Open balls in $mathbb{R}^d$ are Jordan Measurable
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I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.
Show that an open Euclidean
ball $B(x, r) := {y in mathbb{R}^d
: |y − x| < r}$ in $mathbb{R}^d$ is Jordan measurable, with Jordan
measure $c_d r^d$
for some constant $c_d > 0$ depending only on
$d$.
Is there an elementary way to approach this problem?
real-analysis
asked Jan 1 at 17:18
user82261user82261
23517
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add a comment |
$begingroup$
I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.
Show that an open Euclidean
ball $B(x, r) := {y in mathbb{R}^d
: |y − x| < r}$ in $mathbb{R}^d$ is Jordan measurable, with Jordan
measure $c_d r^d$
for some constant $c_d > 0$ depending only on
$d$.
Is there an elementary way to approach this problem?
real-analysis
asked Jan 1 at 17:18
user82261user82261
23517
$endgroup$
add a comment |
$begingroup$
I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.
Show that an open Euclidean
ball $B(x, r) := {y in mathbb{R}^d
: |y − x| < r}$ in $mathbb{R}^d$ is Jordan measurable, with Jordan
measure $c_d r^d$
for some constant $c_d > 0$ depending only on
$d$.
Is there an elementary way to approach this problem?
real-analysis
asked Jan 1 at 17:18
user82261user82261
23517
$endgroup$
I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.
Show that an open Euclidean
ball $B(x, r) := {y in mathbb{R}^d
: |y − x| < r}$ in $mathbb{R}^d$ is Jordan measurable, with Jordan
measure $c_d r^d$
for some constant $c_d > 0$ depending only on
$d$.
Is there an elementary way to approach this problem?
real-analysis
real-analysis
asked Jan 1 at 17:18
user82261user82261
23517
asked Jan 1 at 17:18
user82261user82261
23517
asked Jan 1 at 17:18
user82261user82261
23517
asked Jan 1 at 17:18
user82261user82261
23517
asked Jan 1 at 17:18
user82261user82261
23517
23517
add a comment |
add a comment |
1 Answer
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$begingroup$
I think it's a fairly involved calculation to prove this from scratch (using covers).
On the other hand, we have
$1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),
$2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),
$3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$
so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and
$c(B)=int 1_B=C_npi{(n/2)}r^n.$
Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.
Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.
So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:
Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.
A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.
The result follows.
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
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add a comment |
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$begingroup$
I think it's a fairly involved calculation to prove this from scratch (using covers).
On the other hand, we have
$1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),
$2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),
$3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$
so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and
$c(B)=int 1_B=C_npi{(n/2)}r^n.$
Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.
Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.
So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:
Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.
A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.
The result follows.
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
I think it's a fairly involved calculation to prove this from scratch (using covers).
On the other hand, we have
$1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),
$2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),
$3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$
so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and
$c(B)=int 1_B=C_npi{(n/2)}r^n.$
Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.
Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.
So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:
Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.
A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.
The result follows.
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
I think it's a fairly involved calculation to prove this from scratch (using covers).
On the other hand, we have
$1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),
$2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),
$3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$
so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and
$c(B)=int 1_B=C_npi{(n/2)}r^n.$
Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.
Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.
So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:
Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.
A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.
The result follows.
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
$endgroup$
I think it's a fairly involved calculation to prove this from scratch (using covers).
On the other hand, we have
$1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),
$2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),
$3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$
so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and
$c(B)=int 1_B=C_npi{(n/2)}r^n.$
Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.
Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.
So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:
Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.
A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.
The result follows.
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
edited Jan 2 at 5:04
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
answered Jan 1 at 23:29
MatematletaMatematleta
11.5k2920
11.5k2920
add a comment |
add a comment |
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