Mixing
Show that there exits an open interval $(a,b)$ such that $m(Ecap(a,b)) > frac{1}{2}m(a,b) > 0$ when...
$begingroup$
Let $E subset [0,1]$ be a non-zero measurable set. Show that there exits an open interval $(a,b)subset [0,1]$ such that $m(Ecap(a,b)) > frac{1}{2}m(a,b) > 0$.
Proof: I will do it by contradiction and I want a proof verification.
Assume that
$m(E) > 0, mbox{ and } forall (a,b) mbox{ open interval } m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$
Consider $(a,b) = (0,1)$, then
$$ m(Ecap(0,1)) = m(E) > 0$$
That would contradict that $m(E) leq 0$
I think that there is a mistake in my negation of the inequality because every measure has to be positive. How can I approach this ?
measure-theory proof-verification lebesgue-measure measurable-sets
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
$endgroup$
add a comment |
$begingroup$
Let $E subset [0,1]$ be a non-zero measurable set. Show that there exits an open interval $(a,b)subset [0,1]$ such that $m(Ecap(a,b)) > frac{1}{2}m(a,b) > 0$.
Proof: I will do it by contradiction and I want a proof verification.
Assume that
$m(E) > 0, mbox{ and } forall (a,b) mbox{ open interval } m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$
Consider $(a,b) = (0,1)$, then
$$ m(Ecap(0,1)) = m(E) > 0$$
That would contradict that $m(E) leq 0$
I think that there is a mistake in my negation of the inequality because every measure has to be positive. How can I approach this ?
measure-theory proof-verification lebesgue-measure measurable-sets
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
$endgroup$
$begingroup$
In the contradiction, you should indeed say $m(Ecap (a,b)) leq frac{1}{2}m(a,b)$. But not $m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$ as the measure is non negative.
$endgroup$
– mathcounterexamples.net
Jan 18 at 13:24
$begingroup$
Yeah, the proof is incorrect. Remember that "$u<v<w$" is shorthand for "$u<v$ and $v<w$." The negation is "$ugeq v$ or $vgeq w$."
$endgroup$
– Ben W
Jan 18 at 13:24
$begingroup$
Yeah, can I choose $E = (0,2/3)$ then it contradicts that the measure is less than $1/2$ because it's $2/3$ ?
$endgroup$
– Richard Clare
Jan 18 at 13:27
add a comment |
$begingroup$
Let $E subset [0,1]$ be a non-zero measurable set. Show that there exits an open interval $(a,b)subset [0,1]$ such that $m(Ecap(a,b)) > frac{1}{2}m(a,b) > 0$.
Proof: I will do it by contradiction and I want a proof verification.
Assume that
$m(E) > 0, mbox{ and } forall (a,b) mbox{ open interval } m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$
Consider $(a,b) = (0,1)$, then
$$ m(Ecap(0,1)) = m(E) > 0$$
That would contradict that $m(E) leq 0$
I think that there is a mistake in my negation of the inequality because every measure has to be positive. How can I approach this ?
measure-theory proof-verification lebesgue-measure measurable-sets
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
$endgroup$
Let $E subset [0,1]$ be a non-zero measurable set. Show that there exits an open interval $(a,b)subset [0,1]$ such that $m(Ecap(a,b)) > frac{1}{2}m(a,b) > 0$.
Proof: I will do it by contradiction and I want a proof verification.
Assume that
$m(E) > 0, mbox{ and } forall (a,b) mbox{ open interval } m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$
Consider $(a,b) = (0,1)$, then
$$ m(Ecap(0,1)) = m(E) > 0$$
That would contradict that $m(E) leq 0$
I think that there is a mistake in my negation of the inequality because every measure has to be positive. How can I approach this ?
measure-theory proof-verification lebesgue-measure measurable-sets
measure-theory proof-verification lebesgue-measure measurable-sets
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
asked Jan 18 at 13:17
Richard ClareRichard Clare
1,076314
1,076314
$begingroup$
In the contradiction, you should indeed say $m(Ecap (a,b)) leq frac{1}{2}m(a,b)$. But not $m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$ as the measure is non negative.
$endgroup$
– mathcounterexamples.net
Jan 18 at 13:24
$begingroup$
Yeah, the proof is incorrect. Remember that "$u<v<w$" is shorthand for "$u<v$ and $v<w$." The negation is "$ugeq v$ or $vgeq w$."
