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A piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$

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0

$begingroup$

Definitions:

• A piecewise smooth curve in $mathbbR^2$ is a subset of $mathbbR^2$ that can be written as $boldsymbolvarphi_1[a_1,b_1] cup boldsymbolvarphi_2[a_2,b_2]cup dotscup boldsymbolvarphi_N [a_N,b_N]$, with each $boldsymbolvarphi_i:[a_i,b_i] to mathbbR^2$ a parameterization of a smooth curve, and where each end point $boldsymbolvarphi_i(b_i)$ corresponds with the next starting point $boldsymbolvarphi_i+1(a_i+1)$.



• 
  

  A subset $X subseteq mathbbR^2$ is a null set in $mathbbR^2$ if for each $varepsilon > 0$ there is a sequence $R_1,R_2,dots$ of rectangles with sides parallel to the coordinate axes such that

  
  
  
  1. $X subseteq R_1 cup R_2 cup dots$
  
  
  2. 
     $|R_1|+|R_2| + dots < varepsilon$, with $|R_i|$ the area of rectangle $R_i$.
  
  

My notes proceed with stating that a piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$. I would like to understand to intuiton behind this theorem. Is there an easily understandable explanation behind it?

Thanks in advance.

multivariable-calculus

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

$endgroup$

add a comment | 

0

$begingroup$

Definitions:

• A piecewise smooth curve in $mathbbR^2$ is a subset of $mathbbR^2$ that can be written as $boldsymbolvarphi_1[a_1,b_1] cup boldsymbolvarphi_2[a_2,b_2]cup dotscup boldsymbolvarphi_N [a_N,b_N]$, with each $boldsymbolvarphi_i:[a_i,b_i] to mathbbR^2$ a parameterization of a smooth curve, and where each end point $boldsymbolvarphi_i(b_i)$ corresponds with the next starting point $boldsymbolvarphi_i+1(a_i+1)$.



• 
  

  A subset $X subseteq mathbbR^2$ is a null set in $mathbbR^2$ if for each $varepsilon > 0$ there is a sequence $R_1,R_2,dots$ of rectangles with sides parallel to the coordinate axes such that

  
  
  
  1. $X subseteq R_1 cup R_2 cup dots$
  
  
  2. 
     $|R_1|+|R_2| + dots < varepsilon$, with $|R_i|$ the area of rectangle $R_i$.
  
  

My notes proceed with stating that a piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$. I would like to understand to intuiton behind this theorem. Is there an easily understandable explanation behind it?

Thanks in advance.

multivariable-calculus

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

$endgroup$

add a comment | 

$begingroup$

Definitions:

• A piecewise smooth curve in $mathbbR^2$ is a subset of $mathbbR^2$ that can be written as $boldsymbolvarphi_1[a_1,b_1] cup boldsymbolvarphi_2[a_2,b_2]cup dotscup boldsymbolvarphi_N [a_N,b_N]$, with each $boldsymbolvarphi_i:[a_i,b_i] to mathbbR^2$ a parameterization of a smooth curve, and where each end point $boldsymbolvarphi_i(b_i)$ corresponds with the next starting point $boldsymbolvarphi_i+1(a_i+1)$.



• 
  

  A subset $X subseteq mathbbR^2$ is a null set in $mathbbR^2$ if for each $varepsilon > 0$ there is a sequence $R_1,R_2,dots$ of rectangles with sides parallel to the coordinate axes such that

  
  
  
  1. $X subseteq R_1 cup R_2 cup dots$
  
  
  2. 
     $|R_1|+|R_2| + dots < varepsilon$, with $|R_i|$ the area of rectangle $R_i$.
  
  

My notes proceed with stating that a piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$. I would like to understand to intuiton behind this theorem. Is there an easily understandable explanation behind it?

Thanks in advance.

multivariable-calculus

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

$endgroup$

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

asked yesterday

ZacharyZachary

1799

asked yesterday

ZacharyZachary

1799

1 Answer
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oldest

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0

$begingroup$

The following is not a proof; it just helps the intuition.

It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.

A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.

One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

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0

$begingroup$

The following is not a proof; it just helps the intuition.

It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.

A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.

One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

$endgroup$

add a comment | 

0

$begingroup$

The following is not a proof; it just helps the intuition.

It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.

A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.

One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

$endgroup$

add a comment | 

$begingroup$

The following is not a proof; it just helps the intuition.

It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.

A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.

One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

$endgroup$

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

answered yesterday

Christian BlatterChristian Blatter

175k8115327

answered yesterday

Christian BlatterChristian Blatter

175k8115327