Let $E subset mathbb{R}^n$ be a convex set. Prove there exists an open set $A subset mathbb{R}^n$ equivalent...
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Let $E subset mathbb{R}^n$ be a convex set. I need to prove there exists an open set $A subset mathbb{R}^n$ equivalent to $E$, that means such $mathcal{L}^{n} (A triangle E)= 0$, where $E triangle A = (E setminus A ) cup ( A setminus E) $.
My idea: I know that for all $epsilon >0$ there exist an open set $A supset E$ such that $mathcal{L}^{n} (A setminus E) < epsilon$, but it doesn't help because, in this way, the measure of the symmetric difference $A triangle E$ is less than $epsilon$ but it could be grater than $0$.
My second idea: If I set $A := Int(E)$ then $A$ is open and it holds $mathcal{L}^{n} (A triangle E)= mathcal{L}^{n} (E setminus A) leq mathcal{L}^{n} (partial E)$ (in the case $E$ is closed we have the equality). But in general we do NOT have $mathcal{L}^{n}( partial E) = 0$.
In both attempts I haven't used the fact that $E$ is convext, but I don't know how to use it. Does this fact imply that $mathcal{L}^{n}( partial E) = 0$? If it is true, do you know a simple proof of it?
Thank you
general-topology convex-analysis lebesgue-measure
asked Nov 22 at 14:56
Hermione
599
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Let $E subset mathbb{R}^n$ be a convex set. I need to prove there exists an open set $A subset mathbb{R}^n$ equivalent to $E$, that means such $mathcal{L}^{n} (A triangle E)= 0$, where $E triangle A = (E setminus A ) cup ( A setminus E) $.
My idea: I know that for all $epsilon >0$ there exist an open set $A supset E$ such that $mathcal{L}^{n} (A setminus E) < epsilon$, but it doesn't help because, in this way, the measure of the symmetric difference $A triangle E$ is less than $epsilon$ but it could be grater than $0$.
My second idea: If I set $A := Int(E)$ then $A$ is open and it holds $mathcal{L}^{n} (A triangle E)= mathcal{L}^{n} (E setminus A) leq mathcal{L}^{n} (partial E)$ (in the case $E$ is closed we have the equality). But in general we do NOT have $mathcal{L}^{n}( partial E) = 0$.
In both attempts I haven't used the fact that $E$ is convext, but I don't know how to use it. Does this fact imply that $mathcal{L}^{n}( partial E) = 0$? If it is true, do you know a simple proof of it?
Thank you
general-topology convex-analysis lebesgue-measure
asked Nov 22 at 14:56
Hermione
599
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $E subset mathbb{R}^n$ be a convex set. I need to prove there exists an open set $A subset mathbb{R}^n$ equivalent to $E$, that means such $mathcal{L}^{n} (A triangle E)= 0$, where $E triangle A = (E setminus A ) cup ( A setminus E) $.
My idea: I know that for all $epsilon >0$ there exist an open set $A supset E$ such that $mathcal{L}^{n} (A setminus E) < epsilon$, but it doesn't help because, in this way, the measure of the symmetric difference $A triangle E$ is less than $epsilon$ but it could be grater than $0$.
My second idea: If I set $A := Int(E)$ then $A$ is open and it holds $mathcal{L}^{n} (A triangle E)= mathcal{L}^{n} (E setminus A) leq mathcal{L}^{n} (partial E)$ (in the case $E$ is closed we have the equality). But in general we do NOT have $mathcal{L}^{n}( partial E) = 0$.
In both attempts I haven't used the fact that $E$ is convext, but I don't know how to use it. Does this fact imply that $mathcal{L}^{n}( partial E) = 0$? If it is true, do you know a simple proof of it?
Thank you
general-topology convex-analysis lebesgue-measure
asked Nov 22 at 14:56
Hermione
599
Let $E subset mathbb{R}^n$ be a convex set. I need to prove there exists an open set $A subset mathbb{R}^n$ equivalent to $E$, that means such $mathcal{L}^{n} (A triangle E)= 0$, where $E triangle A = (E setminus A ) cup ( A setminus E) $.
