Mixing
Outer regularity of Lebesgue measure on $mathbb{R}$
$begingroup$
I have proved that the Lebesgue measure, $lambda$, on $mathbb{R}$ is outer regular. By this, it is meant that for any $lambda$-measurable set, $U$ in $mathbb{R}$, we have that $lambda (U)=inf{lambda(A): U subset A ,A text{ is open}}$. However, in the proof I did not use the fact that the set is $lambda$-measurable. Is this necessary, or is this fact true for all subsets of $mathbb{R}$?
measure-theory
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
$endgroup$
add a comment |
$begingroup$
I have proved that the Lebesgue measure, $lambda$, on $mathbb{R}$ is outer regular. By this, it is meant that for any $lambda$-measurable set, $U$ in $mathbb{R}$, we have that $lambda (U)=inf{lambda(A): U subset A ,A text{ is open}}$. However, in the proof I did not use the fact that the set is $lambda$-measurable. Is this necessary, or is this fact true for all subsets of $mathbb{R}$?
measure-theory
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
$endgroup$
$begingroup$
I realise I may have been unclear, sorry. I mean, is the following true, for any subset of the reals, U, and the outer measure $lambda * : lambda * (U)=inf{lambda *(A): U subset A ,A-open}$
$endgroup$
– Max Sykes
Apr 21 '17 at 18:26
add a comment |
$begingroup$
I have proved that the Lebesgue measure, $lambda$, on $mathbb{R}$ is outer regular. By this, it is meant that for any $lambda$-measurable set, $U$ in $mathbb{R}$, we have that $lambda (U)=inf{lambda(A): U subset A ,A text{ is open}}$. However, in the proof I did not use the fact that the set is $lambda$-measurable. Is this necessary, or is this fact true for all subsets of $mathbb{R}$?
measure-theory
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
$endgroup$
I have proved that the Lebesgue measure, $lambda$, on $mathbb{R}$ is outer regular. By this, it is meant that for any $lambda$-measurable set, $U$ in $mathbb{R}$, we have that $lambda (U)=inf{lambda(A): U subset A ,A text{ is open}}$. However, in the proof I did not use the fact that the set is $lambda$-measurable. Is this necessary, or is this fact true for all subsets of $mathbb{R}$?
measure-theory
measure-theory
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
edited Apr 21 '17 at 18:25
DMcMor
2,74521328
2,74521328
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
asked Apr 21 '17 at 18:18
Max SykesMax Sykes
162
162
$begingroup$
I realise I may have been unclear, sorry. I mean, is the following true, for any subset of the reals, U, and the outer measure $lambda * : lambda * (U)=inf{lambda *(A): U subset A ,A-open}$
$endgroup$
– Max Sykes
Apr 21 '17 at 18:26
add a comment |
$begingroup$
I realise I may have been unclear, sorry. I mean, is the following true, for any subset of the reals, U, and the outer measure $lambda * : lambda * (U)=inf{lambda *(A): U subset A ,A-open}$
$endgroup$
– Max Sykes
Apr 21 '17 at 18:26
$begingroup$
I realise I may have been unclear, sorry. I mean, is the following true, for any subset of the reals, U, and the outer measure $lambda * : lambda * (U)=inf{lambda *(A): U subset A ,A-open}$
$endgroup$
– Max Sykes
Apr 21 '17 at 18:26
$begingroup$
I realise I may have been unclear, sorry. I mean, is the following true, for any subset of the reals, U, and the outer measure $lambda * : lambda * (U)=inf{lambda *(A): U subset A ,A-open}$
$endgroup$
– Max Sykes
Apr 21 '17 at 18:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The proof goes as follows;
Let $U$ be a measurable and let $epsilon > 0$ and we first assume that the outer measure of $U$ is finite. By definition of outer measure we have that $exists {I_k}$ a countable collection of open intervals which covers $U$ such that
$sumlimits_{k=1}^{infty}{m(I_k)} < m^*(U)+epsilon$. Define $O=bigcup_k{I_k}$. Then we have that $O$ is open and $U subset O$.
By monotonicity we have that
$m^*(O) leq m^*(U)+ epsilon$, moreover, $m^*(O)-m^*(U) < epsilon$.
By the measurability of $U$ and it having finite measure, by the excision property we have that $m^*(O-U)=m^*(O)-m^*(U)< epsilon$ hence that $m$ is outer regular.
