Mixing
Debunking a false proof
$begingroup$
Consider $E = mathbb{R} times {0}$ in $mathbb{R}^2$. Apparaently this set (the x-axis) has measure zero. However, consider the following set
$$O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $$
and $lim_{n to infty} m(O_n) = lim_{n to infty} frac{2}{n} times infty neq 0$, where $m(*)$ is the Lebesgue measure.
Therefore, we don't have $lim_{n to infty} m(O_n) = m(E)$.
Can you debunk my false proof of $lim_{n to infty} m(O_n) = m(E)$?
Since $E$ is Lebesgue measurable, there is an open set $F$ such that $m(Fsetminus E) < epsilon$. Find $n$ such that $O_n subset F$ and $m(O_n) leq m(E) + epsilon$, because $O_n$ can be made arbitrarily close to $E$. $epsilon$ being arbitrary implies $m(O_n) leq m(E)$ and $n to infty$ shows $lim_{n to infty} m(O_n) leq m(E)$.
The other direction $m(O_n) geq m(E)$ is obvious since $O_n supset E$ for all $n$.
What did I miss?
measure-theory lebesgue-measure
asked Feb 1 at 5:45
user1691278user1691278
49939
$endgroup$
add a comment |
$begingroup$
Consider $E = mathbb{R} times {0}$ in $mathbb{R}^2$. Apparaently this set (the x-axis) has measure zero. However, consider the following set
$$O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $$
and $lim_{n to infty} m(O_n) = lim_{n to infty} frac{2}{n} times infty neq 0$, where $m(*)$ is the Lebesgue measure.
Therefore, we don't have $lim_{n to infty} m(O_n) = m(E)$.
Can you debunk my false proof of $lim_{n to infty} m(O_n) = m(E)$?
Since $E$ is Lebesgue measurable, there is an open set $F$ such that $m(Fsetminus E) < epsilon$. Find $n$ such that $O_n subset F$ and $m(O_n) leq m(E) + epsilon$, because $O_n$ can be made arbitrarily close to $E$. $epsilon$ being arbitrary implies $m(O_n) leq m(E)$ and $n to infty$ shows $lim_{n to infty} m(O_n) leq m(E)$.
The other direction $m(O_n) geq m(E)$ is obvious since $O_n supset E$ for all $n$.
What did I miss?
measure-theory lebesgue-measure
asked Feb 1 at 5:45
user1691278user1691278
49939
$endgroup$
$begingroup$
If $O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $ then $O_n$ is not a subset of $Bbb R^2$. Probably you meant $O_n = mathbb{R} times big(-frac{1}{n}, frac{1}{n}big) $???
$endgroup$
– David C. Ullrich
Feb 1 at 13:27
add a comment |
$begingroup$
Consider $E = mathbb{R} times {0}$ in $mathbb{R}^2$. Apparaently this set (the x-axis) has measure zero. However, consider the following set
$$O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $$
and $lim_{n to infty} m(O_n) = lim_{n to infty} frac{2}{n} times infty neq 0$, where $m(*)$ is the Lebesgue measure.
Therefore, we don't have $lim_{n to infty} m(O_n) = m(E)$.
Can you debunk my false proof of $lim_{n to infty} m(O_n) = m(E)$?
Since $E$ is Lebesgue measurable, there is an open set $F$ such that $m(Fsetminus E) < epsilon$. Find $n$ such that $O_n subset F$ and $m(O_n) leq m(E) + epsilon$, because $O_n$ can be made arbitrarily close to $E$. $epsilon$ being arbitrary implies $m(O_n) leq m(E)$ and $n to infty$ shows $lim_{n to infty} m(O_n) leq m(E)$.
The other direction $m(O_n) geq m(E)$ is obvious since $O_n supset E$ for all $n$.
What did I miss?
measure-theory lebesgue-measure
asked Feb 1 at 5:45
user1691278user1691278
49939
$endgroup$
Consider $E = mathbb{R} times {0}$ in $mathbb{R}^2$. Apparaently this set (the x-axis) has measure zero. However, consider the following set
$$O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $$
and $lim_{n to infty} m(O_n) = lim_{n to infty} frac{2}{n} times infty neq 0$, where $m(*)$ is the Lebesgue measure.
