Find the Lebesgue set w.r.t. to the functions
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:
$f:mathbb R^n to mathbb C$ is continuous
$f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.
$f(x)=lfloor xrfloor$ 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
asked Nov 30 at 0:21
Sank
16511
add a comment |
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:
$f:mathbb R^n to mathbb C$ is continuous
$f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.
$f(x)=lfloor xrfloor$ 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
asked Nov 30 at 0:21
Sank
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
add a comment |
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:
$f:mathbb R^n to mathbb C$ is continuous
$f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.
$f(x)=lfloor xrfloor$ 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
asked Nov 30 at 0:21
Sank
16511
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:
$f:mathbb R^n to mathbb C$ is continuous
$f=chi_{E}$, where $Einmathbb R^n$ is Lebesgue null; $m(E)=0$. $chi$ denotes the indicator function.
$f(x)=lfloor xrfloor$ 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
real-analysis measure-theory
asked Nov 30 at 0:21
Sank
16511
asked Nov 30 at 0:21
Sank
16511
edited Nov 30 at 1:21
asked Nov 30 at 0:21
Sank
16511
asked Nov 30 at 0:21
Sank
16511
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
add a comment |
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
add a comment |
1 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 $0$.
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
add a comment |
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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 $0$.
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
add a comment |
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 $0$.
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
add a comment |
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 $0$.
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
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 $0$.
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
answered Nov 30 at 5:03
Kavi Rama Murthy
49.4k31854
49.4k31854
add a comment |
add a comment |
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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