$endgroup$
– Ben W
Jan 18 at 13:24
$begingroup$
Yeah, can I choose $E = (0,2/3)$ then it contradicts that the measure is less than $1/2$ because it's $2/3$ ?
$endgroup$
– Richard Clare
Jan 18 at 13:27
add a comment |
$begingroup$
In the contradiction, you should indeed say $m(Ecap (a,b)) leq frac{1}{2}m(a,b)$. But not $m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$ as the measure is non negative.
$endgroup$
– mathcounterexamples.net
Jan 18 at 13:24
$begingroup$
Yeah, the proof is incorrect. Remember that "$u<v<w$" is shorthand for "$u<v$ and $v<w$." The negation is "$ugeq v$ or $vgeq w$."
$endgroup$
– Ben W
Jan 18 at 13:24
$begingroup$
Yeah, can I choose $E = (0,2/3)$ then it contradicts that the measure is less than $1/2$ because it's $2/3$ ?
$endgroup$
– Richard Clare
Jan 18 at 13:27
$begingroup$
In the contradiction, you should indeed say $m(Ecap (a,b)) leq frac{1}{2}m(a,b)$. But not $m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$ as the measure is non negative.
$endgroup$
– mathcounterexamples.net
Jan 18 at 13:24
$begingroup$
In the contradiction, you should indeed say $m(Ecap (a,b)) leq frac{1}{2}m(a,b)$. But not $m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$ as the measure is non negative.
$endgroup$
– mathcounterexamples.net
Jan 18 at 13:24
$begingroup$
Yeah, the proof is incorrect. Remember that "$u<v<w$" is shorthand for "$u<v$ and $v<w$." The negation is "$ugeq v$ or $vgeq w$."
$endgroup$
– Ben W
Jan 18 at 13:24
$begingroup$
Yeah, the proof is incorrect. Remember that "$u<v<w$" is shorthand for "$u<v$ and $v<w$." The negation is "$ugeq v$ or $vgeq w$."
$endgroup$
– Ben W
Jan 18 at 13:24
$begingroup$
Yeah, can I choose $E = (0,2/3)$ then it contradicts that the measure is less than $1/2$ because it's $2/3$ ?
$endgroup$
– Richard Clare
Jan 18 at 13:27
$begingroup$
Yeah, can I choose $E = (0,2/3)$ then it contradicts that the measure is less than $1/2$ because it's $2/3$ ?
$endgroup$
– Richard Clare
Jan 18 at 13:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As the above comments point out, your proof is not correct. Here's an approach to the problem. Note that for any given $epsilon>0$, there exists a finite family of disjoint intervals $I_j=(a_j,b_j)$ such that for $S=cup_{j=1}^n I_j$, it holds
$$
m(Etriangle S)<epsilon.
$$ This implies
$$
m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.
$$ On the other hand,
$$begin{eqnarray}
m(Ecap S) = mleft(Ecap left[cup_{j=1}^n I_jright]right)&=&sum_{j=1}^n m(Ecap I_j)\&le& frac{1}{2}sum_{j=1}^n m(I_j)=frac{1}{2}m(S)\&le&frac{1}{2}left[m(E)+m(Etriangle S)right].
end{eqnarray}$$ Combining them, we have
$$
m(E)-epsilonlefrac{1}{2}m(E)+frac{epsilon}{2}.
$$ Since $epsilon>0$ was arbitrary, we have
$$
m(E)le frac{1}{2}m(E)
$$ or $m(E)=0$. This contradicts $m(E)>0$.
answered Jan 18 at 13:36
SongSong
15.6k1737
$endgroup$
$begingroup$
Nice! I would like to see why $m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.$
$endgroup$
– Richard Clare
Jan 18 at 13:42
1
$begingroup$
It came from $E=(Ecap S )cup( Esetminus S)subset (Ecap S) cup (Etriangle S)$ and the fact that $Ecap S$ and $Etriangle S$ are disjoint. This gives $$m(E)le m(Ecap S)+m(Etriangle S).$$
$endgroup$
– Song
Jan 18 at 13:44
add a comment |
Your Answer
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1 Answer
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$begingroup$
As the above comments point out, your proof is not correct. Here's an approach to the problem. Note that for any given $epsilon>0$, there exists a finite family of disjoint intervals $I_j=(a_j,b_j)$ such that for $S=cup_{j=1}^n I_j$, it holds
$$
m(Etriangle S)<epsilon.
