Mixing
Intuition behind inequality for measure of $liminf$ and $limsup$
$begingroup$
For a set $X$ with a $sigma$-algebra $xi subseteq mathcal{P}(X)$ and $sigma$-additive $mu: xi rightarrow [0, infty]$. The following inequality holds for $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$:
$$
mu({liminf}_{n rightarrow infty} A_n) leq {liminf}_{n rightarrow infty} mu(A_n)
$$
and under the assumption of $mu(X) < infty$ we also get:
$$
{limsup}_{n rightarrow infty} mu(A_n)leq mu({limsup}_{n rightarrow infty} A_n)
$$
I was able to prove these inequalities, but can't seem to develop a deeper intuition. Is there any way to better understand why these statements are true?
measure-theory inequality intuition limsup-and-liminf
edited Jan 5 at 15:16
Bernard
119k740113
asked Jan 5 at 15:13
HericksonHerickson
1649
$endgroup$
add a comment |
$begingroup$
For a set $X$ with a $sigma$-algebra $xi subseteq mathcal{P}(X)$ and $sigma$-additive $mu: xi rightarrow [0, infty]$. The following inequality holds for $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$:
$$
mu({liminf}_{n rightarrow infty} A_n) leq {liminf}_{n rightarrow infty} mu(A_n)
$$
and under the assumption of $mu(X) < infty$ we also get:
$$
{limsup}_{n rightarrow infty} mu(A_n)leq mu({limsup}_{n rightarrow infty} A_n)
$$
I was able to prove these inequalities, but can't seem to develop a deeper intuition. Is there any way to better understand why these statements are true?
measure-theory inequality intuition limsup-and-liminf
edited Jan 5 at 15:16
Bernard
119k740113
asked Jan 5 at 15:13
HericksonHerickson
1649
$endgroup$
$begingroup$
At the risk of being rude, I don't think a deep intuition is needed; I think they are just obvious. For example, $liminf_n A_n$ represents the elements that are in all of the $A_n$'s past some point. So clearly the measure of the set of these elements is less than the liminf of the measures of the $A_n$'s.
$endgroup$
– mathworker21
Jan 5 at 15:20
$begingroup$
@mathworker21 I understand, that the $lim inf$ contains all the elements which are in all but finitely many of the $A_n$, but to me it's not obvious why the measure of these elements is always smaller or equal to the $lim inf$ of the measures...
$endgroup$
– Herickson
Jan 5 at 15:23
$begingroup$
Don't think of it as "all but finitely many". Think of it as "past some point". Let's pretend that $liminf_{n to infty} A_n$ happened to be all elements belonging to $A_{10} cap A_{11} cap A_{12} cap dots$. Then $mu(liminf_n A_n) le mu(A_i)$ for each $i ge 10$ and in particular, it is $le liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:26
$begingroup$
In reality, part of $liminf_n A_n$ will also be $A_{12}cap A_{13} cap dots$. But this doesn't matter. Each of these "parts" of $liminf_n A_n$ will have measure at most $liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:27
1
$begingroup$
No. $liminf_n A_n subseteq A_1$ is (in general) false. You should spend some time thinking about these things. In particular, try to connect the proof you have to intuition.
$endgroup$
– mathworker21
Jan 5 at 15:31
add a comment |
$begingroup$
For a set $X$ with a $sigma$-algebra $xi subseteq mathcal{P}(X)$ and $sigma$-additive $mu: xi rightarrow [0, infty]$. The following inequality holds for $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$:
$$
mu({liminf}_{n rightarrow infty} A_n) leq {liminf}_{n rightarrow infty} mu(A_n)
$$
and under the assumption of $mu(X) < infty$ we also get:
$$
{limsup}_{n rightarrow infty} mu(A_n)leq mu({limsup}_{n rightarrow infty} A_n)
$$
I was able to prove these inequalities, but can't seem to develop a deeper intuition. Is there any way to better understand why these statements are true?