This process can easily be extrapolated to the case that U has infinite measure by since $(mathbb{R},mathcal{M},m)$ is a sigma finite measure space. (If you would like I am more than willing to fill in the details).
edited Jan 10 at 16:13
amWhy
1
answered Apr 21 '17 at 18:24
ADAADA
1,272518
$endgroup$
$begingroup$
I did not use this property in my proof. So this is what I did: Take a set A. For all $epsilon$ there exists ${(a_n,b_n)}$ such that $A subset U = bigcup (a_n,b_n)$. So we have $ lambda(U) ge lambda(A) ge sum (b_n - a_n) - epsilon ge lambda(U) - epsilon $. The first inequality follows from monotonicity,the second follows from definition of the Lebesgue outer measure and the third follows from monotonicty.
$endgroup$
– Max Sykes
Apr 21 '17 at 18:40
$begingroup$
Yes so what you are concluding after you move you $epsilon $ around that $m(A)-m(U) < epsilon$ and you have by excision that m(A-U)=m(A)-m(U) since U is measurable, which gives you that outer regularity is equivalent to measurability.
$endgroup$
– ADA
Apr 21 '17 at 18:51
$begingroup$
Thanks for your replies my friend, but I believe that this property of excision has not been used. I have found confirmation from an online source as well that we have $lambda * (U)=inf{lambda *(A): U subset A ,A text{ is open}}$ with $lambda * $Lebesgue outer measure and U any subset from the reals. See prop 6.1 in here: math.ksu.edu/~nagy/real-an/3-06-leb-meas.pdf
$endgroup$
– Max Sykes
Apr 21 '17 at 19:42
$begingroup$
Please see edits. You are more than welcome to upload your full solution and I will tell you from that, but from what you wrote in your comment you are implicitly using this property whether you realize it or not.
$endgroup$
– ADA
Apr 21 '17 at 19:53
$begingroup$
Can one one generalise this proof for the lebesgue measure on $mathbb{R}^d$?
$endgroup$
– Viktor Glombik
Jan 10 at 16:10
add a comment |
Your Answer
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$begingroup$
The proof goes as follows;
Let $U$ be a measurable and let $epsilon > 0$ and we first assume that the outer measure of $U$ is finite. By definition of outer measure we have that $exists {I_k}$ a countable collection of open intervals which covers $U$ such that
$sumlimits_{k=1}^{infty}{m(I_k)} < m^*(U)+epsilon$. Define $O=bigcup_k{I_k}$. Then we have that $O$ is open and $U subset O$.
By monotonicity we have that
$m^*(O) leq m^*(U)+ epsilon$, moreover, $m^*(O)-m^*(U) < epsilon$.
By the measurability of $U$ and it having finite measure, by the excision property we have that $m^*(O-U)=m^*(O)-m^*(U)< epsilon$ hence that $m$ is outer regular.
This process can easily be extrapolated to the case that U has infinite measure by since $(mathbb{R},mathcal{M},m)$ is a sigma finite measure space. (If you would like I am more than willing to fill in the details).
edited Jan 10 at 16:13
amWhy
1
answered Apr 21 '17 at 18:24
ADAADA
1,272518
$endgroup$
$begingroup$
I did not use this property in my proof. So this is what I did: Take a set A. For all $epsilon$ there exists ${(a_n,b_n)}$ such that $A subset U = bigcup (a_n,b_n)$. So we have $ lambda(U) ge lambda(A) ge sum (b_n - a_n) - epsilon ge lambda(U) - epsilon $. The first inequality follows from monotonicity,the second follows from definition of the Lebesgue outer measure and the third follows from monotonicty.
$endgroup$
– Max Sykes
Apr 21 '17 at 18:40
$begingroup$
Yes so what you are concluding after you move you $epsilon $ around that $m(A)-m(U) < epsilon$ and you have by excision that m(A-U)=m(A)-m(U) since U is measurable, which gives you that outer regularity is equivalent to measurability.
$endgroup$
– ADA
Apr 21 '17 at 18:51
$begingroup$
Thanks for your replies my friend, but I believe that this property of excision has not been used. I have found confirmation from an online source as well that we have $lambda * (U)=inf{lambda *(A): U subset A ,A text{ is open}}$ with $lambda * $Lebesgue outer measure and U any subset from the reals. See prop 6.1 in here: math.ksu.edu/~nagy/real-an/3-06-leb-meas.pdf
$endgroup$
– Max Sykes
Apr 21 '17 at 19:42
$begingroup$
Please see edits. You are more than welcome to upload your full solution and I will tell you from that, but from what you wrote in your comment you are implicitly using this property whether you realize it or not.
