Mixing
Sets of measure zero and smooth functions
Clash Royale CLAN TAG#URR8PPP
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I want to prove:
Let $U$ be an open subset of $mathbbR^n$ and $F:Uto mathbbR^n$ a $C^infty$ function. Let $S$ be a subset of $U$ of zero-measure. Then $F(S)$ has zero-measure.
Proof
We can find a countable open cover of $U$, say $mathfrakB$, such that for every $Vin mathfrakB$ we have $overlineVsubseteq U$, and $V$ is an open ball.
Since countable union of subsets of zero-measure has zero-measure, then it is sufficient to prove that $F(Scap V)$ has zero-measure for every $V in mathfrakB$.
We can suppose that $S subseteq V$.
Since $S$ has zero-measure, for every $epsilon >0$ there is $R_j$ countable family of $n-$cubes such that $$Ssubseteqbigcup_j R_j, quad textand quadsum_j textVol(R_j)<epsilon$$
Let $l_j>0$ be the lenght of the side of the $n-$cube $R_j$. We can inscribe $R_j$ in an open ball $B_j$ of radius $r_j=sqrtnfracl_j2$.
Since $F$ is $C^1$ in $overlineV$, then $F$ is Lipschitz in $overlineV$, i.e. there is $c>0$ such that $$lVert F(x)-F(y)rVert leq c lVert x-y rVert$$ for every $x,y in overlineV$.
My book says that we can inscribe $F(B_j)$ in an open ball, say $B'_j$, of radius $r'_j=cr_j$.
My objections are:
1) I don't know if $B_jsubseteq U$ (right?), so I should write $F(B_jcap U)$.
2) If I want to use the Lipschitz of $F$ than I should write $F(B_j cap overlineV)$, right?
3) I don't know if the center of $B_j$ lies in $overlineV$ (right?), so shouldn't be $r'_j=2cr_j$?
Take $x in B_jcap overlineV$ and let $B'_j$ be the $n-$open ball of center $F(x)$ and radius $r'_j=2cr_j$. If $y$ is any point in $B_jcap overlineV$ then we have $lVert x-y rVert <2 r_j$ and then $$lVert F(x)-F(y)rVert leq c lVert x-y rVert <2cr_j$$ so $F(y)in B_j'$, so $F(B_j cap overlineV) subseteq B_j'$. Or is there any way in which it is true that we can inscribe $F(B_j)$ (or, better, $F(B_j cap overlineV)$) in an open ball of radius $r'_j=cr_j$?
For the sake of completeness I will finish the proof.
We can inscribe $B'_j$ in an open $n-$ cube $R'_j$ with length of the side $l'_j=2r'_j=4cr_j=2cl_jsqrtn$. Then we have $$F(S)subseteq bigcup_j R'_j quad textand quad sum_J textVol (R'_j)<2^n c^n n^n/2epsilon $$
real-analysis general-topology smooth-manifolds
asked 5 hours ago
Minato
16910
add a comment |Â
up vote
1
down vote
favorite
I want to prove:
Let $U$ be an open subset of $mathbbR^n$ and $F:Uto mathbbR^n$ a $C^infty$ function. Let $S$ be a subset of $U$ of zero-measure. Then $F(S)$ has zero-measure.
Proof
We can find a countable open cover of $U$, say $mathfrakB$, such that for every $Vin mathfrakB$ we have $overlineVsubseteq U$, and $V$ is an open ball.
Since countable union of subsets of zero-measure has zero-measure, then it is sufficient to prove that $F(Scap V)$ has zero-measure for every $V in mathfrakB$.
We can suppose that $S subseteq V$.
Since $S$ has zero-measure, for every $epsilon >0$ there is $R_j$ countable family of $n-$cubes such that $$Ssubseteqbigcup_j R_j, quad textand quadsum_j textVol(R_j)<epsilon$$
Let $l_j>0$ be the lenght of the side of the $n-$cube $R_j$. We can inscribe $R_j$ in an open ball $B_j$ of radius $r_j=sqrtnfracl_j2$.
