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Non-atomicity of the uniform measure space on $(0, 1]$.
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I was working on the following exercise:
Show that $((0,1], mathcalB_(0, 1], U)$ defines a non-atomic measure space.
Above, $mathcalB$ denotes the Borel $sigma$-algebra, and $U$ denotes Lebesgue measure restricted to this sub-$sigma$-algebra of $mathcalB_mathbfR$. Importantly, all I know about $U$ (for the purposes of this problem) is that (1) it is a measure, and (2) it is the unique measure such that if $A = cup_n=1^N (a_n, b_n] subset (0, 1]$, where $N in mathbfN$ and the intervals are disjoint, then $U(A) = sum_n=1^N (b_n - a_n)$. [In words, "$U$ is finitely-additive over disjoint, finite unions of half-open sub-intervals of $(0, 1]$"]
In working on the exercise, I thought it would be useful to try to prove a property about the measure on the entire $mathcalB_(0, 1]$. Suppose in fact it is true that for any measurable $A$,
$$
U(A) = supleft U(I) : I subseteq A, text$I$ is a disjoint finite union of half-open sub-intervals of $(0, 1]$right
$$
If this statement is true, then the claim follows: if I set $epsilon > 0$, then I get the following approximation result:
$$
exists A supseteq I = cup_n=1^N (a_n, b_n] text (disj.): qquad U(A) - epsilon < U(I) = sum_n=1^N (b_n - a_n).
$$
Suppose now that $U(A) > 0$. If I take $epsilon = U(A)/2$, then
this says there exists a set $C = cup_n=1^N (a_n, b_n] subset A$, such
that $U(C) > U(A)/2$. Additionally $U(C) leq U(A)$ since $C subseteq A$.
Now set $B = cup_n=1^N (a_n/2, b_n/2]$. Then $B subsetneq A$, and since everything has positive measure, the strict inequalities
$$
O < U(B) < U(A)
$$
follow.
So the questions I have are
- Am I on the right track here? Maybe this supremum approximation argument is overkill.
- If this seems like a good approach, I'm struggling a bit to show that the equality holds. In particular if we fix $A$ measurable, and let $S$ denote the RHS (the supremum), then the inequality $U(A) geq S$ is immediate, since the sup is occurring over sets $B$ all of which satisfy $U(A) geq U(B)$ as $B subseteq A$. So the real part that I could use help with is demonstrating $U(A) leq S$. Again if this seems like the right thing to look at, I just want a hint.
probability-theory measure-theory
asked Jul 16 at 18:21
Drew Brady
410112
add a comment |Â
up vote
1
down vote
favorite
I was working on the following exercise:
Show that $((0,1], mathcalB_(0, 1], U)$ defines a non-atomic measure space.
Above, $mathcalB$ denotes the Borel $sigma$-algebra, and $U$ denotes Lebesgue measure restricted to this sub-$sigma$-algebra of $mathcalB_mathbfR$. Importantly, all I know about $U$ (for the purposes of this problem) is that (1) it is a measure, and (2) it is the unique measure such that if $A = cup_n=1^N (a_n, b_n] subset (0, 1]$, where $N in mathbfN$ and the intervals are disjoint, then $U(A) = sum_n=1^N (b_n - a_n)$. [In words, "$U$ is finitely-additive over disjoint, finite unions of half-open sub-intervals of $(0, 1]$"]
In working on the exercise, I thought it would be useful to try to prove a property about the measure on the entire $mathcalB_(0, 1]$. Suppose in fact it is true that for any measurable $A$,
$$
U(A) = supleft U(I) : I subseteq A, text$I$ is a disjoint finite union of half-open sub-intervals of $(0, 1]$right
$$
If this statement is true, then the claim follows: if I set $epsilon > 0$, then I get the following approximation result:
$$
exists A supseteq I = cup_n=1^N (a_n, b_n] text (disj.): qquad U(A) - epsilon < U(I) = sum_n=1^N (b_n - a_n).
$$
Suppose now that $U(A) > 0$. If I take $epsilon = U(A)/2$, then
this says there exists a set $C = cup_n=1^N (a_n, b_n] subset A$, such
that $U(C) > U(A)/2$. Additionally $U(C) leq U(A)$ since $C subseteq A$.
Now set $B = cup_n=1^N (a_n/2, b_n/2]$. Then $B subsetneq A$, and since everything has positive measure, the strict inequalities
$$
O < U(B) < U(A)
$$
follow.
