Proof of monotonicity of Lebesgue measure by contradiction
I want to prove the claim:
$bf{Proposition}$ If $Esubset F subset mathbb{R}^d$, then $m^*(E)leq m^*(F)$, where $m^*(S)$ denotes Lebesque outer measure of set $S$.
I tried to prove it by using "proof of contradiction" as below. Could you check this?
$bf{Proof}$Assume, for the sake of contradiction, that $Esubset FRightarrow m^*(E)>m^*(F)$. Due to the monotonicity of the Jordan outer measure, we can say $m^{*,(J)}(E)<m^{*,(J)}(F)$. Due to the relation of Jordan and Lebsegue outer measure, we can say $m^{*}(E)leq m^{*,(J)}(E)$. Hence, using the assumption, $m^*(F)<m^*(E)leq m^{*,(J)} leq m^{*,(J)}(F)$. However, this contradicts the relation of Jordan and Lebseque outer measure for $F$: $m^*(F)leq m^{*,(J)}(F)$, so the claim follows. $square$
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 26 at 17:49
orematasaburou
345
add a comment |
I want to prove the claim:
$bf{Proposition}$ If $Esubset F subset mathbb{R}^d$, then $m^*(E)leq m^*(F)$, where $m^*(S)$ denotes Lebesque outer measure of set $S$.
I tried to prove it by using "proof of contradiction" as below. Could you check this?
$bf{Proof}$Assume, for the sake of contradiction, that $Esubset FRightarrow m^*(E)>m^*(F)$. Due to the monotonicity of the Jordan outer measure, we can say $m^{*,(J)}(E)<m^{*,(J)}(F)$. Due to the relation of Jordan and Lebsegue outer measure, we can say $m^{*}(E)leq m^{*,(J)}(E)$. Hence, using the assumption, $m^*(F)<m^*(E)leq m^{*,(J)} leq m^{*,(J)}(F)$. However, this contradicts the relation of Jordan and Lebseque outer measure for $F$: $m^*(F)leq m^{*,(J)}(F)$, so the claim follows. $square$
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 26 at 17:49
orematasaburou
345
add a comment |
I want to prove the claim:
$bf{Proposition}$ If $Esubset F subset mathbb{R}^d$, then $m^*(E)leq m^*(F)$, where $m^*(S)$ denotes Lebesque outer measure of set $S$.
I tried to prove it by using "proof of contradiction" as below. Could you check this?
$bf{Proof}$Assume, for the sake of contradiction, that $Esubset FRightarrow m^*(E)>m^*(F)$. Due to the monotonicity of the Jordan outer measure, we can say $m^{*,(J)}(E)<m^{*,(J)}(F)$. Due to the relation of Jordan and Lebsegue outer measure, we can say $m^{*}(E)leq m^{*,(J)}(E)$. Hence, using the assumption, $m^*(F)<m^*(E)leq m^{*,(J)} leq m^{*,(J)}(F)$. However, this contradicts the relation of Jordan and Lebseque outer measure for $F$: $m^*(F)leq m^{*,(J)}(F)$, so the claim follows. $square$
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 26 at 17:49
orematasaburou
345
I want to prove the claim:
$bf{Proposition}$ If $Esubset F subset mathbb{R}^d$, then $m^*(E)leq m^*(F)$, where $m^*(S)$ denotes Lebesque outer measure of set $S$.
I tried to prove it by using "proof of contradiction" as below. Could you check this?
$bf{Proof}$Assume, for the sake of contradiction, that $Esubset FRightarrow m^*(E)>m^*(F)$. Due to the monotonicity of the Jordan outer measure, we can say $m^{*,(J)}(E)<m^{*,(J)}(F)$. Due to the relation of Jordan and Lebsegue outer measure, we can say $m^{*}(E)leq m^{*,(J)}(E)$. Hence, using the assumption, $m^*(F)<m^*(E)leq m^{*,(J)} leq m^{*,(J)}(F)$. However, this contradicts the relation of Jordan and Lebseque outer measure for $F$: $m^*(F)leq m^{*,(J)}(F)$, so the claim follows. $square$
real-analysis measure-theory proof-verification lebesgue-measure
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 26 at 17:49
orematasaburou
345
asked Nov 26 at 17:49
orematasaburou
345
asked Nov 26 at 17:49
orematasaburou
345
asked Nov 26 at 17:49
orematasaburou
345
asked Nov 26 at 17:49
orematasaburou
345
345
add a comment |
add a comment |
1 Answer
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Here are some critiques:
1) You didn't negate the statement correctly. You should assume for the sake of contradiction that there exist sets $E,F$ with $E subset F$ and $m^*(E) > m^*(F)$. There is no implication.
2) I think the proper term for $m^{*,(J)}(E)$ is the Jordan outer content.
3) The best you can do with the Jordan outer content is $m^{*,(J)}(E) le m^{*,(J)}(F)$. You don't have strict inequality.
