Mixing
Infinite union of sets with almost full measure
Clash Royale CLAN TAG#URR8PPP
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This seems to be a simple problem, but I can not come up with a counterexample. I would be happier if it were true!
Set up:
Consider $A_i_iinmathbbN$ a sequence of measurable subsets of $[0,1]$. Fix $epsilonin (0,1/2)$ and assume that the Lebesgue measure of each set is bounded below by $|A_i|>1-epsilon$ for every $iinmathbbN$.
Question:
Can we find an infinite set of indexes $i_n_ninmathbbN$, such that the Lebesgue measure of the intersection of all these sets is bounded below by
$$
left|bigcap_ninmathbbN A_i_nright|>1-2epsilon?
$$
real-analysis analysis measure-theory
edited Jul 18 at 9:42
Asaf Karagila♦
292k31403733
asked Jul 18 at 9:23
Sloth-Meister
34
add a comment |Â
up vote
0
down vote
favorite
This seems to be a simple problem, but I can not come up with a counterexample. I would be happier if it were true!
Set up:
Consider $A_i_iinmathbbN$ a sequence of measurable subsets of $[0,1]$. Fix $epsilonin (0,1/2)$ and assume that the Lebesgue measure of each set is bounded below by $|A_i|>1-epsilon$ for every $iinmathbbN$.
Question:
Can we find an infinite set of indexes $i_n_ninmathbbN$, such that the Lebesgue measure of the intersection of all these sets is bounded below by
$$
left|bigcap_ninmathbbN A_i_nright|>1-2epsilon?
$$
real-analysis analysis measure-theory
edited Jul 18 at 9:42
Asaf Karagila♦
292k31403733
asked Jul 18 at 9:23
Sloth-Meister
34
I'm unclear on your quantifiers. Are you asking if we can find such a sequence for each $A_i$, or if there is an $A_i$ such that we can find such a sequence?
– Mees de Vries
Jul 18 at 9:30
For any sequence $A_i_iin mathbbN$ can we find a subsequence $A_i_n_nin mathbbN$. Does this clear it up?
– Sloth-Meister
Jul 18 at 9:34
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This seems to be a simple problem, but I can not come up with a counterexample. I would be happier if it were true!
Set up:
Consider $A_i_iinmathbbN$ a sequence of measurable subsets of $[0,1]$. Fix $epsilonin (0,1/2)$ and assume that the Lebesgue measure of each set is bounded below by $|A_i|>1-epsilon$ for every $iinmathbbN$.
Question:
Can we find an infinite set of indexes $i_n_ninmathbbN$, such that the Lebesgue measure of the intersection of all these sets is bounded below by
$$
left|bigcap_ninmathbbN A_i_nright|>1-2epsilon?
$$
real-analysis analysis measure-theory
edited Jul 18 at 9:42
Asaf Karagila♦
292k31403733
asked Jul 18 at 9:23
Sloth-Meister
34
This seems to be a simple problem, but I can not come up with a counterexample. I would be happier if it were true!
Set up:
Consider $A_i_iinmathbbN$ a sequence of measurable subsets of $[0,1]$. Fix $epsilonin (0,1/2)$ and assume that the Lebesgue measure of each set is bounded below by $|A_i|>1-epsilon$ for every $iinmathbbN$.
Question:
Can we find an infinite set of indexes $i_n_ninmathbbN$, such that the Lebesgue measure of the intersection of all these sets is bounded below by
$$
left|bigcap_ninmathbbN A_i_nright|>1-2epsilon?
$$
real-analysis analysis measure-theory
edited Jul 18 at 9:42
Asaf Karagila♦
292k31403733
asked Jul 18 at 9:23
Sloth-Meister
34
edited Jul 18 at 9:42
Asaf Karagila♦
292k31403733
edited Jul 18 at 9:42
Asaf Karagila♦
292k31403733
edited Jul 18 at 9:42
Asaf Karagila♦
292k31403733
292k31403733
asked Jul 18 at 9:23
Sloth-Meister
34
asked Jul 18 at 9:23
Sloth-Meister
34
asked Jul 18 at 9:23
Sloth-Meister
34
34
I'm unclear on your quantifiers. Are you asking if we can find such a sequence for each $A_i$, or if there is an $A_i$ such that we can find such a sequence?
– Mees de Vries
Jul 18 at 9:30
For any sequence $A_i_iin mathbbN$ can we find a subsequence $A_i_n_nin mathbbN$. Does this clear it up?
– Sloth-Meister
Jul 18 at 9:34
add a comment |Â
I'm unclear on your quantifiers. Are you asking if we can find such a sequence for each $A_i$, or if there is an $A_i$ such that we can find such a sequence?
– Mees de Vries
Jul 18 at 9:30
For any sequence $A_i_iin mathbbN$ can we find a subsequence $A_i_n_nin mathbbN$. Does this clear it up?
– Sloth-Meister
Jul 18 at 9:34
I'm unclear on your quantifiers. Are you asking if we can find such a sequence for each $A_i$, or if there is an $A_i$ such that we can find such a sequence?
– Mees de Vries
Jul 18 at 9:30
I'm unclear on your quantifiers. Are you asking if we can find such a sequence for each $A_i$, or if there is an $A_i$ such that we can find such a sequence?