measure-theory inequality intuition limsup-and-liminf
edited Jan 5 at 15:16
Bernard
119k740113
asked Jan 5 at 15:13
HericksonHerickson
1649
$endgroup$
For a set $X$ with a $sigma$-algebra $xi subseteq mathcal{P}(X)$ and $sigma$-additive $mu: xi rightarrow [0, infty]$. The following inequality holds for $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$:
$$
mu({liminf}_{n rightarrow infty} A_n) leq {liminf}_{n rightarrow infty} mu(A_n)
$$
and under the assumption of $mu(X) < infty$ we also get:
$$
{limsup}_{n rightarrow infty} mu(A_n)leq mu({limsup}_{n rightarrow infty} A_n)
$$
I was able to prove these inequalities, but can't seem to develop a deeper intuition. Is there any way to better understand why these statements are true?
measure-theory inequality intuition limsup-and-liminf
measure-theory inequality intuition limsup-and-liminf
edited Jan 5 at 15:16
Bernard
119k740113
asked Jan 5 at 15:13
HericksonHerickson
1649
edited Jan 5 at 15:16
Bernard
119k740113
asked Jan 5 at 15:13
HericksonHerickson
1649
edited Jan 5 at 15:16
Bernard
119k740113
edited Jan 5 at 15:16
Bernard
119k740113
edited Jan 5 at 15:16
Bernard
119k740113
119k740113
asked Jan 5 at 15:13
HericksonHerickson
1649
asked Jan 5 at 15:13
HericksonHerickson
1649
asked Jan 5 at 15:13
HericksonHerickson
1649
1649
$begingroup$
At the risk of being rude, I don't think a deep intuition is needed; I think they are just obvious. For example, $liminf_n A_n$ represents the elements that are in all of the $A_n$'s past some point. So clearly the measure of the set of these elements is less than the liminf of the measures of the $A_n$'s.
$endgroup$
– mathworker21
Jan 5 at 15:20
$begingroup$
@mathworker21 I understand, that the $lim inf$ contains all the elements which are in all but finitely many of the $A_n$, but to me it's not obvious why the measure of these elements is always smaller or equal to the $lim inf$ of the measures...
$endgroup$
– Herickson
Jan 5 at 15:23
$begingroup$
Don't think of it as "all but finitely many". Think of it as "past some point". Let's pretend that $liminf_{n to infty} A_n$ happened to be all elements belonging to $A_{10} cap A_{11} cap A_{12} cap dots$. Then $mu(liminf_n A_n) le mu(A_i)$ for each $i ge 10$ and in particular, it is $le liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:26
$begingroup$
In reality, part of $liminf_n A_n$ will also be $A_{12}cap A_{13} cap dots$. But this doesn't matter. Each of these "parts" of $liminf_n A_n$ will have measure at most $liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:27
1
$begingroup$
No. $liminf_n A_n subseteq A_1$ is (in general) false. You should spend some time thinking about these things. In particular, try to connect the proof you have to intuition.
$endgroup$
– mathworker21
Jan 5 at 15:31
add a comment |
$begingroup$
At the risk of being rude, I don't think a deep intuition is needed; I think they are just obvious. For example, $liminf_n A_n$ represents the elements that are in all of the $A_n$'s past some point. So clearly the measure of the set of these elements is less than the liminf of the measures of the $A_n$'s.
$endgroup$
– mathworker21
Jan 5 at 15:20
$begingroup$
@mathworker21 I understand, that the $lim inf$ contains all the elements which are in all but finitely many of the $A_n$, but to me it's not obvious why the measure of these elements is always smaller or equal to the $lim inf$ of the measures...