$$ This implies
$$
m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.
$$ On the other hand,
$$begin{eqnarray}
m(Ecap S) = mleft(Ecap left[cup_{j=1}^n I_jright]right)&=&sum_{j=1}^n m(Ecap I_j)\&le& frac{1}{2}sum_{j=1}^n m(I_j)=frac{1}{2}m(S)\&le&frac{1}{2}left[m(E)+m(Etriangle S)right].
end{eqnarray}$$ Combining them, we have
$$
m(E)-epsilonlefrac{1}{2}m(E)+frac{epsilon}{2}.
$$ Since $epsilon>0$ was arbitrary, we have
$$
m(E)le frac{1}{2}m(E)
$$ or $m(E)=0$. This contradicts $m(E)>0$.
answered Jan 18 at 13:36
SongSong
15.6k1737
$endgroup$
$begingroup$
Nice! I would like to see why $m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.$
$endgroup$
– Richard Clare
Jan 18 at 13:42
1
$begingroup$
It came from $E=(Ecap S )cup( Esetminus S)subset (Ecap S) cup (Etriangle S)$ and the fact that $Ecap S$ and $Etriangle S$ are disjoint. This gives $$m(E)le m(Ecap S)+m(Etriangle S).$$
$endgroup$
– Song
Jan 18 at 13:44
add a comment |
$begingroup$
As the above comments point out, your proof is not correct. Here's an approach to the problem. Note that for any given $epsilon>0$, there exists a finite family of disjoint intervals $I_j=(a_j,b_j)$ such that for $S=cup_{j=1}^n I_j$, it holds
$$
m(Etriangle S)<epsilon.
$$ This implies
$$
m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.
$$ On the other hand,
$$begin{eqnarray}
m(Ecap S) = mleft(Ecap left[cup_{j=1}^n I_jright]right)&=&sum_{j=1}^n m(Ecap I_j)\&le& frac{1}{2}sum_{j=1}^n m(I_j)=frac{1}{2}m(S)\&le&frac{1}{2}left[m(E)+m(Etriangle S)right].
end{eqnarray}$$ Combining them, we have
$$
m(E)-epsilonlefrac{1}{2}m(E)+frac{epsilon}{2}.
$$ Since $epsilon>0$ was arbitrary, we have
$$
m(E)le frac{1}{2}m(E)
$$ or $m(E)=0$. This contradicts $m(E)>0$.
answered Jan 18 at 13:36
SongSong
15.6k1737
$endgroup$
$begingroup$
Nice! I would like to see why $m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.$
$endgroup$
– Richard Clare
Jan 18 at 13:42
1
$begingroup$
It came from $E=(Ecap S )cup( Esetminus S)subset (Ecap S) cup (Etriangle S)$ and the fact that $Ecap S$ and $Etriangle S$ are disjoint. This gives $$m(E)le m(Ecap S)+m(Etriangle S).$$
$endgroup$
– Song
Jan 18 at 13:44
add a comment |
$begingroup$
As the above comments point out, your proof is not correct. Here's an approach to the problem. Note that for any given $epsilon>0$, there exists a finite family of disjoint intervals $I_j=(a_j,b_j)$ such that for $S=cup_{j=1}^n I_j$, it holds
$$
m(Etriangle S)<epsilon.
$$ This implies
$$
m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.
$$ On the other hand,
$$begin{eqnarray}
m(Ecap S) = mleft(Ecap left[cup_{j=1}^n I_jright]right)&=&sum_{j=1}^n m(Ecap I_j)\&le& frac{1}{2}sum_{j=1}^n m(I_j)=frac{1}{2}m(S)\&le&frac{1}{2}left[m(E)+m(Etriangle S)right].
end{eqnarray}$$ Combining them, we have
$$
m(E)-epsilonlefrac{1}{2}m(E)+frac{epsilon}{2}.
$$ Since $epsilon>0$ was arbitrary, we have
$$
m(E)le frac{1}{2}m(E)
$$ or $m(E)=0$. This contradicts $m(E)>0$.
