Property of a supremum of a measure
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Prove that if the Measure Space is Finite ($mu(X)<infty)$, then $limsup mu(A_n)le mu(limsup A_n)$ Where $limsup A_n=cap_{n=1}^{infty}cup_{k=n}^{infty}A_k$
I understand the solution, what I don't get is why do we have to demand the space to be finite? Yet we don't demand that in the similar equation for the infimum: $mu(liminf A_n)le liminfmu(A_n)$ Where $liminf A_n=cup_{n=1}^{infty}cap_{k=n}^{infty}A_k$
For if $mu(A_n)=infty$ surely $mu(limsup A_n)=infty$ right? so the equation will hold.
Thanks,
measure-theory
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
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add a comment |
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Prove that if the Measure Space is Finite ($mu(X)<infty)$, then $limsup mu(A_n)le mu(limsup A_n)$ Where $limsup A_n=cap_{n=1}^{infty}cup_{k=n}^{infty}A_k$
I understand the solution, what I don't get is why do we have to demand the space to be finite? Yet we don't demand that in the similar equation for the infimum: $mu(liminf A_n)le liminfmu(A_n)$ Where $liminf A_n=cup_{n=1}^{infty}cap_{k=n}^{infty}A_k$
For if $mu(A_n)=infty$ surely $mu(limsup A_n)=infty$ right? so the equation will hold.
Thanks,
measure-theory
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
$endgroup$
$begingroup$
Try sets $[n,n+1)$ on the real line.
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– user25959
Dec 7 '18 at 5:32
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That's pretty cool. Why wouldn't there be such an example for the Infimum?
$endgroup$
– SlyxBrd
Dec 7 '18 at 5:35
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$mu(A_n) = infty$ will not imply that $mu(limsup A_n) = infty$, since the limit superior consists of points that belong in infinitely many of the sets. In particular, since we can easily find infinitely many disjoint sets on the real line with infinite measure, the limit superior of such a sequence will be the empty set. This cannot be done with a set of finite measure, since for every $epsilon > 0$ you cannot have infinitely many disjoint sets of measure greater than $epsilon$.
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– астон вілла олоф мэллбэрг
Dec 7 '18 at 5:40
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For the liminf case it "feels" like there shouldn't be a problem with infinite measures because the terms that could potentially go to infinity are on the greater side of the inequality. But of course proof and counterexample are the way to go.
$endgroup$
– user25959
Dec 7 '18 at 5:42
add a comment |
$begingroup$
Prove that if the Measure Space is Finite ($mu(X)<infty)$, then $limsup mu(A_n)le mu(limsup A_n)$ Where $limsup A_n=cap_{n=1}^{infty}cup_{k=n}^{infty}A_k$
I understand the solution, what I don't get is why do we have to demand the space to be finite? Yet we don't demand that in the similar equation for the infimum: $mu(liminf A_n)le liminfmu(A_n)$ Where $liminf A_n=cup_{n=1}^{infty}cap_{k=n}^{infty}A_k$
For if $mu(A_n)=infty$ surely $mu(limsup A_n)=infty$ right? so the equation will hold.
Thanks,
measure-theory
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
$endgroup$
Prove that if the Measure Space is Finite ($mu(X)<infty)$, then $limsup mu(A_n)le mu(limsup A_n)$ Where $limsup A_n=cap_{n=1}^{infty}cup_{k=n}^{infty}A_k$
I understand the solution, what I don't get is why do we have to demand the space to be finite? Yet we don't demand that in the similar equation for the infimum: $mu(liminf A_n)le liminfmu(A_n)$ Where $liminf A_n=cup_{n=1}^{infty}cap_{k=n}^{infty}A_k$
For if $mu(A_n)=infty$ surely $mu(limsup A_n)=infty$ right? so the equation will hold.
Thanks,
measure-theory
measure-theory
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
edited Dec 7 '18 at 5:29
SlyxBrd
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
asked Dec 7 '18 at 5:23
SlyxBrdSlyxBrd
646
646
$begingroup$
Try sets $[n,n+1)$ on the real line.
$endgroup$
– user25959
Dec 7 '18 at 5:32
$begingroup$
That's pretty cool. Why wouldn't there be such an example for the Infimum?
$endgroup$
– SlyxBrd
Dec 7 '18 at 5:35
$begingroup$
$mu(A_n) = infty$ will not imply that $mu(limsup A_n) = infty$, since the limit superior consists of points that belong in infinitely many of the sets. In particular, since we can easily find infinitely many disjoint sets on the real line with infinite measure, the limit superior of such a sequence will be the empty set. This cannot be done with a set of finite measure, since for every $epsilon > 0$ you cannot have infinitely many disjoint sets of measure greater than $epsilon$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 7 '18 at 5:40
$begingroup$
For the liminf case it "feels" like there shouldn't be a problem with infinite measures because the terms that could potentially go to infinity are on the greater side of the inequality. But of course proof and counterexample are the way to go.