So the questions I have are
- Am I on the right track here? Maybe this supremum approximation argument is overkill.
- If this seems like a good approach, I'm struggling a bit to show that the equality holds. In particular if we fix $A$ measurable, and let $S$ denote the RHS (the supremum), then the inequality $U(A) geq S$ is immediate, since the sup is occurring over sets $B$ all of which satisfy $U(A) geq U(B)$ as $B subseteq A$. So the real part that I could use help with is demonstrating $U(A) leq S$. Again if this seems like the right thing to look at, I just want a hint.
probability-theory measure-theory
asked Jul 16 at 18:21
Drew Brady
410112
1
Unfortunately your expression for $U(A)$ is incorrect. If $A$ has no interior points the supremum will be zero.
– Umberto P.
Jul 16 at 18:26
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was working on the following exercise:
Show that $((0,1], mathcalB_(0, 1], U)$ defines a non-atomic measure space.
Above, $mathcalB$ denotes the Borel $sigma$-algebra, and $U$ denotes Lebesgue measure restricted to this sub-$sigma$-algebra of $mathcalB_mathbfR$. Importantly, all I know about $U$ (for the purposes of this problem) is that (1) it is a measure, and (2) it is the unique measure such that if $A = cup_n=1^N (a_n, b_n] subset (0, 1]$, where $N in mathbfN$ and the intervals are disjoint, then $U(A) = sum_n=1^N (b_n - a_n)$. [In words, "$U$ is finitely-additive over disjoint, finite unions of half-open sub-intervals of $(0, 1]$"]
In working on the exercise, I thought it would be useful to try to prove a property about the measure on the entire $mathcalB_(0, 1]$. Suppose in fact it is true that for any measurable $A$,
$$
U(A) = supleft U(I) : I subseteq A, text$I$ is a disjoint finite union of half-open sub-intervals of $(0, 1]$right
$$
If this statement is true, then the claim follows: if I set $epsilon > 0$, then I get the following approximation result:
$$
exists A supseteq I = cup_n=1^N (a_n, b_n] text (disj.): qquad U(A) - epsilon < U(I) = sum_n=1^N (b_n - a_n).
$$
Suppose now that $U(A) > 0$. If I take $epsilon = U(A)/2$, then
this says there exists a set $C = cup_n=1^N (a_n, b_n] subset A$, such
that $U(C) > U(A)/2$. Additionally $U(C) leq U(A)$ since $C subseteq A$.
Now set $B = cup_n=1^N (a_n/2, b_n/2]$. Then $B subsetneq A$, and since everything has positive measure, the strict inequalities
$$
O < U(B) < U(A)
$$
follow.
So the questions I have are
- Am I on the right track here? Maybe this supremum approximation argument is overkill.
- If this seems like a good approach, I'm struggling a bit to show that the equality holds. In particular if we fix $A$ measurable, and let $S$ denote the RHS (the supremum), then the inequality $U(A) geq S$ is immediate, since the sup is occurring over sets $B$ all of which satisfy $U(A) geq U(B)$ as $B subseteq A$. So the real part that I could use help with is demonstrating $U(A) leq S$. Again if this seems like the right thing to look at, I just want a hint.
probability-theory measure-theory
asked Jul 16 at 18:21
Drew Brady
410112
I was working on the following exercise:
Show that $((0,1], mathcalB_(0, 1], U)$ defines a non-atomic measure space.
Above, $mathcalB$ denotes the Borel $sigma$-algebra, and $U$ denotes Lebesgue measure restricted to this sub-$sigma$-algebra of $mathcalB_mathbfR$. Importantly, all I know about $U$ (for the purposes of this problem) is that (1) it is a measure, and (2) it is the unique measure such that if $A = cup_n=1^N (a_n, b_n] subset (0, 1]$, where $N in mathbfN$ and the intervals are disjoint, then $U(A) = sum_n=1^N (b_n - a_n)$. [In words, "$U$ is finitely-additive over disjoint, finite unions of half-open sub-intervals of $(0, 1]$"]
In working on the exercise, I thought it would be useful to try to prove a property about the measure on the entire $mathcalB_(0, 1]$. Suppose in fact it is true that for any measurable $A$,
$$
U(A) = supleft U(I) : I subseteq A, text$I$ is a disjoint finite union of half-open sub-intervals of $(0, 1]$right
$$
If this statement is true, then the claim follows: if I set $epsilon > 0$, then I get the following approximation result:
$$
exists A supseteq I = cup_n=1^N (a_n, b_n] text (disj.): qquad U(A) - epsilon < U(I) = sum_n=1^N (b_n - a_n).
$$
Suppose now that $U(A) > 0$. If I take $epsilon = U(A)/2$, then
this says there exists a set $C = cup_n=1^N (a_n, b_n] subset A$, such
that $U(C) > U(A)/2$. Additionally $U(C) leq U(A)$ since $C subseteq A$.