answered Jan 18 at 13:36
SongSong
15.6k1737
$endgroup$
As the above comments point out, your proof is not correct. Here's an approach to the problem. Note that for any given $epsilon>0$, there exists a finite family of disjoint intervals $I_j=(a_j,b_j)$ such that for $S=cup_{j=1}^n I_j$, it holds
$$
m(Etriangle S)<epsilon.
$$ This implies
$$
m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.
$$ On the other hand,
$$begin{eqnarray}
m(Ecap S) = mleft(Ecap left[cup_{j=1}^n I_jright]right)&=&sum_{j=1}^n m(Ecap I_j)\&le& frac{1}{2}sum_{j=1}^n m(I_j)=frac{1}{2}m(S)\&le&frac{1}{2}left[m(E)+m(Etriangle S)right].
end{eqnarray}$$ Combining them, we have
$$
m(E)-epsilonlefrac{1}{2}m(E)+frac{epsilon}{2}.
$$ Since $epsilon>0$ was arbitrary, we have
$$
m(E)le frac{1}{2}m(E)
$$ or $m(E)=0$. This contradicts $m(E)>0$.
answered Jan 18 at 13:36
SongSong
15.6k1737
answered Jan 18 at 13:36
SongSong
15.6k1737
answered Jan 18 at 13:36
SongSong
15.6k1737
answered Jan 18 at 13:36
SongSong
15.6k1737
15.6k1737
$begingroup$
Nice! I would like to see why $m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.$
$endgroup$
– Richard Clare
Jan 18 at 13:42
1
$begingroup$
It came from $E=(Ecap S )cup( Esetminus S)subset (Ecap S) cup (Etriangle S)$ and the fact that $Ecap S$ and $Etriangle S$ are disjoint. This gives $$m(E)le m(Ecap S)+m(Etriangle S).$$
$endgroup$
– Song
Jan 18 at 13:44
add a comment |
$begingroup$
Nice! I would like to see why $m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.$
$endgroup$
– Richard Clare
Jan 18 at 13:42
1
$begingroup$
It came from $E=(Ecap S )cup( Esetminus S)subset (Ecap S) cup (Etriangle S)$ and the fact that $Ecap S$ and $Etriangle S$ are disjoint. This gives $$m(E)le m(Ecap S)+m(Etriangle S).$$
$endgroup$
– Song
Jan 18 at 13:44
$begingroup$
Nice! I would like to see why $m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.$
$endgroup$
– Richard Clare
Jan 18 at 13:42
$begingroup$
Nice! I would like to see why $m(Ecap S)ge m(E)-m(Etriangle S)>m(E)-epsilon.$
$endgroup$
– Richard Clare
Jan 18 at 13:42
1
1
$begingroup$
It came from $E=(Ecap S )cup( Esetminus S)subset (Ecap S) cup (Etriangle S)$ and the fact that $Ecap S$ and $Etriangle S$ are disjoint. This gives $$m(E)le m(Ecap S)+m(Etriangle S).$$
$endgroup$
– Song
Jan 18 at 13:44
$begingroup$
It came from $E=(Ecap S )cup( Esetminus S)subset (Ecap S) cup (Etriangle S)$ and the fact that $Ecap S$ and $Etriangle S$ are disjoint. This gives $$m(E)le m(Ecap S)+m(Etriangle S).$$
$endgroup$
– Song
Jan 18 at 13:44
add a comment |
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$begingroup$
In the contradiction, you should indeed say $m(Ecap (a,b)) leq frac{1}{2}m(a,b)$. But not $m(Ecap (a,b)) leq frac{1}{2}m(a,b) leq 0$ as the measure is non negative.
$endgroup$
– mathcounterexamples.net
Jan 18 at 13:24
$begingroup$
Yeah, the proof is incorrect. Remember that "$u<v<w$" is shorthand for "$u<v$ and $v<w$." The negation is "$ugeq v$ or $vgeq w$."
$endgroup$
– Ben W
Jan 18 at 13:24
$begingroup$
Yeah, can I choose $E = (0,2/3)$ then it contradicts that the measure is less than $1/2$ because it's $2/3$ ?
$endgroup$
– Richard Clare
Jan 18 at 13:27