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The set of points in $A_n$ for infinitely many $n$ is measurable.
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This is part of an exercise from "Real analysis for graduate students" by Richard Bass: Let $m$ be a Lebesgue measure. Suppose for each $n$, $A_n$ is a Lebesgue measurable subset of $[0,1]$. Let $B$ consist of those points $x$ that are in infinitely many of the $A_n$. Show that $B$ is Lebesgue measurable.
I'm in general undecided about what to use to prove that something is Lebesgue measurable. Is it a good way to use outer measure definition and try to show $m^*(E)=m^*(Ecap B)+ m(Ecap B^c)$, for instance. It seemed hard to prove this way for me, and I was wondering what is the best way. Thanks!
real-analysis measure-theory lebesgue-measure
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
asked May 24 '15 at 20:25
vgmath
895413
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
up vote
3
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This is part of an exercise from "Real analysis for graduate students" by Richard Bass: Let $m$ be a Lebesgue measure. Suppose for each $n$, $A_n$ is a Lebesgue measurable subset of $[0,1]$. Let $B$ consist of those points $x$ that are in infinitely many of the $A_n$. Show that $B$ is Lebesgue measurable.
I'm in general undecided about what to use to prove that something is Lebesgue measurable. Is it a good way to use outer measure definition and try to show $m^*(E)=m^*(Ecap B)+ m(Ecap B^c)$, for instance. It seemed hard to prove this way for me, and I was wondering what is the best way. Thanks!
real-analysis measure-theory lebesgue-measure
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
asked May 24 '15 at 20:25
vgmath
895413
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Write $B=bigcaplimits_{n=1}^infty bigcuplimits_{k=n}^infty A_n$.
– David Mitra
May 24 '15 at 20:45
Should be $A_k$ above...
– David Mitra
May 24 '15 at 20:54
@vgmath: As a general piece of advice: Note that for a set $B$ to be $m^*$-measurable it suffices to show $m^*(E) geq m^*(Ecap B)+ m^*(Ecap B^c)$, since the other inequality is fulfilled always if $m^*$ is an outer measure.
– el_tenedor
May 24 '15 at 21:04
@David: Thanks a lot, I learned a new nice trick, thanks to you.
– vgmath
May 24 '15 at 22:13
@ el_tenedor: Thanks for the advice. Yes, actually I've already seen that idea from the text I've been reading.
– vgmath
May 24 '15 at 22:15
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is part of an exercise from "Real analysis for graduate students" by Richard Bass: Let $m$ be a Lebesgue measure. Suppose for each $n$, $A_n$ is a Lebesgue measurable subset of $[0,1]$. Let $B$ consist of those points $x$ that are in infinitely many of the $A_n$. Show that $B$ is Lebesgue measurable.
I'm in general undecided about what to use to prove that something is Lebesgue measurable. Is it a good way to use outer measure definition and try to show $m^*(E)=m^*(Ecap B)+ m(Ecap B^c)$, for instance. It seemed hard to prove this way for me, and I was wondering what is the best way. Thanks!
real-analysis measure-theory lebesgue-measure
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
asked May 24 '15 at 20:25
vgmath
895413
This is part of an exercise from "Real analysis for graduate students" by Richard Bass: Let $m$ be a Lebesgue measure. Suppose for each $n$, $A_n$ is a Lebesgue measurable subset of $[0,1]$. Let $B$ consist of those points $x$ that are in infinitely many of the $A_n$. Show that $B$ is Lebesgue measurable.
I'm in general undecided about what to use to prove that something is Lebesgue measurable. Is it a good way to use outer measure definition and try to show $m^*(E)=m^*(Ecap B)+ m(Ecap B^c)$, for instance. It seemed hard to prove this way for me, and I was wondering what is the best way. Thanks!
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
asked May 24 '15 at 20:25
vgmath
895413
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
asked May 24 '15 at 20:25
vgmath
895413
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
edited Jun 26 '15 at 14:44
Michael Albanese
62.5k1598299
62.5k1598299
asked May 24 '15 at 20:25
vgmath
895413
asked May 24 '15 at 20:25
vgmath
895413
asked May 24 '15 at 20:25
vgmath
895413
895413
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Write $B=bigcaplimits_{n=1}^infty bigcuplimits_{k=n}^infty A_n$.
– David Mitra
May 24 '15 at 20:45
Should be $A_k$ above...
– David Mitra
May 24 '15 at 20:54
@vgmath: As a general piece of advice: Note that for a set $B$ to be $m^*$-measurable it suffices to show $m^*(E) geq m^*(Ecap B)+ m^*(Ecap B^c)$, since the other inequality is fulfilled always if $m^*$ is an outer measure.
– el_tenedor
May 24 '15 at 21:04
@David: Thanks a lot, I learned a new nice trick, thanks to you.