Now set $B = cup_n=1^N (a_n/2, b_n/2]$. Then $B subsetneq A$, and since everything has positive measure, the strict inequalities
$$
O < U(B) < U(A)
$$
follow.
So the questions I have are
- Am I on the right track here? Maybe this supremum approximation argument is overkill.
- If this seems like a good approach, I'm struggling a bit to show that the equality holds. In particular if we fix $A$ measurable, and let $S$ denote the RHS (the supremum), then the inequality $U(A) geq S$ is immediate, since the sup is occurring over sets $B$ all of which satisfy $U(A) geq U(B)$ as $B subseteq A$. So the real part that I could use help with is demonstrating $U(A) leq S$. Again if this seems like the right thing to look at, I just want a hint.
probability-theory measure-theory
asked Jul 16 at 18:21
Drew Brady
410112
asked Jul 16 at 18:21
Drew Brady
410112
asked Jul 16 at 18:21
Drew Brady
410112
asked Jul 16 at 18:21
Drew Brady
410112
410112
1
Unfortunately your expression for $U(A)$ is incorrect. If $A$ has no interior points the supremum will be zero.
– Umberto P.
Jul 16 at 18:26
add a comment |Â
1
Unfortunately your expression for $U(A)$ is incorrect. If $A$ has no interior points the supremum will be zero.
– Umberto P.
Jul 16 at 18:26
1
1
Unfortunately your expression for $U(A)$ is incorrect. If $A$ has no interior points the supremum will be zero.
– Umberto P.
Jul 16 at 18:26
Unfortunately your expression for $U(A)$ is incorrect. If $A$ has no interior points the supremum will be zero.
– Umberto P.
Jul 16 at 18:26
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
You are not on the right track.
For instance observe that the supremum takes value $0$ on a sets like $(0,1]-mathbb Q$ and also on sets like $mathbb Q$.
For $uin(0,1]$ and $epsilon>0$ small enough we have $uin(u-epsilon,u]subseteq(0,1]$ implying that: $$U(u)leq U((u-epsilon,u])=epsilon$$
Since we can take $epsilon>0$ as small as we like this can only be true if:$$U(u)=0$$
answered Jul 16 at 18:55
drhab
86.6k541118
add a comment |Â
up vote
0
down vote
Suppose for a contradiction that the set $A$ would be an atom. We may partition $(0,1]$ into intervals
$$
B_k,n := [(k-1)/n,k/n] ~~~(1 le k le n)
$$
for $n = 1, 2, ldots$. Suppose that $mu(A) = epsilon$. Then choose $n$ sufficiently small so that $1/n < epsilon$ (Achimedean axiom) and observe that one of the sets
$$
B_k, n cap A
$$
has positive measure (otherwise $mu(A) = 0$), but also measure less than $epsilon$.
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
Very clever. How'd you come up with it or is it sort of just a standard analysis argument?
– Drew Brady
Jul 16 at 19:16
(It kinda looks like the pigeonhole principle)
– Drew Brady
Jul 16 at 19:17
(The idea being that if the set A lied strictly inside any of these intervals, its mass would be too little, so it must have nontrivial intersection with at least two of these intervals).
– Drew Brady
Jul 16 at 19:19
1
I came up with it using intuition for the real world: If a set has measure, it must be localised somewhere (imagine it to be a weight or a density). Now an atom cannot be cut into tiny pieces, but the real line can, and the atom is a subset of the real line.
– AlgebraicsAnonymous
Jul 17 at 9:48
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You are not on the right track.
For instance observe that the supremum takes value $0$ on a sets like $(0,1]-mathbb Q$ and also on sets like $mathbb Q$.