$endgroup$
– Herickson
Jan 5 at 15:23
$begingroup$
Don't think of it as "all but finitely many". Think of it as "past some point". Let's pretend that $liminf_{n to infty} A_n$ happened to be all elements belonging to $A_{10} cap A_{11} cap A_{12} cap dots$. Then $mu(liminf_n A_n) le mu(A_i)$ for each $i ge 10$ and in particular, it is $le liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:26
$begingroup$
In reality, part of $liminf_n A_n$ will also be $A_{12}cap A_{13} cap dots$. But this doesn't matter. Each of these "parts" of $liminf_n A_n$ will have measure at most $liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:27
1
$begingroup$
No. $liminf_n A_n subseteq A_1$ is (in general) false. You should spend some time thinking about these things. In particular, try to connect the proof you have to intuition.
$endgroup$
– mathworker21
Jan 5 at 15:31
$begingroup$
At the risk of being rude, I don't think a deep intuition is needed; I think they are just obvious. For example, $liminf_n A_n$ represents the elements that are in all of the $A_n$'s past some point. So clearly the measure of the set of these elements is less than the liminf of the measures of the $A_n$'s.
$endgroup$
– mathworker21
Jan 5 at 15:20
$begingroup$
At the risk of being rude, I don't think a deep intuition is needed; I think they are just obvious. For example, $liminf_n A_n$ represents the elements that are in all of the $A_n$'s past some point. So clearly the measure of the set of these elements is less than the liminf of the measures of the $A_n$'s.
$endgroup$
– mathworker21
Jan 5 at 15:20
$begingroup$
@mathworker21 I understand, that the $lim inf$ contains all the elements which are in all but finitely many of the $A_n$, but to me it's not obvious why the measure of these elements is always smaller or equal to the $lim inf$ of the measures...
$endgroup$
– Herickson
Jan 5 at 15:23
$begingroup$
@mathworker21 I understand, that the $lim inf$ contains all the elements which are in all but finitely many of the $A_n$, but to me it's not obvious why the measure of these elements is always smaller or equal to the $lim inf$ of the measures...
$endgroup$
– Herickson
Jan 5 at 15:23
$begingroup$
Don't think of it as "all but finitely many". Think of it as "past some point". Let's pretend that $liminf_{n to infty} A_n$ happened to be all elements belonging to $A_{10} cap A_{11} cap A_{12} cap dots$. Then $mu(liminf_n A_n) le mu(A_i)$ for each $i ge 10$ and in particular, it is $le liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:26
$begingroup$
Don't think of it as "all but finitely many". Think of it as "past some point". Let's pretend that $liminf_{n to infty} A_n$ happened to be all elements belonging to $A_{10} cap A_{11} cap A_{12} cap dots$. Then $mu(liminf_n A_n) le mu(A_i)$ for each $i ge 10$ and in particular, it is $le liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:26
$begingroup$
In reality, part of $liminf_n A_n$ will also be $A_{12}cap A_{13} cap dots$. But this doesn't matter. Each of these "parts" of $liminf_n A_n$ will have measure at most $liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:27
$begingroup$
In reality, part of $liminf_n A_n$ will also be $A_{12}cap A_{13} cap dots$. But this doesn't matter. Each of these "parts" of $liminf_n A_n$ will have measure at most $liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:27
1
1
$begingroup$
No. $liminf_n A_n subseteq A_1$ is (in general) false. You should spend some time thinking about these things. In particular, try to connect the proof you have to intuition.
$endgroup$
– mathworker21
Jan 5 at 15:31
$begingroup$
No. $liminf_n A_n subseteq A_1$ is (in general) false. You should spend some time thinking about these things. In particular, try to connect the proof you have to intuition.
