A continuous function for which Lebesgue’s Differentiation Theorem doesn’t hold for Rectangles?
Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But this paper says that if we restrict things to continuous functions with compact support, then Lebesgue’s differentiation holds true for rectangles. My question is, what is an example of a continuous function (without compact support) for which Lebesgue’s theorem doesn’t hold for rectangles?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
add a comment |
Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But this paper says that if we restrict things to continuous functions with compact support, then Lebesgue’s differentiation holds true for rectangles. My question is, what is an example of a continuous function (without compact support) for which Lebesgue’s theorem doesn’t hold for rectangles?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
You have some misunderstandings. Read about sets ''shrinking nicely to a point'' in Rudin's RCA.
– Kavi Rama Murthy
Nov 24 at 5:52
@KaviRamaMurthy What misunderstandings do I have?
– Keshav Srinivasan
Nov 24 at 7:00
add a comment |
Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But this paper says that if we restrict things to continuous functions with compact support, then Lebesgue’s differentiation holds true for rectangles. My question is, what is an example of a continuous function (without compact support) for which Lebesgue’s theorem doesn’t hold for rectangles?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.
But this paper says that if we restrict things to continuous functions with compact support, then Lebesgue’s differentiation holds true for rectangles. My question is, what is an example of a continuous function (without compact support) for which Lebesgue’s theorem doesn’t hold for rectangles?
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples
measure-theory lebesgue-integral lebesgue-measure examples-counterexamples
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
asked Nov 24 at 4:34
Keshav Srinivasan
2,05711440
2,05711440
You have some misunderstandings. Read about sets ''shrinking nicely to a point'' in Rudin's RCA.
– Kavi Rama Murthy
Nov 24 at 5:52
@KaviRamaMurthy What misunderstandings do I have?
– Keshav Srinivasan
Nov 24 at 7:00
add a comment |
You have some misunderstandings. Read about sets ''shrinking nicely to a point'' in Rudin's RCA.
– Kavi Rama Murthy
Nov 24 at 5:52
@KaviRamaMurthy What misunderstandings do I have?
– Keshav Srinivasan
Nov 24 at 7:00
You have some misunderstandings. Read about sets ''shrinking nicely to a point'' in Rudin's RCA.
– Kavi Rama Murthy
Nov 24 at 5:52
You have some misunderstandings. Read about sets ''shrinking nicely to a point'' in Rudin's RCA.
– Kavi Rama Murthy
Nov 24 at 5:52
@KaviRamaMurthy What misunderstandings do I have?
– Keshav Srinivasan
Nov 24 at 7:00
@KaviRamaMurthy What misunderstandings do I have?
– Keshav Srinivasan
Nov 24 at 7:00
add a comment |
1 Answer
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As already pointed out in the comments you need to take sets which are "shrinking nicely to a point", i.e. $(A_n(x))_{n in mathbb{N}}$ is measurable and satisfies $A_n(x) subset B_{r_n}(x)$ with $r_n rightarrow 0$ and $lambda(A_n(x)) ge c lambda (B_{r_n}(x))$ for some $c>0$ independent of $n$.
If you take for example $A_n = [-n^{-2},n^{-2}] times [- n^{-1},n^{-1}]$, then $(A_n)_{n in mathbb{N}}$ has diameter shrinking to zero, but is not shrinking nicelty to zero. (The scaling in the different directions is not propotional!) For example take $$ Delta:= { (x,y) in [-1,1]^2 : |y| le |x| },$$
then $$ lambda(A_n)^{-1} int_{A_n} 1_{Delta} , d lambda^2 = frac{n^3}{4} lambda(Delta cap A_n) = frac{n^3}{4} lambda([-n^{-2},n^{-2}]^2) = frac{1}{n} rightarrow 0,$$
i.e. $ne 1_{Delta}(0,0) =1$.
answered Nov 28 at 10:00
p4sch
4,760217
add a comment |
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1 Answer
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As already pointed out in the comments you need to take sets which are "shrinking nicely to a point", i.e. $(A_n(x))_{n in mathbb{N}}$ is measurable and satisfies $A_n(x) subset B_{r_n}(x)$ with $r_n rightarrow 0$ and $lambda(A_n(x)) ge c lambda (B_{r_n}(x))$ for some $c>0$ independent of $n$.