Since $F$ is $C^1$ in $overlineV$, then $F$ is Lipschitz in $overlineV$, i.e. there is $c>0$ such that $$lVert F(x)-F(y)rVert leq c lVert x-y rVert$$ for every $x,y in overlineV$.
My book says that we can inscribe $F(B_j)$ in an open ball, say $B'_j$, of radius $r'_j=cr_j$.
My objections are:
1) I don't know if $B_jsubseteq U$ (right?), so I should write $F(B_jcap U)$.
2) If I want to use the Lipschitz of $F$ than I should write $F(B_j cap overlineV)$, right?
3) I don't know if the center of $B_j$ lies in $overlineV$ (right?), so shouldn't be $r'_j=2cr_j$?
Take $x in B_jcap overlineV$ and let $B'_j$ be the $n-$open ball of center $F(x)$ and radius $r'_j=2cr_j$. If $y$ is any point in $B_jcap overlineV$ then we have $lVert x-y rVert <2 r_j$ and then $$lVert F(x)-F(y)rVert leq c lVert x-y rVert <2cr_j$$ so $F(y)in B_j'$, so $F(B_j cap overlineV) subseteq B_j'$. Or is there any way in which it is true that we can inscribe $F(B_j)$ (or, better, $F(B_j cap overlineV)$) in an open ball of radius $r'_j=cr_j$?
For the sake of completeness I will finish the proof.
We can inscribe $B'_j$ in an open $n-$ cube $R'_j$ with length of the side $l'_j=2r'_j=4cr_j=2cl_jsqrtn$. Then we have $$F(S)subseteq bigcup_j R'_j quad textand quad sum_J textVol (R'_j)<2^n c^n n^n/2epsilon $$
real-analysis general-topology smooth-manifolds
asked 5 hours ago
Minato
16910
It's "measure" not "misure"
– zhw.
4 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to prove:
Let $U$ be an open subset of $mathbbR^n$ and $F:Uto mathbbR^n$ a $C^infty$ function. Let $S$ be a subset of $U$ of zero-measure. Then $F(S)$ has zero-measure.
Proof
We can find a countable open cover of $U$, say $mathfrakB$, such that for every $Vin mathfrakB$ we have $overlineVsubseteq U$, and $V$ is an open ball.
Since countable union of subsets of zero-measure has zero-measure, then it is sufficient to prove that $F(Scap V)$ has zero-measure for every $V in mathfrakB$.
We can suppose that $S subseteq V$.
Since $S$ has zero-measure, for every $epsilon >0$ there is $R_j$ countable family of $n-$cubes such that $$Ssubseteqbigcup_j R_j, quad textand quadsum_j textVol(R_j)<epsilon$$
Let $l_j>0$ be the lenght of the side of the $n-$cube $R_j$. We can inscribe $R_j$ in an open ball $B_j$ of radius $r_j=sqrtnfracl_j2$.
Since $F$ is $C^1$ in $overlineV$, then $F$ is Lipschitz in $overlineV$, i.e. there is $c>0$ such that $$lVert F(x)-F(y)rVert leq c lVert x-y rVert$$ for every $x,y in overlineV$.
My book says that we can inscribe $F(B_j)$ in an open ball, say $B'_j$, of radius $r'_j=cr_j$.
My objections are:
1) I don't know if $B_jsubseteq U$ (right?), so I should write $F(B_jcap U)$.
2) If I want to use the Lipschitz of $F$ than I should write $F(B_j cap overlineV)$, right?
3) I don't know if the center of $B_j$ lies in $overlineV$ (right?), so shouldn't be $r'_j=2cr_j$?
Take $x in B_jcap overlineV$ and let $B'_j$ be the $n-$open ball of center $F(x)$ and radius $r'_j=2cr_j$. If $y$ is any point in $B_jcap overlineV$ then we have $lVert x-y rVert <2 r_j$ and then $$lVert F(x)-F(y)rVert leq c lVert x-y rVert <2cr_j$$ so $F(y)in B_j'$, so $F(B_j cap overlineV) subseteq B_j'$. Or is there any way in which it is true that we can inscribe $F(B_j)$ (or, better, $F(B_j cap overlineV)$) in an open ball of radius $r'_j=cr_j$?