Therefore, we don't have $lim_{n to infty} m(O_n) = m(E)$.
Can you debunk my false proof of $lim_{n to infty} m(O_n) = m(E)$?
Since $E$ is Lebesgue measurable, there is an open set $F$ such that $m(Fsetminus E) < epsilon$. Find $n$ such that $O_n subset F$ and $m(O_n) leq m(E) + epsilon$, because $O_n$ can be made arbitrarily close to $E$. $epsilon$ being arbitrary implies $m(O_n) leq m(E)$ and $n to infty$ shows $lim_{n to infty} m(O_n) leq m(E)$.
The other direction $m(O_n) geq m(E)$ is obvious since $O_n supset E$ for all $n$.
What did I miss?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Feb 1 at 5:45
user1691278user1691278
49939
asked Feb 1 at 5:45
user1691278user1691278
49939
edited Feb 1 at 7:21
user1691278
asked Feb 1 at 5:45
user1691278user1691278
49939
asked Feb 1 at 5:45
user1691278user1691278
49939
asked Feb 1 at 5:45
user1691278user1691278
49939
49939
$begingroup$
If $O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $ then $O_n$ is not a subset of $Bbb R^2$. Probably you meant $O_n = mathbb{R} times big(-frac{1}{n}, frac{1}{n}big) $???
$endgroup$
– David C. Ullrich
Feb 1 at 13:27
add a comment |
$begingroup$
If $O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $ then $O_n$ is not a subset of $Bbb R^2$. Probably you meant $O_n = mathbb{R} times big(-frac{1}{n}, frac{1}{n}big) $???
$endgroup$
– David C. Ullrich
Feb 1 at 13:27
$begingroup$
If $O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $ then $O_n$ is not a subset of $Bbb R^2$. Probably you meant $O_n = mathbb{R} times big(-frac{1}{n}, frac{1}{n}big) $???
$endgroup$
– David C. Ullrich
Feb 1 at 13:27
$begingroup$
If $O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $ then $O_n$ is not a subset of $Bbb R^2$. Probably you meant $O_n = mathbb{R} times big(-frac{1}{n}, frac{1}{n}big) $???
$endgroup$
– David C. Ullrich
Feb 1 at 13:27
add a comment |
1 Answer
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$begingroup$
There is not necessarily any $n$ such that $O_nsubset F$, since $F$ might shrink arbitrarily close to $E$ as $x$ grows. For instance, $F$ might be ${(x,y):|y|<1/(x^2+1)}$.
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is not necessarily any $n$ such that $O_nsubset F$, since $F$ might shrink arbitrarily close to $E$ as $x$ grows. For instance, $F$ might be ${(x,y):|y|<1/(x^2+1)}$.
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
$endgroup$
add a comment |
$begingroup$
There is not necessarily any $n$ such that $O_nsubset F$, since $F$ might shrink arbitrarily close to $E$ as $x$ grows. For instance, $F$ might be ${(x,y):|y|<1/(x^2+1)}$.
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
$endgroup$
add a comment |
$begingroup$
There is not necessarily any $n$ such that $O_nsubset F$, since $F$ might shrink arbitrarily close to $E$ as $x$ grows. For instance, $F$ might be ${(x,y):|y|<1/(x^2+1)}$.
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
$endgroup$
There is not necessarily any $n$ such that $O_nsubset F$, since $F$ might shrink arbitrarily close to $E$ as $x$ grows. For instance, $F$ might be ${(x,y):|y|<1/(x^2+1)}$.
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
answered Feb 1 at 6:36
Eric WofseyEric Wofsey
192k14220352
192k14220352
add a comment |
add a comment |
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$begingroup$
If $O_n = big{mathbb{R} times big(-frac{1}{n}, frac{1}{n}big)big} $ then $O_n$ is not a subset of $Bbb R^2$. Probably you meant $O_n = mathbb{R} times big(-frac{1}{n}, frac{1}{n}big) $???
$endgroup$
– David C. Ullrich
Feb 1 at 13:27