Is there an outer measure on $mathbb R$ whose only measurable sets are $mathbb R, emptyset$?
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I.e. an outer measure $mu^*$ on the reals such that if for some $A subset mathbb R$, we have that for all $B subset R$ $mu^*(B) = mu^*(B cap A) + mu^*(B setminus A)$, then $A=mathbb R$ or $A = emptyset$.
measure-theory outer-measure
asked Dec 8 '18 at 11:18
D GD G
1628
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add a comment |
$begingroup$
I.e. an outer measure $mu^*$ on the reals such that if for some $A subset mathbb R$, we have that for all $B subset R$ $mu^*(B) = mu^*(B cap A) + mu^*(B setminus A)$, then $A=mathbb R$ or $A = emptyset$.
measure-theory outer-measure
asked Dec 8 '18 at 11:18
D GD G
1628
$endgroup$
$begingroup$
Suggestion: $mu$ is usually used for measures and $mu^star$ for outer measures. So it will make less confusion if you add that $^star$.
$endgroup$
– Yanko
Dec 8 '18 at 11:21
add a comment |
$begingroup$
I.e. an outer measure $mu^*$ on the reals such that if for some $A subset mathbb R$, we have that for all $B subset R$ $mu^*(B) = mu^*(B cap A) + mu^*(B setminus A)$, then $A=mathbb R$ or $A = emptyset$.
measure-theory outer-measure
asked Dec 8 '18 at 11:18
D GD G
1628
$endgroup$
I.e. an outer measure $mu^*$ on the reals such that if for some $A subset mathbb R$, we have that for all $B subset R$ $mu^*(B) = mu^*(B cap A) + mu^*(B setminus A)$, then $A=mathbb R$ or $A = emptyset$.
measure-theory outer-measure
measure-theory outer-measure
asked Dec 8 '18 at 11:18
D GD G
1628
asked Dec 8 '18 at 11:18
D GD G
1628
edited Dec 8 '18 at 11:30
D G
asked Dec 8 '18 at 11:18
D GD G
1628
asked Dec 8 '18 at 11:18
D GD G
1628
asked Dec 8 '18 at 11:18
D GD G
1628
1628
$begingroup$
Suggestion: $mu$ is usually used for measures and $mu^star$ for outer measures. So it will make less confusion if you add that $^star$.
$endgroup$
– Yanko
Dec 8 '18 at 11:21
add a comment |
$begingroup$
Suggestion: $mu$ is usually used for measures and $mu^star$ for outer measures. So it will make less confusion if you add that $^star$.
$endgroup$
– Yanko
Dec 8 '18 at 11:21
$begingroup$
Suggestion: $mu$ is usually used for measures and $mu^star$ for outer measures. So it will make less confusion if you add that $^star$.
$endgroup$
– Yanko
Dec 8 '18 at 11:21
$begingroup$
Suggestion: $mu$ is usually used for measures and $mu^star$ for outer measures. So it will make less confusion if you add that $^star$.
$endgroup$
– Yanko
Dec 8 '18 at 11:21
add a comment |
1 Answer
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$begingroup$
Define $mu^*$ by $mu^*(emptyset) = 0$ and $mu^*(A) = 1$ for all $A ne emptyset$. It is trivial to check that this is an outer measure.
Now if $A$ is a set which is neither $emptyset$ nor $mathbb{R}$, taking $B = mathbb{R}$, we have that $B, Bcap A$ and $B setminus A$ are all nonempty, so $mu^*(B) = 1$ and $mu^*(B cap A) + mu^*(B setminus A) = 2$. Hence $A$ is not measurable.
The same works if you replace $mathbb{R}$ by any other set.
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
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1 Answer
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$begingroup$
Define $mu^*$ by $mu^*(emptyset) = 0$ and $mu^*(A) = 1$ for all $A ne emptyset$. It is trivial to check that this is an outer measure.
Now if $A$ is a set which is neither $emptyset$ nor $mathbb{R}$, taking $B = mathbb{R}$, we have that $B, Bcap A$ and $B setminus A$ are all nonempty, so $mu^*(B) = 1$ and $mu^*(B cap A) + mu^*(B setminus A) = 2$. Hence $A$ is not measurable.
The same works if you replace $mathbb{R}$ by any other set.
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
$endgroup$
add a comment |
$begingroup$
Define $mu^*$ by $mu^*(emptyset) = 0$ and $mu^*(A) = 1$ for all $A ne emptyset$. It is trivial to check that this is an outer measure.
Now if $A$ is a set which is neither $emptyset$ nor $mathbb{R}$, taking $B = mathbb{R}$, we have that $B, Bcap A$ and $B setminus A$ are all nonempty, so $mu^*(B) = 1$ and $mu^*(B cap A) + mu^*(B setminus A) = 2$. Hence $A$ is not measurable.
The same works if you replace $mathbb{R}$ by any other set.
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
$endgroup$
add a comment |
$begingroup$
Define $mu^*$ by $mu^*(emptyset) = 0$ and $mu^*(A) = 1$ for all $A ne emptyset$. It is trivial to check that this is an outer measure.
Now if $A$ is a set which is neither $emptyset$ nor $mathbb{R}$, taking $B = mathbb{R}$, we have that $B, Bcap A$ and $B setminus A$ are all nonempty, so $mu^*(B) = 1$ and $mu^*(B cap A) + mu^*(B setminus A) = 2$. Hence $A$ is not measurable.
The same works if you replace $mathbb{R}$ by any other set.
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
$endgroup$
Define $mu^*$ by $mu^*(emptyset) = 0$ and $mu^*(A) = 1$ for all $A ne emptyset$. It is trivial to check that this is an outer measure.
Now if $A$ is a set which is neither $emptyset$ nor $mathbb{R}$, taking $B = mathbb{R}$, we have that $B, Bcap A$ and $B setminus A$ are all nonempty, so $mu^*(B) = 1$ and $mu^*(B cap A) + mu^*(B setminus A) = 2$. Hence $A$ is not measurable.
The same works if you replace $mathbb{R}$ by any other set.
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
edited Dec 8 '18 at 17:31
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
answered Dec 8 '18 at 17:23
Nate EldredgeNate Eldredge
64.2k682174
64.2k682174
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$begingroup$
Suggestion: $mu$ is usually used for measures and $mu^star$ for outer measures. So it will make less confusion if you add that $^star$.
$endgroup$
– Yanko
Dec 8 '18 at 11:21