– vgmath
May 24 '15 at 22:13
@ el_tenedor: Thanks for the advice. Yes, actually I've already seen that idea from the text I've been reading.
– vgmath
May 24 '15 at 22:15
add a comment |
1
Write $B=bigcaplimits_{n=1}^infty bigcuplimits_{k=n}^infty A_n$.
– David Mitra
May 24 '15 at 20:45
Should be $A_k$ above...
– David Mitra
May 24 '15 at 20:54
@vgmath: As a general piece of advice: Note that for a set $B$ to be $m^*$-measurable it suffices to show $m^*(E) geq m^*(Ecap B)+ m^*(Ecap B^c)$, since the other inequality is fulfilled always if $m^*$ is an outer measure.
– el_tenedor
May 24 '15 at 21:04
@David: Thanks a lot, I learned a new nice trick, thanks to you.
– vgmath
May 24 '15 at 22:13
@ el_tenedor: Thanks for the advice. Yes, actually I've already seen that idea from the text I've been reading.
– vgmath
May 24 '15 at 22:15
1
1
Write $B=bigcaplimits_{n=1}^infty bigcuplimits_{k=n}^infty A_n$.
– David Mitra
May 24 '15 at 20:45
Write $B=bigcaplimits_{n=1}^infty bigcuplimits_{k=n}^infty A_n$.
– David Mitra
May 24 '15 at 20:45
Should be $A_k$ above...
– David Mitra
May 24 '15 at 20:54
Should be $A_k$ above...
– David Mitra
May 24 '15 at 20:54
@vgmath: As a general piece of advice: Note that for a set $B$ to be $m^*$-measurable it suffices to show $m^*(E) geq m^*(Ecap B)+ m^*(Ecap B^c)$, since the other inequality is fulfilled always if $m^*$ is an outer measure.
– el_tenedor
May 24 '15 at 21:04
@vgmath: As a general piece of advice: Note that for a set $B$ to be $m^*$-measurable it suffices to show $m^*(E) geq m^*(Ecap B)+ m^*(Ecap B^c)$, since the other inequality is fulfilled always if $m^*$ is an outer measure.
– el_tenedor
May 24 '15 at 21:04
@David: Thanks a lot, I learned a new nice trick, thanks to you.
– vgmath
May 24 '15 at 22:13
@David: Thanks a lot, I learned a new nice trick, thanks to you.
– vgmath
May 24 '15 at 22:13
@ el_tenedor: Thanks for the advice. Yes, actually I've already seen that idea from the text I've been reading.
– vgmath
May 24 '15 at 22:15
@ el_tenedor: Thanks for the advice. Yes, actually I've already seen that idea from the text I've been reading.
– vgmath
May 24 '15 at 22:15
add a comment |
1 Answer
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Let ${A_n}$ be a countable collection of sets and let $B$ denote the set of elements which are in $A_n$ for infinitely many $n$. Then
$$B = bigcap_{n=1}^{infty}bigcup_{k = n}^{infty}A_k.$$
To see this, note that $x in bigcaplimits_{n=1}^{infty}bigcuplimits_{k = n}^{infty}A_k$ if and only if $x in bigcuplimits_{k=n}^{infty}A_k$ for all $n$. The latter occurs if and only if $x in A_n$ for infinitely many $n$; if $x$ were only in finitely many of the sets, say $A_{n_1}, dots, A_{n_j}$, then $x notin bigcuplimits_{k=n_j+1}^{infty}A_k$.
As the collection of Lebesgue measurable sets is closed under countable unions and countable intersections (it is a $sigma$-algebra), $B$ is Lebesgue measurable.
A similar exercise is to show that the set of elements which belong to $A_n$ for all but finitely many $n$ is measurable. Denoting this set by $C$, you can show that $C$ is Lebesgue measurable by using the fact that
$$C = bigcup_{n=1}^{infty}bigcap_{k = n}^{infty}A_k.$$
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let ${A_n}$ be a countable collection of sets and let $B$ denote the set of elements which are in $A_n$ for infinitely many $n$. Then
$$B = bigcap_{n=1}^{infty}bigcup_{k = n}^{infty}A_k.$$
To see this, note that $x in bigcaplimits_{n=1}^{infty}bigcuplimits_{k = n}^{infty}A_k$ if and only if $x in bigcuplimits_{k=n}^{infty}A_k$ for all $n$. The latter occurs if and only if $x in A_n$ for infinitely many $n$; if $x$ were only in finitely many of the sets, say $A_{n_1}, dots, A_{n_j}$, then $x notin bigcuplimits_{k=n_j+1}^{infty}A_k$.
As the collection of Lebesgue measurable sets is closed under countable unions and countable intersections (it is a $sigma$-algebra), $B$ is Lebesgue measurable.