For $uin(0,1]$ and $epsilon>0$ small enough we have $uin(u-epsilon,u]subseteq(0,1]$ implying that: $$U(u)leq U((u-epsilon,u])=epsilon$$
Since we can take $epsilon>0$ as small as we like this can only be true if:$$U(u)=0$$
answered Jul 16 at 18:55
drhab
86.6k541118
add a comment |Â
up vote
1
down vote
You are not on the right track.
For instance observe that the supremum takes value $0$ on a sets like $(0,1]-mathbb Q$ and also on sets like $mathbb Q$.
For $uin(0,1]$ and $epsilon>0$ small enough we have $uin(u-epsilon,u]subseteq(0,1]$ implying that: $$U(u)leq U((u-epsilon,u])=epsilon$$
Since we can take $epsilon>0$ as small as we like this can only be true if:$$U(u)=0$$
answered Jul 16 at 18:55
drhab
86.6k541118
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are not on the right track.
For instance observe that the supremum takes value $0$ on a sets like $(0,1]-mathbb Q$ and also on sets like $mathbb Q$.
For $uin(0,1]$ and $epsilon>0$ small enough we have $uin(u-epsilon,u]subseteq(0,1]$ implying that: $$U(u)leq U((u-epsilon,u])=epsilon$$
Since we can take $epsilon>0$ as small as we like this can only be true if:$$U(u)=0$$
answered Jul 16 at 18:55
drhab
86.6k541118
You are not on the right track.
For instance observe that the supremum takes value $0$ on a sets like $(0,1]-mathbb Q$ and also on sets like $mathbb Q$.
For $uin(0,1]$ and $epsilon>0$ small enough we have $uin(u-epsilon,u]subseteq(0,1]$ implying that: $$U(u)leq U((u-epsilon,u])=epsilon$$
Since we can take $epsilon>0$ as small as we like this can only be true if:$$U(u)=0$$
answered Jul 16 at 18:55
drhab
86.6k541118
edited Jul 16 at 20:07
answered Jul 16 at 18:55
drhab
86.6k541118
answered Jul 16 at 18:55
drhab
86.6k541118
answered Jul 16 at 18:55
drhab
86.6k541118
86.6k541118
add a comment |Â
add a comment |Â
up vote
0
down vote
Suppose for a contradiction that the set $A$ would be an atom. We may partition $(0,1]$ into intervals
$$
B_k,n := [(k-1)/n,k/n] ~~~(1 le k le n)
$$
for $n = 1, 2, ldots$. Suppose that $mu(A) = epsilon$. Then choose $n$ sufficiently small so that $1/n < epsilon$ (Achimedean axiom) and observe that one of the sets
$$
B_k, n cap A
$$
has positive measure (otherwise $mu(A) = 0$), but also measure less than $epsilon$.
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
Very clever. How'd you come up with it or is it sort of just a standard analysis argument?
– Drew Brady
Jul 16 at 19:16
(It kinda looks like the pigeonhole principle)
– Drew Brady
Jul 16 at 19:17
(The idea being that if the set A lied strictly inside any of these intervals, its mass would be too little, so it must have nontrivial intersection with at least two of these intervals).
– Drew Brady
Jul 16 at 19:19
1
I came up with it using intuition for the real world: If a set has measure, it must be localised somewhere (imagine it to be a weight or a density). Now an atom cannot be cut into tiny pieces, but the real line can, and the atom is a subset of the real line.
– AlgebraicsAnonymous
Jul 17 at 9:48
add a comment |Â
up vote
0
down vote
Suppose for a contradiction that the set $A$ would be an atom. We may partition $(0,1]$ into intervals
$$
B_k,n := [(k-1)/n,k/n] ~~~(1 le k le n)
$$
for $n = 1, 2, ldots$. Suppose that $mu(A) = epsilon$. Then choose $n$ sufficiently small so that $1/n < epsilon$ (Achimedean axiom) and observe that one of the sets
$$
B_k, n cap A
$$
has positive measure (otherwise $mu(A) = 0$), but also measure less than $epsilon$.
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
Very clever. How'd you come up with it or is it sort of just a standard analysis argument?
– Drew Brady
Jul 16 at 19:16
(It kinda looks like the pigeonhole principle)
– Drew Brady
Jul 16 at 19:17
(The idea being that if the set A lied strictly inside any of these intervals, its mass would be too little, so it must have nontrivial intersection with at least two of these intervals).