$endgroup$
– mathworker21
Jan 5 at 15:31
add a comment |
1 Answer
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$begingroup$
Well, let's focus on the first inequality. The idea is simple: on the left-hand-side one is taking the intersection of the sets from some moment, while on the right-hand-side one is taking the smallest size of the sets from some moment. If you consider $X := X_1cap cdots cap X_n$, there is no way that $|X| > |X_i|$ for some $i$, right? The same reason here, and all you need to do is to fit it in a limit and measure theoretic setting.
answered Jan 8 at 15:48
AtugoAtugo
363
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add a comment |
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$begingroup$
Well, let's focus on the first inequality. The idea is simple: on the left-hand-side one is taking the intersection of the sets from some moment, while on the right-hand-side one is taking the smallest size of the sets from some moment. If you consider $X := X_1cap cdots cap X_n$, there is no way that $|X| > |X_i|$ for some $i$, right? The same reason here, and all you need to do is to fit it in a limit and measure theoretic setting.
answered Jan 8 at 15:48
AtugoAtugo
363
$endgroup$
add a comment |
$begingroup$
Well, let's focus on the first inequality. The idea is simple: on the left-hand-side one is taking the intersection of the sets from some moment, while on the right-hand-side one is taking the smallest size of the sets from some moment. If you consider $X := X_1cap cdots cap X_n$, there is no way that $|X| > |X_i|$ for some $i$, right? The same reason here, and all you need to do is to fit it in a limit and measure theoretic setting.
answered Jan 8 at 15:48
AtugoAtugo
363
$endgroup$
add a comment |
$begingroup$
Well, let's focus on the first inequality. The idea is simple: on the left-hand-side one is taking the intersection of the sets from some moment, while on the right-hand-side one is taking the smallest size of the sets from some moment. If you consider $X := X_1cap cdots cap X_n$, there is no way that $|X| > |X_i|$ for some $i$, right? The same reason here, and all you need to do is to fit it in a limit and measure theoretic setting.
answered Jan 8 at 15:48
AtugoAtugo
363
$endgroup$
Well, let's focus on the first inequality. The idea is simple: on the left-hand-side one is taking the intersection of the sets from some moment, while on the right-hand-side one is taking the smallest size of the sets from some moment. If you consider $X := X_1cap cdots cap X_n$, there is no way that $|X| > |X_i|$ for some $i$, right? The same reason here, and all you need to do is to fit it in a limit and measure theoretic setting.
answered Jan 8 at 15:48
AtugoAtugo
363
answered Jan 8 at 15:48
AtugoAtugo
363
answered Jan 8 at 15:48
AtugoAtugo
363
answered Jan 8 at 15:48
AtugoAtugo
363
363
add a comment |
add a comment |
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$begingroup$
At the risk of being rude, I don't think a deep intuition is needed; I think they are just obvious. For example, $liminf_n A_n$ represents the elements that are in all of the $A_n$'s past some point. So clearly the measure of the set of these elements is less than the liminf of the measures of the $A_n$'s.
$endgroup$
– mathworker21
Jan 5 at 15:20
$begingroup$
@mathworker21 I understand, that the $lim inf$ contains all the elements which are in all but finitely many of the $A_n$, but to me it's not obvious why the measure of these elements is always smaller or equal to the $lim inf$ of the measures...
$endgroup$
– Herickson
Jan 5 at 15:23
$begingroup$
Don't think of it as "all but finitely many". Think of it as "past some point". Let's pretend that $liminf_{n to infty} A_n$ happened to be all elements belonging to $A_{10} cap A_{11} cap A_{12} cap dots$. Then $mu(liminf_n A_n) le mu(A_i)$ for each $i ge 10$ and in particular, it is $le liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:26
$begingroup$
In reality, part of $liminf_n A_n$ will also be $A_{12}cap A_{13} cap dots$. But this doesn't matter. Each of these "parts" of $liminf_n A_n$ will have measure at most $liminf_i mu(A_i)$.
$endgroup$
– mathworker21
Jan 5 at 15:27
1
$begingroup$
No. $liminf_n A_n subseteq A_1$ is (in general) false. You should spend some time thinking about these things. In particular, try to connect the proof you have to intuition.
$endgroup$
– mathworker21
Jan 5 at 15:31