A similar exercise is to show that the set of elements which belong to $A_n$ for all but finitely many $n$ is measurable. Denoting this set by $C$, you can show that $C$ is Lebesgue measurable by using the fact that
$$C = bigcup_{n=1}^{infty}bigcap_{k = n}^{infty}A_k.$$
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
add a comment |
up vote
0
down vote
Let ${A_n}$ be a countable collection of sets and let $B$ denote the set of elements which are in $A_n$ for infinitely many $n$. Then
$$B = bigcap_{n=1}^{infty}bigcup_{k = n}^{infty}A_k.$$
To see this, note that $x in bigcaplimits_{n=1}^{infty}bigcuplimits_{k = n}^{infty}A_k$ if and only if $x in bigcuplimits_{k=n}^{infty}A_k$ for all $n$. The latter occurs if and only if $x in A_n$ for infinitely many $n$; if $x$ were only in finitely many of the sets, say $A_{n_1}, dots, A_{n_j}$, then $x notin bigcuplimits_{k=n_j+1}^{infty}A_k$.
As the collection of Lebesgue measurable sets is closed under countable unions and countable intersections (it is a $sigma$-algebra), $B$ is Lebesgue measurable.
A similar exercise is to show that the set of elements which belong to $A_n$ for all but finitely many $n$ is measurable. Denoting this set by $C$, you can show that $C$ is Lebesgue measurable by using the fact that
$$C = bigcup_{n=1}^{infty}bigcap_{k = n}^{infty}A_k.$$
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
add a comment |
up vote
0
down vote
up vote
0
down vote
Let ${A_n}$ be a countable collection of sets and let $B$ denote the set of elements which are in $A_n$ for infinitely many $n$. Then
$$B = bigcap_{n=1}^{infty}bigcup_{k = n}^{infty}A_k.$$
To see this, note that $x in bigcaplimits_{n=1}^{infty}bigcuplimits_{k = n}^{infty}A_k$ if and only if $x in bigcuplimits_{k=n}^{infty}A_k$ for all $n$. The latter occurs if and only if $x in A_n$ for infinitely many $n$; if $x$ were only in finitely many of the sets, say $A_{n_1}, dots, A_{n_j}$, then $x notin bigcuplimits_{k=n_j+1}^{infty}A_k$.
As the collection of Lebesgue measurable sets is closed under countable unions and countable intersections (it is a $sigma$-algebra), $B$ is Lebesgue measurable.
A similar exercise is to show that the set of elements which belong to $A_n$ for all but finitely many $n$ is measurable. Denoting this set by $C$, you can show that $C$ is Lebesgue measurable by using the fact that
$$C = bigcup_{n=1}^{infty}bigcap_{k = n}^{infty}A_k.$$
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
Let ${A_n}$ be a countable collection of sets and let $B$ denote the set of elements which are in $A_n$ for infinitely many $n$. Then
$$B = bigcap_{n=1}^{infty}bigcup_{k = n}^{infty}A_k.$$
To see this, note that $x in bigcaplimits_{n=1}^{infty}bigcuplimits_{k = n}^{infty}A_k$ if and only if $x in bigcuplimits_{k=n}^{infty}A_k$ for all $n$. The latter occurs if and only if $x in A_n$ for infinitely many $n$; if $x$ were only in finitely many of the sets, say $A_{n_1}, dots, A_{n_j}$, then $x notin bigcuplimits_{k=n_j+1}^{infty}A_k$.
As the collection of Lebesgue measurable sets is closed under countable unions and countable intersections (it is a $sigma$-algebra), $B$ is Lebesgue measurable.
A similar exercise is to show that the set of elements which belong to $A_n$ for all but finitely many $n$ is measurable. Denoting this set by $C$, you can show that $C$ is Lebesgue measurable by using the fact that
$$C = bigcup_{n=1}^{infty}bigcap_{k = n}^{infty}A_k.$$
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
answered Jun 26 '15 at 7:29
Michael Albanese
62.5k1598299
62.5k1598299
add a comment |
add a comment |
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1
Write $B=bigcaplimits_{n=1}^infty bigcuplimits_{k=n}^infty A_n$.
– David Mitra
May 24 '15 at 20:45
Should be $A_k$ above...
– David Mitra
May 24 '15 at 20:54
@vgmath: As a general piece of advice: Note that for a set $B$ to be $m^*$-measurable it suffices to show $m^*(E) geq m^*(Ecap B)+ m^*(Ecap B^c)$, since the other inequality is fulfilled always if $m^*$ is an outer measure.
– el_tenedor
May 24 '15 at 21:04
@David: Thanks a lot, I learned a new nice trick, thanks to you.
– vgmath
May 24 '15 at 22:13
@ el_tenedor: Thanks for the advice. Yes, actually I've already seen that idea from the text I've been reading.
– vgmath
May 24 '15 at 22:15