (There was mistake in my comment. I meant $x in E^{c}$ not $x in E$). In 3) all points except integers are points of continuity and hence they are Lebesgue points. At integer points it is easy to see that the limit in the definition equals $frac 1 2$, not $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 at 5:03









        Kavi Rama Murthy

        49.4k31854




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