Mixing
Show that Lebesgue measure is translation-invariant.
0
$begingroup$
Let $A subset mathbb{R^n}$ and $x in mathbb{R^n}$ and $A + x= :{u: u - x in A}$. If $A$ is measurable, then $A + x$ is also measurable and $mu(A) = mu(A+x)$. This means that the Lebesgue measure is translation-invariant.
How can I start the proof? Any help is appreciated.
Thank you!
measure-theory lebesgue-measure
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
asked Jan 9 at 16:51
bozcanbozcan
25818
$endgroup$
add a comment |
0
$begingroup$
Let $A subset mathbb{R^n}$ and $x in mathbb{R^n}$ and $A + x= :{u: u - x in A}$. If $A$ is measurable, then $A + x$ is also measurable and $mu(A) = mu(A+x)$. This means that the Lebesgue measure is translation-invariant.
How can I start the proof? Any help is appreciated.
Thank you!
measure-theory lebesgue-measure
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
asked Jan 9 at 16:51
bozcanbozcan
25818
$endgroup$
$begingroup$
How did you define Lebesgue measure in the first place? If it's the restriction of the Lebesgue outer measure to the measurable sets, then you can pretty much just expand the definition of the Lebesgue outer measure (convert a cover of $A$ into a cover of $A+x$ by just translating it), and everything will work out.
$endgroup$
– Ian
Jan 9 at 16:55
$begingroup$
this is almost immediate from the Lebesgue outer measure
$endgroup$
– Masacroso
Jan 9 at 16:55
$begingroup$
There are quite a few elementary proofs of this but one way to avoid the proof from scratch is to observe that $Emapsto x+E$ defines a measure $nu$ on the algebra $mathscr A$ of half-open cubes in $mathbb R^n$, that agrees with the Lebesgue measure on $mathscr A$, hence on all of $mathscr B(mathbb R^n)$. The result now follows because it also is $0$ on the Lebesgue -null sets.
$endgroup$
– Matematleta
Jan 9 at 17:50
add a comment |
0
0
0
$begingroup$
Let $A subset mathbb{R^n}$ and $x in mathbb{R^n}$ and $A + x= :{u: u - x in A}$. If $A$ is measurable, then $A + x$ is also measurable and $mu(A) = mu(A+x)$. This means that the Lebesgue measure is translation-invariant.
How can I start the proof? Any help is appreciated.
Thank you!
measure-theory lebesgue-measure
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
asked Jan 9 at 16:51
bozcanbozcan
25818
$endgroup$
Let $A subset mathbb{R^n}$ and $x in mathbb{R^n}$ and $A + x= :{u: u - x in A}$. If $A$ is measurable, then $A + x$ is also measurable and $mu(A) = mu(A+x)$. This means that the Lebesgue measure is translation-invariant.
How can I start the proof? Any help is appreciated.
Thank you!
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
asked Jan 9 at 16:51
bozcanbozcan
25818
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
asked Jan 9 at 16:51
bozcanbozcan
25818
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
edited Jan 9 at 16:55
Giuseppe Negro
17k330124
17k330124
asked Jan 9 at 16:51
bozcanbozcan
25818
asked Jan 9 at 16:51
bozcanbozcan
25818
asked Jan 9 at 16:51
bozcanbozcan
25818
25818
$begingroup$
How did you define Lebesgue measure in the first place? If it's the restriction of the Lebesgue outer measure to the measurable sets, then you can pretty much just expand the definition of the Lebesgue outer measure (convert a cover of $A$ into a cover of $A+x$ by just translating it), and everything will work out.
$endgroup$
– Ian
Jan 9 at 16:55
$begingroup$
this is almost immediate from the Lebesgue outer measure
$endgroup$
– Masacroso
Jan 9 at 16:55
$begingroup$
There are quite a few elementary proofs of this but one way to avoid the proof from scratch is to observe that $Emapsto x+E$ defines a measure $nu$ on the algebra $mathscr A$ of half-open cubes in $mathbb R^n$, that agrees with the Lebesgue measure on $mathscr A$, hence on all of $mathscr B(mathbb R^n)$. The result now follows because it also is $0$ on the Lebesgue -null sets.
