Determine $lambda^{2}({(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]})$
$begingroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
$endgroup$
add a comment |
$begingroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
$endgroup$
add a comment |
$begingroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
$endgroup$
Define $M:={(x,y) in mathbb R^{2}: x in mathbb Qcap [0,1], y in [0,1]}$
Find $lambda^{2}(M)$
Ideas:
Well, first I should show that $M in mathcal{B}^2$
Note that for $x in mathbb Q cap [0,1]$
${x}times[0,1]inmathcal{B}^2$
And so it follows: $(mathbb Q cap [0,1])times[0,1]=bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1]$ which is $in mathcal{B}^2$
Now onto $lambda^{2}(M)$:
As proven above:
$lambda^{2}(M)=lambda^{2}(bigcup_{x in mathbb Q cap [0,1]}{x}times[0,1])=sum_{x in mathbb Q cap [0,1]}lambda^{2}({x}times[0,1])$
I assume that ${x}times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $dim ({x}times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A in mathcal{B}^{2}$, with $lambda^{2}(A)=c,$ where $c in mathbb R_{>0}$
what can I say about $lambda(A)$ as well as $lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $lambda^{3}(A)=0$ as $A=Atimes {0}in mathcal{B}^3$ is a hyperplane in $mathbb R^{3}$, is it not?
real-analysis measure-theory lebesgue-integral lebesgue-measure
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
asked Jan 13 at 13:58
MinaThumaMinaThuma
798
798
add a comment |
add a comment |
1 Answer
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$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
$endgroup$
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
$endgroup$
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
$endgroup$
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
$begingroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
$endgroup$
For the first question, ${x}times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane ${x}timesBbb R$ in $Bbb R^2$. The dimension of ${x}timesBbb R$ is $1$.
Regarding the second part (I'm assuming that $lambda^n$ denotes the Lebesgue measure on $Bbb R^n$), if $lambda^n(A) >0$ then we have $lambda^m(A)=0$ for all $m>n$
because $A$ would be contained in a hyperplane of $Bbb R^m$
.
On the other hand, $lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $Bbb R^n$ which is not contained in $Bbb R^r$.
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
answered Jan 13 at 14:08
BigbearZzzBigbearZzz
8,58221652
8,58221652
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
$begingroup$
Thank you for the clarification: Is ${x}times mathbb R$ a hyperplane in $mathbb R^{2}$ because $dim mathbb R=1$ and $(x,0)+mathbb R={x}times mathbb R$?
$endgroup$
– MinaThuma
Jan 13 at 14:29
1
1
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
$begingroup$
Basically yes, but note that in your expression $(x,0)+mathbb R$ here $Bbb R$ should be identified with ${ (0,y): yinBbb R }$.
$endgroup$
– BigbearZzz
Jan 13 at 14:32
add a comment |
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