Mixing
Is there a non-zero measure on $mathbb{R}^2$ such that all subsets of $mathbb{R}^2$ are measurable?
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1) Everywhere one can find many things about Lebesgue measure on $mathbb{R}^2$.
But what about arbitary measure? Is there a non-zero measure on $mathbb{R}^2$ such that all subsets of $mathbb{R}^2$ are measurable?
2) Another variant of the question:
Is there a measure on $mathbb{R}^2$ such that the measure of a unit square is one, the measure is invariant under isometries and all subsets of $mathbb{R}^2$ are measurable?
EDITED: I miss one important condition that a mesure of any bounded set should be finite.
measure-theory
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
$endgroup$
|
show 1 more comment
$begingroup$
1) Everywhere one can find many things about Lebesgue measure on $mathbb{R}^2$.
But what about arbitary measure? Is there a non-zero measure on $mathbb{R}^2$ such that all subsets of $mathbb{R}^2$ are measurable?
2) Another variant of the question:
Is there a measure on $mathbb{R}^2$ such that the measure of a unit square is one, the measure is invariant under isometries and all subsets of $mathbb{R}^2$ are measurable?
EDITED: I miss one important condition that a mesure of any bounded set should be finite.
measure-theory
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
$endgroup$
1
$begingroup$
To answer 1): Let $m(varnothing)=0$ and $m(A)=infty$ for all $Aneqvarnothing$.
$endgroup$
– Christoph
Jan 21 at 11:21
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2) can be easily answered using an analogue of Vitali set.
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– Wojowu
Jan 21 at 11:23
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Okay, I mean something non-trivial for the question 1.
$endgroup$
– Alex-omsk
Jan 21 at 11:30
2
$begingroup$
As non-trivial measure for 1), there is the counting measure
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– Bermudes
Jan 21 at 11:31
2
$begingroup$
An answer to 1) with unit square having measure $1$ would be any dirac measure of a point in the unit square. (Let $p$ be such a point and $m(A)=1$ iff $pin A$, $0$ otherwise.)
$endgroup$
– Christoph
Jan 21 at 11:36
|
show 1 more comment
$begingroup$
1) Everywhere one can find many things about Lebesgue measure on $mathbb{R}^2$.
But what about arbitary measure? Is there a non-zero measure on $mathbb{R}^2$ such that all subsets of $mathbb{R}^2$ are measurable?
2) Another variant of the question:
Is there a measure on $mathbb{R}^2$ such that the measure of a unit square is one, the measure is invariant under isometries and all subsets of $mathbb{R}^2$ are measurable?
EDITED: I miss one important condition that a mesure of any bounded set should be finite.
measure-theory
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
$endgroup$
1) Everywhere one can find many things about Lebesgue measure on $mathbb{R}^2$.
But what about arbitary measure? Is there a non-zero measure on $mathbb{R}^2$ such that all subsets of $mathbb{R}^2$ are measurable?
2) Another variant of the question:
Is there a measure on $mathbb{R}^2$ such that the measure of a unit square is one, the measure is invariant under isometries and all subsets of $mathbb{R}^2$ are measurable?
EDITED: I miss one important condition that a mesure of any bounded set should be finite.
measure-theory
measure-theory
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
edited Jan 21 at 11:36
Alex-omsk
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
asked Jan 21 at 11:18
Alex-omskAlex-omsk
1558
1558
1
$begingroup$
To answer 1): Let $m(varnothing)=0$ and $m(A)=infty$ for all $Aneqvarnothing$.
$endgroup$
– Christoph
Jan 21 at 11:21
$begingroup$
2) can be easily answered using an analogue of Vitali set.
$endgroup$
– Wojowu
Jan 21 at 11:23
$begingroup$
Okay, I mean something non-trivial for the question 1.