$endgroup$
– Matematleta
Jan 9 at 17:50
add a comment |
$begingroup$
How did you define Lebesgue measure in the first place? If it's the restriction of the Lebesgue outer measure to the measurable sets, then you can pretty much just expand the definition of the Lebesgue outer measure (convert a cover of $A$ into a cover of $A+x$ by just translating it), and everything will work out.
$endgroup$
– Ian
Jan 9 at 16:55
$begingroup$
this is almost immediate from the Lebesgue outer measure
$endgroup$
– Masacroso
Jan 9 at 16:55
$begingroup$
There are quite a few elementary proofs of this but one way to avoid the proof from scratch is to observe that $Emapsto x+E$ defines a measure $nu$ on the algebra $mathscr A$ of half-open cubes in $mathbb R^n$, that agrees with the Lebesgue measure on $mathscr A$, hence on all of $mathscr B(mathbb R^n)$. The result now follows because it also is $0$ on the Lebesgue -null sets.
$endgroup$
– Matematleta
Jan 9 at 17:50
$begingroup$
How did you define Lebesgue measure in the first place? If it's the restriction of the Lebesgue outer measure to the measurable sets, then you can pretty much just expand the definition of the Lebesgue outer measure (convert a cover of $A$ into a cover of $A+x$ by just translating it), and everything will work out.
$endgroup$
– Ian
Jan 9 at 16:55
$begingroup$
How did you define Lebesgue measure in the first place? If it's the restriction of the Lebesgue outer measure to the measurable sets, then you can pretty much just expand the definition of the Lebesgue outer measure (convert a cover of $A$ into a cover of $A+x$ by just translating it), and everything will work out.
$endgroup$
– Ian
Jan 9 at 16:55
$begingroup$
this is almost immediate from the Lebesgue outer measure
$endgroup$
– Masacroso
Jan 9 at 16:55
$begingroup$
this is almost immediate from the Lebesgue outer measure
$endgroup$
– Masacroso
Jan 9 at 16:55
$begingroup$
There are quite a few elementary proofs of this but one way to avoid the proof from scratch is to observe that $Emapsto x+E$ defines a measure $nu$ on the algebra $mathscr A$ of half-open cubes in $mathbb R^n$, that agrees with the Lebesgue measure on $mathscr A$, hence on all of $mathscr B(mathbb R^n)$. The result now follows because it also is $0$ on the Lebesgue -null sets.
$endgroup$
– Matematleta
Jan 9 at 17:50
$begingroup$
There are quite a few elementary proofs of this but one way to avoid the proof from scratch is to observe that $Emapsto x+E$ defines a measure $nu$ on the algebra $mathscr A$ of half-open cubes in $mathbb R^n$, that agrees with the Lebesgue measure on $mathscr A$, hence on all of $mathscr B(mathbb R^n)$. The result now follows because it also is $0$ on the Lebesgue -null sets.
$endgroup$
– Matematleta
Jan 9 at 17:50
add a comment |
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$begingroup$
How did you define Lebesgue measure in the first place? If it's the restriction of the Lebesgue outer measure to the measurable sets, then you can pretty much just expand the definition of the Lebesgue outer measure (convert a cover of $A$ into a cover of $A+x$ by just translating it), and everything will work out.
$endgroup$
– Ian
Jan 9 at 16:55
$begingroup$
this is almost immediate from the Lebesgue outer measure
$endgroup$
– Masacroso
Jan 9 at 16:55
$begingroup$
There are quite a few elementary proofs of this but one way to avoid the proof from scratch is to observe that $Emapsto x+E$ defines a measure $nu$ on the algebra $mathscr A$ of half-open cubes in $mathbb R^n$, that agrees with the Lebesgue measure on $mathscr A$, hence on all of $mathscr B(mathbb R^n)$. The result now follows because it also is $0$ on the Lebesgue -null sets.
$endgroup$
– Matematleta
Jan 9 at 17:50