$endgroup$
– user25959
Dec 7 '18 at 5:42
add a comment |
$begingroup$
Try sets $[n,n+1)$ on the real line.
$endgroup$
– user25959
Dec 7 '18 at 5:32
$begingroup$
That's pretty cool. Why wouldn't there be such an example for the Infimum?
$endgroup$
– SlyxBrd
Dec 7 '18 at 5:35
$begingroup$
$mu(A_n) = infty$ will not imply that $mu(limsup A_n) = infty$, since the limit superior consists of points that belong in infinitely many of the sets. In particular, since we can easily find infinitely many disjoint sets on the real line with infinite measure, the limit superior of such a sequence will be the empty set. This cannot be done with a set of finite measure, since for every $epsilon > 0$ you cannot have infinitely many disjoint sets of measure greater than $epsilon$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 7 '18 at 5:40
$begingroup$
For the liminf case it "feels" like there shouldn't be a problem with infinite measures because the terms that could potentially go to infinity are on the greater side of the inequality. But of course proof and counterexample are the way to go.
$endgroup$
– user25959
Dec 7 '18 at 5:42
$begingroup$
Try sets $[n,n+1)$ on the real line.
$endgroup$
– user25959
Dec 7 '18 at 5:32
$begingroup$
Try sets $[n,n+1)$ on the real line.
$endgroup$
– user25959
Dec 7 '18 at 5:32
$begingroup$
That's pretty cool. Why wouldn't there be such an example for the Infimum?
$endgroup$
– SlyxBrd
Dec 7 '18 at 5:35
$begingroup$
That's pretty cool. Why wouldn't there be such an example for the Infimum?
$endgroup$
– SlyxBrd
Dec 7 '18 at 5:35
$begingroup$
$mu(A_n) = infty$ will not imply that $mu(limsup A_n) = infty$, since the limit superior consists of points that belong in infinitely many of the sets. In particular, since we can easily find infinitely many disjoint sets on the real line with infinite measure, the limit superior of such a sequence will be the empty set. This cannot be done with a set of finite measure, since for every $epsilon > 0$ you cannot have infinitely many disjoint sets of measure greater than $epsilon$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 7 '18 at 5:40
$begingroup$
$mu(A_n) = infty$ will not imply that $mu(limsup A_n) = infty$, since the limit superior consists of points that belong in infinitely many of the sets. In particular, since we can easily find infinitely many disjoint sets on the real line with infinite measure, the limit superior of such a sequence will be the empty set. This cannot be done with a set of finite measure, since for every $epsilon > 0$ you cannot have infinitely many disjoint sets of measure greater than $epsilon$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 7 '18 at 5:40
$begingroup$
For the liminf case it "feels" like there shouldn't be a problem with infinite measures because the terms that could potentially go to infinity are on the greater side of the inequality. But of course proof and counterexample are the way to go.
$endgroup$
– user25959
Dec 7 '18 at 5:42
$begingroup$
For the liminf case it "feels" like there shouldn't be a problem with infinite measures because the terms that could potentially go to infinity are on the greater side of the inequality. But of course proof and counterexample are the way to go.
$endgroup$
– user25959
Dec 7 '18 at 5:42
add a comment |
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$begingroup$
Try sets $[n,n+1)$ on the real line.
$endgroup$
– user25959
Dec 7 '18 at 5:32
$begingroup$
That's pretty cool. Why wouldn't there be such an example for the Infimum?
$endgroup$
– SlyxBrd
Dec 7 '18 at 5:35
$begingroup$
$mu(A_n) = infty$ will not imply that $mu(limsup A_n) = infty$, since the limit superior consists of points that belong in infinitely many of the sets. In particular, since we can easily find infinitely many disjoint sets on the real line with infinite measure, the limit superior of such a sequence will be the empty set. This cannot be done with a set of finite measure, since for every $epsilon > 0$ you cannot have infinitely many disjoint sets of measure greater than $epsilon$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 7 '18 at 5:40
$begingroup$
For the liminf case it "feels" like there shouldn't be a problem with infinite measures because the terms that could potentially go to infinity are on the greater side of the inequality. But of course proof and counterexample are the way to go.
$endgroup$
– user25959
Dec 7 '18 at 5:42