– Drew Brady
Jul 16 at 19:19
1
I came up with it using intuition for the real world: If a set has measure, it must be localised somewhere (imagine it to be a weight or a density). Now an atom cannot be cut into tiny pieces, but the real line can, and the atom is a subset of the real line.
– AlgebraicsAnonymous
Jul 17 at 9:48
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Suppose for a contradiction that the set $A$ would be an atom. We may partition $(0,1]$ into intervals
$$
B_k,n := [(k-1)/n,k/n] ~~~(1 le k le n)
$$
for $n = 1, 2, ldots$. Suppose that $mu(A) = epsilon$. Then choose $n$ sufficiently small so that $1/n < epsilon$ (Achimedean axiom) and observe that one of the sets
$$
B_k, n cap A
$$
has positive measure (otherwise $mu(A) = 0$), but also measure less than $epsilon$.
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
Suppose for a contradiction that the set $A$ would be an atom. We may partition $(0,1]$ into intervals
$$
B_k,n := [(k-1)/n,k/n] ~~~(1 le k le n)
$$
for $n = 1, 2, ldots$. Suppose that $mu(A) = epsilon$. Then choose $n$ sufficiently small so that $1/n < epsilon$ (Achimedean axiom) and observe that one of the sets
$$
B_k, n cap A
$$
has positive measure (otherwise $mu(A) = 0$), but also measure less than $epsilon$.
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
answered Jul 16 at 18:31
AlgebraicsAnonymous
69111
69111
Very clever. How'd you come up with it or is it sort of just a standard analysis argument?
– Drew Brady
Jul 16 at 19:16
(It kinda looks like the pigeonhole principle)
– Drew Brady
Jul 16 at 19:17
(The idea being that if the set A lied strictly inside any of these intervals, its mass would be too little, so it must have nontrivial intersection with at least two of these intervals).
– Drew Brady
Jul 16 at 19:19
1
I came up with it using intuition for the real world: If a set has measure, it must be localised somewhere (imagine it to be a weight or a density). Now an atom cannot be cut into tiny pieces, but the real line can, and the atom is a subset of the real line.
– AlgebraicsAnonymous
Jul 17 at 9:48
add a comment |Â
Very clever. How'd you come up with it or is it sort of just a standard analysis argument?
– Drew Brady
Jul 16 at 19:16
(It kinda looks like the pigeonhole principle)
– Drew Brady
Jul 16 at 19:17
(The idea being that if the set A lied strictly inside any of these intervals, its mass would be too little, so it must have nontrivial intersection with at least two of these intervals).
– Drew Brady
Jul 16 at 19:19
1
I came up with it using intuition for the real world: If a set has measure, it must be localised somewhere (imagine it to be a weight or a density). Now an atom cannot be cut into tiny pieces, but the real line can, and the atom is a subset of the real line.
– AlgebraicsAnonymous
Jul 17 at 9:48
Very clever. How'd you come up with it or is it sort of just a standard analysis argument?
– Drew Brady
Jul 16 at 19:16
Very clever. How'd you come up with it or is it sort of just a standard analysis argument?
– Drew Brady
Jul 16 at 19:16
(It kinda looks like the pigeonhole principle)
– Drew Brady
Jul 16 at 19:17
(It kinda looks like the pigeonhole principle)
– Drew Brady
Jul 16 at 19:17
(The idea being that if the set A lied strictly inside any of these intervals, its mass would be too little, so it must have nontrivial intersection with at least two of these intervals).
– Drew Brady
Jul 16 at 19:19
(The idea being that if the set A lied strictly inside any of these intervals, its mass would be too little, so it must have nontrivial intersection with at least two of these intervals).
– Drew Brady
Jul 16 at 19:19
1
1
I came up with it using intuition for the real world: If a set has measure, it must be localised somewhere (imagine it to be a weight or a density). Now an atom cannot be cut into tiny pieces, but the real line can, and the atom is a subset of the real line.
– AlgebraicsAnonymous
Jul 17 at 9:48
I came up with it using intuition for the real world: If a set has measure, it must be localised somewhere (imagine it to be a weight or a density). Now an atom cannot be cut into tiny pieces, but the real line can, and the atom is a subset of the real line.
– AlgebraicsAnonymous
Jul 17 at 9:48
add a comment |Â
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1
Unfortunately your expression for $U(A)$ is incorrect. If $A$ has no interior points the supremum will be zero.
– Umberto P.
Jul 16 at 18:26