4) You concluded that $m^*(F) < m^{*,(J)}(F)$. This is in no way whatsoever contradictory with $m^*(F) le m^{*,(J)}(F)$.
For example, $1 < 2$ does not contradict $1 le 2$.
Points 1-3 are minor and easily corrected. Point 4 is serious and can't be easily corrected.
answered Nov 26 at 18:20
Umberto P.
38.5k13064
I somehow didn't even noticed point 4. I ashamed of myself, and want to delete this question.
– orematasaburou
Nov 26 at 18:27
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Here are some critiques:
1) You didn't negate the statement correctly. You should assume for the sake of contradiction that there exist sets $E,F$ with $E subset F$ and $m^*(E) > m^*(F)$. There is no implication.
2) I think the proper term for $m^{*,(J)}(E)$ is the Jordan outer content.
3) The best you can do with the Jordan outer content is $m^{*,(J)}(E) le m^{*,(J)}(F)$. You don't have strict inequality.
4) You concluded that $m^*(F) < m^{*,(J)}(F)$. This is in no way whatsoever contradictory with $m^*(F) le m^{*,(J)}(F)$.
For example, $1 < 2$ does not contradict $1 le 2$.
Points 1-3 are minor and easily corrected. Point 4 is serious and can't be easily corrected.
answered Nov 26 at 18:20
Umberto P.
38.5k13064
I somehow didn't even noticed point 4. I ashamed of myself, and want to delete this question.
– orematasaburou
Nov 26 at 18:27
add a comment |
Here are some critiques:
1) You didn't negate the statement correctly. You should assume for the sake of contradiction that there exist sets $E,F$ with $E subset F$ and $m^*(E) > m^*(F)$. There is no implication.
2) I think the proper term for $m^{*,(J)}(E)$ is the Jordan outer content.
3) The best you can do with the Jordan outer content is $m^{*,(J)}(E) le m^{*,(J)}(F)$. You don't have strict inequality.
4) You concluded that $m^*(F) < m^{*,(J)}(F)$. This is in no way whatsoever contradictory with $m^*(F) le m^{*,(J)}(F)$.
For example, $1 < 2$ does not contradict $1 le 2$.
Points 1-3 are minor and easily corrected. Point 4 is serious and can't be easily corrected.
answered Nov 26 at 18:20
Umberto P.
38.5k13064
I somehow didn't even noticed point 4. I ashamed of myself, and want to delete this question.
– orematasaburou
Nov 26 at 18:27
add a comment |
Here are some critiques:
1) You didn't negate the statement correctly. You should assume for the sake of contradiction that there exist sets $E,F$ with $E subset F$ and $m^*(E) > m^*(F)$. There is no implication.
2) I think the proper term for $m^{*,(J)}(E)$ is the Jordan outer content.
3) The best you can do with the Jordan outer content is $m^{*,(J)}(E) le m^{*,(J)}(F)$. You don't have strict inequality.
4) You concluded that $m^*(F) < m^{*,(J)}(F)$. This is in no way whatsoever contradictory with $m^*(F) le m^{*,(J)}(F)$.
For example, $1 < 2$ does not contradict $1 le 2$.
Points 1-3 are minor and easily corrected. Point 4 is serious and can't be easily corrected.
answered Nov 26 at 18:20
Umberto P.
38.5k13064
Here are some critiques:
1) You didn't negate the statement correctly. You should assume for the sake of contradiction that there exist sets $E,F$ with $E subset F$ and $m^*(E) > m^*(F)$. There is no implication.
2) I think the proper term for $m^{*,(J)}(E)$ is the Jordan outer content.
3) The best you can do with the Jordan outer content is $m^{*,(J)}(E) le m^{*,(J)}(F)$. You don't have strict inequality.
4) You concluded that $m^*(F) < m^{*,(J)}(F)$. This is in no way whatsoever contradictory with $m^*(F) le m^{*,(J)}(F)$.
For example, $1 < 2$ does not contradict $1 le 2$.
Points 1-3 are minor and easily corrected. Point 4 is serious and can't be easily corrected.
answered Nov 26 at 18:20
Umberto P.
38.5k13064
answered Nov 26 at 18:20
Umberto P.
38.5k13064
answered Nov 26 at 18:20
Umberto P.
38.5k13064
answered Nov 26 at 18:20
Umberto P.
38.5k13064
38.5k13064
I somehow didn't even noticed point 4. I ashamed of myself, and want to delete this question.
– orematasaburou
Nov 26 at 18:27
add a comment |
I somehow didn't even noticed point 4. I ashamed of myself, and want to delete this question.
– orematasaburou
Nov 26 at 18:27
I somehow didn't even noticed point 4. I ashamed of myself, and want to delete this question.
– orematasaburou
Nov 26 at 18:27
I somehow didn't even noticed point 4. I ashamed of myself, and want to delete this question.
– orematasaburou
Nov 26 at 18:27
add a comment |
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