– Mees de Vries
Jul 18 at 9:30
For any sequence $A_i_iin mathbbN$ can we find a subsequence $A_i_n_nin mathbbN$. Does this clear it up?
– Sloth-Meister
Jul 18 at 9:34
For any sequence $A_i_iin mathbbN$ can we find a subsequence $A_i_n_nin mathbbN$. Does this clear it up?
– Sloth-Meister
Jul 18 at 9:34
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
No, you cannot do it in general.
Take $1/3 < epsilon < 1/2$. Let $A_i$ be the subset of elements of $[0, 1]$ whose ternary expansion does not have a $2$ at the $i$th position (in case of two possible decimal expansions, always use the one which terminates). Trivially $mu(A_i) = 2/3 > 1 - epsilon$, and each of the $A_i$ are independent, in the sense that for any finite set $I subseteq mathbb N$ we have
$$
muleft(bigcap_i in I A_iright) = prod_i in Imu(A_i).
$$
Therefore, for any infinite $I subseteq mathbb N$ we have that
$$
muleft(bigcap_i in I A_iright) = 0 < 1 - 2epsilon.
$$
Edit: And obviously this approach can be generalized to any $epsilon > 0$, you just have to take $n$-ary expansions for sufficiently large $n$ that $epsilon n > 1$.
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No, you cannot do it in general.
Take $1/3 < epsilon < 1/2$. Let $A_i$ be the subset of elements of $[0, 1]$ whose ternary expansion does not have a $2$ at the $i$th position (in case of two possible decimal expansions, always use the one which terminates). Trivially $mu(A_i) = 2/3 > 1 - epsilon$, and each of the $A_i$ are independent, in the sense that for any finite set $I subseteq mathbb N$ we have
$$
muleft(bigcap_i in I A_iright) = prod_i in Imu(A_i).
$$
Therefore, for any infinite $I subseteq mathbb N$ we have that
$$
muleft(bigcap_i in I A_iright) = 0 < 1 - 2epsilon.
$$
Edit: And obviously this approach can be generalized to any $epsilon > 0$, you just have to take $n$-ary expansions for sufficiently large $n$ that $epsilon n > 1$.
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
add a comment |Â
up vote
2
down vote
accepted
No, you cannot do it in general.
Take $1/3 < epsilon < 1/2$. Let $A_i$ be the subset of elements of $[0, 1]$ whose ternary expansion does not have a $2$ at the $i$th position (in case of two possible decimal expansions, always use the one which terminates). Trivially $mu(A_i) = 2/3 > 1 - epsilon$, and each of the $A_i$ are independent, in the sense that for any finite set $I subseteq mathbb N$ we have
$$
muleft(bigcap_i in I A_iright) = prod_i in Imu(A_i).
$$
Therefore, for any infinite $I subseteq mathbb N$ we have that
$$
muleft(bigcap_i in I A_iright) = 0 < 1 - 2epsilon.
$$
Edit: And obviously this approach can be generalized to any $epsilon > 0$, you just have to take $n$-ary expansions for sufficiently large $n$ that $epsilon n > 1$.
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No, you cannot do it in general.
Take $1/3 < epsilon < 1/2$. Let $A_i$ be the subset of elements of $[0, 1]$ whose ternary expansion does not have a $2$ at the $i$th position (in case of two possible decimal expansions, always use the one which terminates). Trivially $mu(A_i) = 2/3 > 1 - epsilon$, and each of the $A_i$ are independent, in the sense that for any finite set $I subseteq mathbb N$ we have
$$
muleft(bigcap_i in I A_iright) = prod_i in Imu(A_i).
$$
Therefore, for any infinite $I subseteq mathbb N$ we have that
$$
muleft(bigcap_i in I A_iright) = 0 < 1 - 2epsilon.
$$
Edit: And obviously this approach can be generalized to any $epsilon > 0$, you just have to take $n$-ary expansions for sufficiently large $n$ that $epsilon n > 1$.
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
No, you cannot do it in general.
Take $1/3 < epsilon < 1/2$. Let $A_i$ be the subset of elements of $[0, 1]$ whose ternary expansion does not have a $2$ at the $i$th position (in case of two possible decimal expansions, always use the one which terminates). Trivially $mu(A_i) = 2/3 > 1 - epsilon$, and each of the $A_i$ are independent, in the sense that for any finite set $I subseteq mathbb N$ we have
$$
muleft(bigcap_i in I A_iright) = prod_i in Imu(A_i).
$$
Therefore, for any infinite $I subseteq mathbb N$ we have that
$$
muleft(bigcap_i in I A_iright) = 0 < 1 - 2epsilon.
$$
Edit: And obviously this approach can be generalized to any $epsilon > 0$, you just have to take $n$-ary expansions for sufficiently large $n$ that $epsilon n > 1$.
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
answered Jul 18 at 9:36
Mees de Vries
13.7k12345
13.7k12345
add a comment |Â
add a comment |Â
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I'm unclear on your quantifiers. Are you asking if we can find such a sequence for each $A_i$, or if there is an $A_i$ such that we can find such a sequence?
– Mees de Vries
Jul 18 at 9:30
For any sequence $A_i_iin mathbbN$ can we find a subsequence $A_i_n_nin mathbbN$. Does this clear it up?
– Sloth-Meister
Jul 18 at 9:34