For the sake of completeness I will finish the proof.
We can inscribe $B'_j$ in an open $n-$ cube $R'_j$ with length of the side $l'_j=2r'_j=4cr_j=2cl_jsqrtn$. Then we have $$F(S)subseteq bigcup_j R'_j quad textand quad sum_J textVol (R'_j)<2^n c^n n^n/2epsilon $$
real-analysis general-topology smooth-manifolds
asked 5 hours ago
Minato
16910
I want to prove:
Let $U$ be an open subset of $mathbbR^n$ and $F:Uto mathbbR^n$ a $C^infty$ function. Let $S$ be a subset of $U$ of zero-measure. Then $F(S)$ has zero-measure.
Proof
We can find a countable open cover of $U$, say $mathfrakB$, such that for every $Vin mathfrakB$ we have $overlineVsubseteq U$, and $V$ is an open ball.
Since countable union of subsets of zero-measure has zero-measure, then it is sufficient to prove that $F(Scap V)$ has zero-measure for every $V in mathfrakB$.
We can suppose that $S subseteq V$.
Since $S$ has zero-measure, for every $epsilon >0$ there is $R_j$ countable family of $n-$cubes such that $$Ssubseteqbigcup_j R_j, quad textand quadsum_j textVol(R_j)<epsilon$$
Let $l_j>0$ be the lenght of the side of the $n-$cube $R_j$. We can inscribe $R_j$ in an open ball $B_j$ of radius $r_j=sqrtnfracl_j2$.
Since $F$ is $C^1$ in $overlineV$, then $F$ is Lipschitz in $overlineV$, i.e. there is $c>0$ such that $$lVert F(x)-F(y)rVert leq c lVert x-y rVert$$ for every $x,y in overlineV$.
My book says that we can inscribe $F(B_j)$ in an open ball, say $B'_j$, of radius $r'_j=cr_j$.
My objections are:
1) I don't know if $B_jsubseteq U$ (right?), so I should write $F(B_jcap U)$.
2) If I want to use the Lipschitz of $F$ than I should write $F(B_j cap overlineV)$, right?
3) I don't know if the center of $B_j$ lies in $overlineV$ (right?), so shouldn't be $r'_j=2cr_j$?
Take $x in B_jcap overlineV$ and let $B'_j$ be the $n-$open ball of center $F(x)$ and radius $r'_j=2cr_j$. If $y$ is any point in $B_jcap overlineV$ then we have $lVert x-y rVert <2 r_j$ and then $$lVert F(x)-F(y)rVert leq c lVert x-y rVert <2cr_j$$ so $F(y)in B_j'$, so $F(B_j cap overlineV) subseteq B_j'$. Or is there any way in which it is true that we can inscribe $F(B_j)$ (or, better, $F(B_j cap overlineV)$) in an open ball of radius $r'_j=cr_j$?
For the sake of completeness I will finish the proof.
We can inscribe $B'_j$ in an open $n-$ cube $R'_j$ with length of the side $l'_j=2r'_j=4cr_j=2cl_jsqrtn$. Then we have $$F(S)subseteq bigcup_j R'_j quad textand quad sum_J textVol (R'_j)<2^n c^n n^n/2epsilon $$
real-analysis general-topology smooth-manifolds
asked 5 hours ago
Minato
16910
edited 12 mins ago
asked 5 hours ago
Minato
16910
asked 5 hours ago
Minato
16910
asked 5 hours ago
Minato
16910
16910
It's "measure" not "misure"
– zhw.
4 hours ago
add a comment |Â
It's "measure" not "misure"
– zhw.
4 hours ago
It's "measure" not "misure"
– zhw.
4 hours ago
It's "measure" not "misure"
– zhw.
4 hours ago
add a comment |Â
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It's "measure" not "misure"
– zhw.
4 hours ago