$endgroup$
– ADA
Apr 21 '17 at 19:53
$begingroup$
Can one one generalise this proof for the lebesgue measure on $mathbb{R}^d$?
$endgroup$
– Viktor Glombik
Jan 10 at 16:10
add a comment |
$begingroup$
The proof goes as follows;
Let $U$ be a measurable and let $epsilon > 0$ and we first assume that the outer measure of $U$ is finite. By definition of outer measure we have that $exists {I_k}$ a countable collection of open intervals which covers $U$ such that
$sumlimits_{k=1}^{infty}{m(I_k)} < m^*(U)+epsilon$. Define $O=bigcup_k{I_k}$. Then we have that $O$ is open and $U subset O$.
By monotonicity we have that
$m^*(O) leq m^*(U)+ epsilon$, moreover, $m^*(O)-m^*(U) < epsilon$.
By the measurability of $U$ and it having finite measure, by the excision property we have that $m^*(O-U)=m^*(O)-m^*(U)< epsilon$ hence that $m$ is outer regular.
This process can easily be extrapolated to the case that U has infinite measure by since $(mathbb{R},mathcal{M},m)$ is a sigma finite measure space. (If you would like I am more than willing to fill in the details).
edited Jan 10 at 16:13
amWhy
1
answered Apr 21 '17 at 18:24
ADAADA
1,272518
$endgroup$
$begingroup$
I did not use this property in my proof. So this is what I did: Take a set A. For all $epsilon$ there exists ${(a_n,b_n)}$ such that $A subset U = bigcup (a_n,b_n)$. So we have $ lambda(U) ge lambda(A) ge sum (b_n - a_n) - epsilon ge lambda(U) - epsilon $. The first inequality follows from monotonicity,the second follows from definition of the Lebesgue outer measure and the third follows from monotonicty.
$endgroup$
– Max Sykes
Apr 21 '17 at 18:40
$begingroup$
Yes so what you are concluding after you move you $epsilon $ around that $m(A)-m(U) < epsilon$ and you have by excision that m(A-U)=m(A)-m(U) since U is measurable, which gives you that outer regularity is equivalent to measurability.
$endgroup$
– ADA
Apr 21 '17 at 18:51
$begingroup$
Thanks for your replies my friend, but I believe that this property of excision has not been used. I have found confirmation from an online source as well that we have $lambda * (U)=inf{lambda *(A): U subset A ,A text{ is open}}$ with $lambda * $Lebesgue outer measure and U any subset from the reals. See prop 6.1 in here: math.ksu.edu/~nagy/real-an/3-06-leb-meas.pdf
$endgroup$
– Max Sykes
Apr 21 '17 at 19:42
$begingroup$
Please see edits. You are more than welcome to upload your full solution and I will tell you from that, but from what you wrote in your comment you are implicitly using this property whether you realize it or not.
$endgroup$
– ADA
Apr 21 '17 at 19:53
$begingroup$
Can one one generalise this proof for the lebesgue measure on $mathbb{R}^d$?
$endgroup$
– Viktor Glombik
Jan 10 at 16:10
add a comment |
$begingroup$
The proof goes as follows;
Let $U$ be a measurable and let $epsilon > 0$ and we first assume that the outer measure of $U$ is finite. By definition of outer measure we have that $exists {I_k}$ a countable collection of open intervals which covers $U$ such that
$sumlimits_{k=1}^{infty}{m(I_k)} < m^*(U)+epsilon$. Define $O=bigcup_k{I_k}$. Then we have that $O$ is open and $U subset O$.
By monotonicity we have that
$m^*(O) leq m^*(U)+ epsilon$, moreover, $m^*(O)-m^*(U) < epsilon$.
By the measurability of $U$ and it having finite measure, by the excision property we have that $m^*(O-U)=m^*(O)-m^*(U)< epsilon$ hence that $m$ is outer regular.
