Intuition behind inequality for measure of $liminf$ and $limsup$












1














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







share|cite|improve this question
























  • 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.
    – mathworker21
    2 days ago












  • @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...
    – Herickson
    2 days ago










  • 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 $le liminf_i mu(A_i)$.
    – mathworker21
    2 days ago










  • 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 $liminf_i mu(A_i)$.
    – mathworker21
    2 days ago






  • 1




    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.
    – mathworker21
    2 days ago
















1














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







share|cite|improve this question
























  • 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.
    – mathworker21
    2 days ago












  • @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...
    – Herickson
    2 days ago










  • 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 $le liminf_i mu(A_i)$.
    – mathworker21
    2 days ago










  • 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 $liminf_i mu(A_i)$.
    – mathworker21
    2 days ago






  • 1




    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.
    – mathworker21
    2 days ago














1












1








1


1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited 2 days ago









Bernard

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asked 2 days ago









HericksonHerickson

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1649












  • 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.
    – mathworker21
    2 days ago












  • @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...
    – Herickson
    2 days ago










  • 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 $le liminf_i mu(A_i)$.
    – mathworker21
    2 days ago










  • 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 $liminf_i mu(A_i)$.
    – mathworker21
    2 days ago






  • 1




    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.
    – mathworker21
    2 days ago


















  • 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.
    – mathworker21
    2 days ago












  • @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...
    – Herickson
    2 days ago










  • 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 $le liminf_i mu(A_i)$.
    – mathworker21
    2 days ago










  • 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 $liminf_i mu(A_i)$.
    – mathworker21
    2 days ago






  • 1




    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.
    – mathworker21
    2 days ago
















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.
– mathworker21
2 days ago






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.
– mathworker21
2 days ago














@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...
– Herickson
2 days ago




@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...
– Herickson
2 days ago












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 $le liminf_i mu(A_i)$.
– mathworker21
2 days ago




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 $le liminf_i mu(A_i)$.
– mathworker21
2 days ago












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 $liminf_i mu(A_i)$.
– mathworker21
2 days ago




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 $liminf_i mu(A_i)$.
– mathworker21
2 days ago




1




1




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.
– mathworker21
2 days ago




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.
– mathworker21
2 days ago










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