Prove that a set of norm p is of measure zero
$begingroup$
The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.
($||x||_p$ is the $p$ -norm).
I know i need to prove that the boundary of $A$ is of measure $0$.
And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.
I started by declaring the function $f(x) = (||x||_p)^p -1$.
So $B=$ {$x in R^n || f(x)=0$}.
My direction was to prove that the graph of the function is of measure zero
real-analysis measure-theory multivariable-calculus
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
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add a comment |
$begingroup$
The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.
($||x||_p$ is the $p$ -norm).
I know i need to prove that the boundary of $A$ is of measure $0$.
And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.
I started by declaring the function $f(x) = (||x||_p)^p -1$.
So $B=$ {$x in R^n || f(x)=0$}.
My direction was to prove that the graph of the function is of measure zero
real-analysis measure-theory multivariable-calculus
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
$endgroup$
add a comment |
$begingroup$
The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.
($||x||_p$ is the $p$ -norm).
I know i need to prove that the boundary of $A$ is of measure $0$.
And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.
I started by declaring the function $f(x) = (||x||_p)^p -1$.
So $B=$ {$x in R^n || f(x)=0$}.
My direction was to prove that the graph of the function is of measure zero
real-analysis measure-theory multivariable-calculus
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
$endgroup$
The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.
($||x||_p$ is the $p$ -norm).
I know i need to prove that the boundary of $A$ is of measure $0$.
And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.
I started by declaring the function $f(x) = (||x||_p)^p -1$.
So $B=$ {$x in R^n || f(x)=0$}.
My direction was to prove that the graph of the function is of measure zero
real-analysis measure-theory multivariable-calculus
real-analysis measure-theory multivariable-calculus
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
asked Dec 20 '18 at 11:28
Gabi GGabi G
463110
463110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
$endgroup$
$begingroup$
Nice answer, but I am looking for something that will use the graph of the function I wrote
$endgroup$
– Gabi G
Dec 20 '18 at 14:52
add a comment |
$begingroup$
To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$
Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$
Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
$endgroup$
$begingroup$
Nice answer, but I am looking for something that will use the graph of the function I wrote
$endgroup$
– Gabi G
Dec 20 '18 at 14:52
add a comment |
$begingroup$
The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
$endgroup$
$begingroup$
Nice answer, but I am looking for something that will use the graph of the function I wrote
$endgroup$
– Gabi G
Dec 20 '18 at 14:52
add a comment |
$begingroup$
The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
$endgroup$
The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
answered Dec 20 '18 at 12:45
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,607917
6,607917
$begingroup$
Nice answer, but I am looking for something that will use the graph of the function I wrote
$endgroup$
– Gabi G
Dec 20 '18 at 14:52
add a comment |
$begingroup$
Nice answer, but I am looking for something that will use the graph of the function I wrote
$endgroup$
– Gabi G
Dec 20 '18 at 14:52
$begingroup$
Nice answer, but I am looking for something that will use the graph of the function I wrote
$endgroup$
– Gabi G
Dec 20 '18 at 14:52
$begingroup$
Nice answer, but I am looking for something that will use the graph of the function I wrote
$endgroup$
– Gabi G
Dec 20 '18 at 14:52
add a comment |
$begingroup$
To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$
Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$
Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
$endgroup$
add a comment |
$begingroup$
To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$
Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$
Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
$endgroup$
add a comment |
$begingroup$
To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$
Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$
Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
$endgroup$
To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$
Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$
Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
edited Dec 20 '18 at 15:34
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
answered Dec 20 '18 at 15:19
MatematletaMatematleta
11.6k2920
11.6k2920
add a comment |
add a comment |
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