This process can easily be extrapolated to the case that U has infinite measure by since $(mathbb{R},mathcal{M},m)$ is a sigma finite measure space. (If you would like I am more than willing to fill in the details).
edited Jan 10 at 16:13
amWhy
1
answered Apr 21 '17 at 18:24
ADAADA
1,272518
$endgroup$
The proof goes as follows;
Let $U$ be a measurable and let $epsilon > 0$ and we first assume that the outer measure of $U$ is finite. By definition of outer measure we have that $exists {I_k}$ a countable collection of open intervals which covers $U$ such that
$sumlimits_{k=1}^{infty}{m(I_k)} < m^*(U)+epsilon$. Define $O=bigcup_k{I_k}$. Then we have that $O$ is open and $U subset O$.
By monotonicity we have that
$m^*(O) leq m^*(U)+ epsilon$, moreover, $m^*(O)-m^*(U) < epsilon$.
By the measurability of $U$ and it having finite measure, by the excision property we have that $m^*(O-U)=m^*(O)-m^*(U)< epsilon$ hence that $m$ is outer regular.
This process can easily be extrapolated to the case that U has infinite measure by since $(mathbb{R},mathcal{M},m)$ is a sigma finite measure space. (If you would like I am more than willing to fill in the details).
edited Jan 10 at 16:13
amWhy
1
answered Apr 21 '17 at 18:24
ADAADA
1,272518
edited Jan 10 at 16:13
amWhy
1
edited Jan 10 at 16:13
amWhy
1
edited Jan 10 at 16:13
amWhy
1
1
answered Apr 21 '17 at 18:24
ADAADA
1,272518
answered Apr 21 '17 at 18:24
ADAADA
1,272518
answered Apr 21 '17 at 18:24
ADAADA
1,272518
1,272518
$begingroup$
I did not use this property in my proof. So this is what I did: Take a set A. For all $epsilon$ there exists ${(a_n,b_n)}$ such that $A subset U = bigcup (a_n,b_n)$. So we have $ lambda(U) ge lambda(A) ge sum (b_n - a_n) - epsilon ge lambda(U) - epsilon $. The first inequality follows from monotonicity,the second follows from definition of the Lebesgue outer measure and the third follows from monotonicty.
$endgroup$
– Max Sykes
Apr 21 '17 at 18:40
$begingroup$
Yes so what you are concluding after you move you $epsilon $ around that $m(A)-m(U) < epsilon$ and you have by excision that m(A-U)=m(A)-m(U) since U is measurable, which gives you that outer regularity is equivalent to measurability.
$endgroup$
– ADA
Apr 21 '17 at 18:51
$begingroup$
Thanks for your replies my friend, but I believe that this property of excision has not been used. I have found confirmation from an online source as well that we have $lambda * (U)=inf{lambda *(A): U subset A ,A text{ is open}}$ with $lambda * $Lebesgue outer measure and U any subset from the reals. See prop 6.1 in here: math.ksu.edu/~nagy/real-an/3-06-leb-meas.pdf
$endgroup$
– Max Sykes
Apr 21 '17 at 19:42
$begingroup$
Please see edits. You are more than welcome to upload your full solution and I will tell you from that, but from what you wrote in your comment you are implicitly using this property whether you realize it or not.
$endgroup$
– ADA
Apr 21 '17 at 19:53
$begingroup$
Can one one generalise this proof for the lebesgue measure on $mathbb{R}^d$?
$endgroup$
– Viktor Glombik
Jan 10 at 16:10
add a comment |
$begingroup$
I did not use this property in my proof. So this is what I did: Take a set A. For all $epsilon$ there exists ${(a_n,b_n)}$ such that $A subset U = bigcup (a_n,b_n)$. So we have $ lambda(U) ge lambda(A) ge sum (b_n - a_n) - epsilon ge lambda(U) - epsilon $. The first inequality follows from monotonicity,the second follows from definition of the Lebesgue outer measure and the third follows from monotonicty.
$endgroup$
– Max Sykes
Apr 21 '17 at 18:40
$begingroup$
Yes so what you are concluding after you move you $epsilon $ around that $m(A)-m(U) < epsilon$ and you have by excision that m(A-U)=m(A)-m(U) since U is measurable, which gives you that outer regularity is equivalent to measurability.