My idea: I know that for all $epsilon >0$ there exist an open set $A supset E$ such that $mathcal{L}^{n} (A setminus E) < epsilon$, but it doesn't help because, in this way, the measure of the symmetric difference $A triangle E$ is less than $epsilon$ but it could be grater than $0$.
My second idea: If I set $A := Int(E)$ then $A$ is open and it holds $mathcal{L}^{n} (A triangle E)= mathcal{L}^{n} (E setminus A) leq mathcal{L}^{n} (partial E)$ (in the case $E$ is closed we have the equality). But in general we do NOT have $mathcal{L}^{n}( partial E) = 0$.
In both attempts I haven't used the fact that $E$ is convext, but I don't know how to use it. Does this fact imply that $mathcal{L}^{n}( partial E) = 0$? If it is true, do you know a simple proof of it?
Thank you
general-topology convex-analysis lebesgue-measure
general-topology convex-analysis lebesgue-measure
asked Nov 22 at 14:56
Hermione
599
asked Nov 22 at 14:56
Hermione
599
asked Nov 22 at 14:56
Hermione
599
asked Nov 22 at 14:56
Hermione
599
asked Nov 22 at 14:56
Hermione
599
599
add a comment |
add a comment |
1 Answer
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Claim: if $E$ is convex, then $mathcal{L}^n(partial E) = 0$.
Proof: Suppose otherwise. Then there is $x in partial E$ such that $limsup_{r downarrow 0} frac{|partial E cap B_r(x)|}{|B_r(x)|} = 1$. That is, $partial E$ contains most points of small balls around $x$. But by convexity, this means $E$ contains a small ball around $x$. This contradicts $x in partial E$.
answered Nov 22 at 15:05
mathworker21
8,2761827
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Claim: if $E$ is convex, then $mathcal{L}^n(partial E) = 0$.
Proof: Suppose otherwise. Then there is $x in partial E$ such that $limsup_{r downarrow 0} frac{|partial E cap B_r(x)|}{|B_r(x)|} = 1$. That is, $partial E$ contains most points of small balls around $x$. But by convexity, this means $E$ contains a small ball around $x$. This contradicts $x in partial E$.
answered Nov 22 at 15:05
mathworker21
8,2761827
add a comment |
up vote
0
down vote
accepted
Claim: if $E$ is convex, then $mathcal{L}^n(partial E) = 0$.
Proof: Suppose otherwise. Then there is $x in partial E$ such that $limsup_{r downarrow 0} frac{|partial E cap B_r(x)|}{|B_r(x)|} = 1$. That is, $partial E$ contains most points of small balls around $x$. But by convexity, this means $E$ contains a small ball around $x$. This contradicts $x in partial E$.
answered Nov 22 at 15:05
mathworker21
8,2761827
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Claim: if $E$ is convex, then $mathcal{L}^n(partial E) = 0$.
Proof: Suppose otherwise. Then there is $x in partial E$ such that $limsup_{r downarrow 0} frac{|partial E cap B_r(x)|}{|B_r(x)|} = 1$. That is, $partial E$ contains most points of small balls around $x$. But by convexity, this means $E$ contains a small ball around $x$. This contradicts $x in partial E$.
answered Nov 22 at 15:05
mathworker21
8,2761827
Claim: if $E$ is convex, then $mathcal{L}^n(partial E) = 0$.
Proof: Suppose otherwise. Then there is $x in partial E$ such that $limsup_{r downarrow 0} frac{|partial E cap B_r(x)|}{|B_r(x)|} = 1$. That is, $partial E$ contains most points of small balls around $x$. But by convexity, this means $E$ contains a small ball around $x$. This contradicts $x in partial E$.
answered Nov 22 at 15:05
mathworker21
8,2761827
answered Nov 22 at 15:05
mathworker21
8,2761827
answered Nov 22 at 15:05
mathworker21
8,2761827
answered Nov 22 at 15:05
mathworker21
8,2761827
8,2761827
add a comment |
add a comment |
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