If you take for example $A_n = [-n^{-2},n^{-2}] times [- n^{-1},n^{-1}]$, then $(A_n)_{n in mathbb{N}}$ has diameter shrinking to zero, but is not shrinking nicelty to zero. (The scaling in the different directions is not propotional!) For example take $$ Delta:= { (x,y) in [-1,1]^2 : |y| le |x| },$$
then $$ lambda(A_n)^{-1} int_{A_n} 1_{Delta} , d lambda^2 = frac{n^3}{4} lambda(Delta cap A_n) = frac{n^3}{4} lambda([-n^{-2},n^{-2}]^2) = frac{1}{n} rightarrow 0,$$
i.e. $ne 1_{Delta}(0,0) =1$.
answered Nov 28 at 10:00
p4sch
4,760217
add a comment |
As already pointed out in the comments you need to take sets which are "shrinking nicely to a point", i.e. $(A_n(x))_{n in mathbb{N}}$ is measurable and satisfies $A_n(x) subset B_{r_n}(x)$ with $r_n rightarrow 0$ and $lambda(A_n(x)) ge c lambda (B_{r_n}(x))$ for some $c>0$ independent of $n$.
If you take for example $A_n = [-n^{-2},n^{-2}] times [- n^{-1},n^{-1}]$, then $(A_n)_{n in mathbb{N}}$ has diameter shrinking to zero, but is not shrinking nicelty to zero. (The scaling in the different directions is not propotional!) For example take $$ Delta:= { (x,y) in [-1,1]^2 : |y| le |x| },$$
then $$ lambda(A_n)^{-1} int_{A_n} 1_{Delta} , d lambda^2 = frac{n^3}{4} lambda(Delta cap A_n) = frac{n^3}{4} lambda([-n^{-2},n^{-2}]^2) = frac{1}{n} rightarrow 0,$$
i.e. $ne 1_{Delta}(0,0) =1$.
answered Nov 28 at 10:00
p4sch
4,760217
add a comment |
As already pointed out in the comments you need to take sets which are "shrinking nicely to a point", i.e. $(A_n(x))_{n in mathbb{N}}$ is measurable and satisfies $A_n(x) subset B_{r_n}(x)$ with $r_n rightarrow 0$ and $lambda(A_n(x)) ge c lambda (B_{r_n}(x))$ for some $c>0$ independent of $n$.
If you take for example $A_n = [-n^{-2},n^{-2}] times [- n^{-1},n^{-1}]$, then $(A_n)_{n in mathbb{N}}$ has diameter shrinking to zero, but is not shrinking nicelty to zero. (The scaling in the different directions is not propotional!) For example take $$ Delta:= { (x,y) in [-1,1]^2 : |y| le |x| },$$
then $$ lambda(A_n)^{-1} int_{A_n} 1_{Delta} , d lambda^2 = frac{n^3}{4} lambda(Delta cap A_n) = frac{n^3}{4} lambda([-n^{-2},n^{-2}]^2) = frac{1}{n} rightarrow 0,$$
i.e. $ne 1_{Delta}(0,0) =1$.
answered Nov 28 at 10:00
p4sch
4,760217
As already pointed out in the comments you need to take sets which are "shrinking nicely to a point", i.e. $(A_n(x))_{n in mathbb{N}}$ is measurable and satisfies $A_n(x) subset B_{r_n}(x)$ with $r_n rightarrow 0$ and $lambda(A_n(x)) ge c lambda (B_{r_n}(x))$ for some $c>0$ independent of $n$.
If you take for example $A_n = [-n^{-2},n^{-2}] times [- n^{-1},n^{-1}]$, then $(A_n)_{n in mathbb{N}}$ has diameter shrinking to zero, but is not shrinking nicelty to zero. (The scaling in the different directions is not propotional!) For example take $$ Delta:= { (x,y) in [-1,1]^2 : |y| le |x| },$$
then $$ lambda(A_n)^{-1} int_{A_n} 1_{Delta} , d lambda^2 = frac{n^3}{4} lambda(Delta cap A_n) = frac{n^3}{4} lambda([-n^{-2},n^{-2}]^2) = frac{1}{n} rightarrow 0,$$
i.e. $ne 1_{Delta}(0,0) =1$.
answered Nov 28 at 10:00
p4sch
4,760217
answered Nov 28 at 10:00
p4sch
4,760217
answered Nov 28 at 10:00
p4sch
4,760217
answered Nov 28 at 10:00
p4sch
4,760217
4,760217
add a comment |
add a comment |
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You have some misunderstandings. Read about sets ''shrinking nicely to a point'' in Rudin's RCA.
– Kavi Rama Murthy
Nov 24 at 5:52
@KaviRamaMurthy What misunderstandings do I have?
– Keshav Srinivasan
Nov 24 at 7:00