$endgroup$
– ADA
Apr 21 '17 at 18:51
$begingroup$
Thanks for your replies my friend, but I believe that this property of excision has not been used. I have found confirmation from an online source as well that we have $lambda * (U)=inf{lambda *(A): U subset A ,A text{ is open}}$ with $lambda * $Lebesgue outer measure and U any subset from the reals. See prop 6.1 in here: math.ksu.edu/~nagy/real-an/3-06-leb-meas.pdf
$endgroup$
– Max Sykes
Apr 21 '17 at 19:42
$begingroup$
Please see edits. You are more than welcome to upload your full solution and I will tell you from that, but from what you wrote in your comment you are implicitly using this property whether you realize it or not.
$endgroup$
– ADA
Apr 21 '17 at 19:53
$begingroup$
Can one one generalise this proof for the lebesgue measure on $mathbb{R}^d$?
$endgroup$
– Viktor Glombik
Jan 10 at 16:10
$begingroup$
I did not use this property in my proof. So this is what I did: Take a set A. For all $epsilon$ there exists ${(a_n,b_n)}$ such that $A subset U = bigcup (a_n,b_n)$. So we have $ lambda(U) ge lambda(A) ge sum (b_n - a_n) - epsilon ge lambda(U) - epsilon $. The first inequality follows from monotonicity,the second follows from definition of the Lebesgue outer measure and the third follows from monotonicty.
$endgroup$
– Max Sykes
Apr 21 '17 at 18:40
$begingroup$
I did not use this property in my proof. So this is what I did: Take a set A. For all $epsilon$ there exists ${(a_n,b_n)}$ such that $A subset U = bigcup (a_n,b_n)$. So we have $ lambda(U) ge lambda(A) ge sum (b_n - a_n) - epsilon ge lambda(U) - epsilon $. The first inequality follows from monotonicity,the second follows from definition of the Lebesgue outer measure and the third follows from monotonicty.
$endgroup$
– Max Sykes
Apr 21 '17 at 18:40
$begingroup$
Yes so what you are concluding after you move you $epsilon $ around that $m(A)-m(U) < epsilon$ and you have by excision that m(A-U)=m(A)-m(U) since U is measurable, which gives you that outer regularity is equivalent to measurability.
$endgroup$
– ADA
Apr 21 '17 at 18:51
$begingroup$
Yes so what you are concluding after you move you $epsilon $ around that $m(A)-m(U) < epsilon$ and you have by excision that m(A-U)=m(A)-m(U) since U is measurable, which gives you that outer regularity is equivalent to measurability.
$endgroup$
– ADA
Apr 21 '17 at 18:51
$begingroup$
Thanks for your replies my friend, but I believe that this property of excision has not been used. I have found confirmation from an online source as well that we have $lambda * (U)=inf{lambda *(A): U subset A ,A text{ is open}}$ with $lambda * $Lebesgue outer measure and U any subset from the reals. See prop 6.1 in here: math.ksu.edu/~nagy/real-an/3-06-leb-meas.pdf
$endgroup$
– Max Sykes
Apr 21 '17 at 19:42
$begingroup$
Thanks for your replies my friend, but I believe that this property of excision has not been used. I have found confirmation from an online source as well that we have $lambda * (U)=inf{lambda *(A): U subset A ,A text{ is open}}$ with $lambda * $Lebesgue outer measure and U any subset from the reals. See prop 6.1 in here: math.ksu.edu/~nagy/real-an/3-06-leb-meas.pdf
$endgroup$
– Max Sykes
Apr 21 '17 at 19:42
$begingroup$
Please see edits. You are more than welcome to upload your full solution and I will tell you from that, but from what you wrote in your comment you are implicitly using this property whether you realize it or not.
$endgroup$
– ADA
Apr 21 '17 at 19:53
$begingroup$
Please see edits. You are more than welcome to upload your full solution and I will tell you from that, but from what you wrote in your comment you are implicitly using this property whether you realize it or not.
$endgroup$
– ADA
Apr 21 '17 at 19:53
$begingroup$
Can one one generalise this proof for the lebesgue measure on $mathbb{R}^d$?
$endgroup$
– Viktor Glombik
Jan 10 at 16:10
$begingroup$
Can one one generalise this proof for the lebesgue measure on $mathbb{R}^d$?
$endgroup$
– Viktor Glombik
Jan 10 at 16:10
add a comment |
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$begingroup$
I realise I may have been unclear, sorry. I mean, is the following true, for any subset of the reals, U, and the outer measure $lambda * : lambda * (U)=inf{lambda *(A): U subset A ,A-open}$
$endgroup$
– Max Sykes
Apr 21 '17 at 18:26