$endgroup$
– Alex-omsk
Jan 21 at 11:30
2
$begingroup$
As non-trivial measure for 1), there is the counting measure
$endgroup$
– Bermudes
Jan 21 at 11:31
2
$begingroup$
An answer to 1) with unit square having measure $1$ would be any dirac measure of a point in the unit square. (Let $p$ be such a point and $m(A)=1$ iff $pin A$, $0$ otherwise.)
$endgroup$
– Christoph
Jan 21 at 11:36
|
show 1 more comment
1
$begingroup$
To answer 1): Let $m(varnothing)=0$ and $m(A)=infty$ for all $Aneqvarnothing$.
$endgroup$
– Christoph
Jan 21 at 11:21
$begingroup$
2) can be easily answered using an analogue of Vitali set.
$endgroup$
– Wojowu
Jan 21 at 11:23
$begingroup$
Okay, I mean something non-trivial for the question 1.
$endgroup$
– Alex-omsk
Jan 21 at 11:30
2
$begingroup$
As non-trivial measure for 1), there is the counting measure
$endgroup$
– Bermudes
Jan 21 at 11:31
2
$begingroup$
An answer to 1) with unit square having measure $1$ would be any dirac measure of a point in the unit square. (Let $p$ be such a point and $m(A)=1$ iff $pin A$, $0$ otherwise.)
$endgroup$
– Christoph
Jan 21 at 11:36
1
1
$begingroup$
To answer 1): Let $m(varnothing)=0$ and $m(A)=infty$ for all $Aneqvarnothing$.
$endgroup$
– Christoph
Jan 21 at 11:21
$begingroup$
To answer 1): Let $m(varnothing)=0$ and $m(A)=infty$ for all $Aneqvarnothing$.
$endgroup$
– Christoph
Jan 21 at 11:21
$begingroup$
2) can be easily answered using an analogue of Vitali set.
$endgroup$
– Wojowu
Jan 21 at 11:23
$begingroup$
2) can be easily answered using an analogue of Vitali set.
$endgroup$
– Wojowu
Jan 21 at 11:23
$begingroup$
Okay, I mean something non-trivial for the question 1.
$endgroup$
– Alex-omsk
Jan 21 at 11:30
$begingroup$
Okay, I mean something non-trivial for the question 1.
$endgroup$
– Alex-omsk
Jan 21 at 11:30
2
2
$begingroup$
As non-trivial measure for 1), there is the counting measure
$endgroup$
– Bermudes
Jan 21 at 11:31
$begingroup$
As non-trivial measure for 1), there is the counting measure
$endgroup$
– Bermudes
Jan 21 at 11:31
2
2
$begingroup$
An answer to 1) with unit square having measure $1$ would be any dirac measure of a point in the unit square. (Let $p$ be such a point and $m(A)=1$ iff $pin A$, $0$ otherwise.)
$endgroup$
– Christoph
Jan 21 at 11:36
$begingroup$
An answer to 1) with unit square having measure $1$ would be any dirac measure of a point in the unit square. (Let $p$ be such a point and $m(A)=1$ iff $pin A$, $0$ otherwise.)
$endgroup$
– Christoph
Jan 21 at 11:36
|
show 1 more comment
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1
$begingroup$
To answer 1): Let $m(varnothing)=0$ and $m(A)=infty$ for all $Aneqvarnothing$.
$endgroup$
– Christoph
Jan 21 at 11:21
$begingroup$
2) can be easily answered using an analogue of Vitali set.
$endgroup$
– Wojowu
Jan 21 at 11:23
$begingroup$
Okay, I mean something non-trivial for the question 1.
$endgroup$
– Alex-omsk
Jan 21 at 11:30
2
$begingroup$
As non-trivial measure for 1), there is the counting measure
$endgroup$
– Bermudes
Jan 21 at 11:31
2
$begingroup$
An answer to 1) with unit square having measure $1$ would be any dirac measure of a point in the unit square. (Let $p$ be such a point and $m(A)=1$ iff $pin A$, $0$ otherwise.)
$endgroup$
– Christoph
Jan 21 at 11:36