(Use the Vitali covering lemma) to show that the union of closed intervals of real numbers is measurableConsequence Vitali covering lemmaVariant of the Vitali Covering LemmaVitali CoveringOn why the Vitali Covering Lemma does not apply when the covering collection contains degenerate closed intervalsShow that the union over a collection of compact cubes in $mathbbR^n$ is Lebesgue measurableVersion of the Vitali Covering LemmaShow (using Vitali Covering Lemma) that the union of any collection of closed, bounded non-generate intervals is measurable.$int_bigcup_n=1^inftyE_nf=sum_n=1^inftyint_E_nf$ given $f$ positive and measurableQuestion about Vitali Covering (from a Lemma in Royden and Fitzpatrick's book)Covering of Intervals
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(Use the Vitali covering lemma) to show that the union of closed intervals of real numbers is measurable
Consequence Vitali covering lemmaVariant of the Vitali Covering LemmaVitali CoveringOn why the Vitali Covering Lemma does not apply when the covering collection contains degenerate closed intervalsShow that the union over a collection of compact cubes in $mathbbR^n$ is Lebesgue measurableVersion of the Vitali Covering LemmaShow (using Vitali Covering Lemma) that the union of any collection of closed, bounded non-generate intervals is measurable.$int_bigcup_n=1^inftyE_nf=sum_n=1^inftyint_E_nf$ given $f$ positive and measurableQuestion about Vitali Covering (from a Lemma in Royden and Fitzpatrick's book)Covering of Intervals
$begingroup$
I have seen similar questions before but the replies given were less than satisfactory.
Let $mathcalF = F_alpha_alpha in J$ be an arbitrary family of closed intervals of real numbers. Let $E = displaystyle bigcup_alpha in J F_alpha$, and let $E_n = E cap [-n, n]$. Let $V$ be a Vitali covering of $E$; then $V$ is a Vitali covering of $E_n$. By the Vitali Lemma, we can find a finite, pairwise disjoint collection $V_i_i = 1^m$ of $V$ such that $m^ast(E_n smallsetminus displaystyle bigcup_i = 1^m V_i) < varepsilon$. Since each $V_i$ is an interval of real numbers, and intervals are Lebesgue measurable, it follows that $displaystyle bigcup_i = 1^m V_i$ is measurable. Thus
$$bigcup_n = 1^ infty E_n$$
is a countable collection of measurable sets and is therefore measurable. Moreover, the Lebesgue measure of this union is equal to the Lebasgue measure of $E$, which is what we sought to prove.
Does this make any sense?
real-analysis measure-theory proof-verification
asked Mar 20 at 18:21
JunglemathJunglemath
184
$endgroup$
add a comment |
$begingroup$
I have seen similar questions before but the replies given were less than satisfactory.
Let $mathcalF = F_alpha_alpha in J$ be an arbitrary family of closed intervals of real numbers. Let $E = displaystyle bigcup_alpha in J F_alpha$, and let $E_n = E cap [-n, n]$. Let $V$ be a Vitali covering of $E$; then $V$ is a Vitali covering of $E_n$. By the Vitali Lemma, we can find a finite, pairwise disjoint collection $V_i_i = 1^m$ of $V$ such that $m^ast(E_n smallsetminus displaystyle bigcup_i = 1^m V_i) < varepsilon$. Since each $V_i$ is an interval of real numbers, and intervals are Lebesgue measurable, it follows that $displaystyle bigcup_i = 1^m V_i$ is measurable. Thus
$$bigcup_n = 1^ infty E_n$$
is a countable collection of measurable sets and is therefore measurable. Moreover, the Lebesgue measure of this union is equal to the Lebasgue measure of $E$, which is what we sought to prove.
Does this make any sense?
real-analysis measure-theory proof-verification
asked Mar 20 at 18:21
JunglemathJunglemath
184
$endgroup$
$begingroup$
You don't really need Vitali's covering lemma for this: just note that an arbitrary union of (nontrivial) closed intervals has countably many connected components (because they have nonempty disjoint interiors), and every connected subset of reals is an interval, and thus measurable.
$endgroup$
– tomasz
Mar 20 at 18:37
$begingroup$
@tomasz If my collection of closed intervals is uncountable, why are there only countably many connected components in the union? Why must the interiors be disjoint?
$endgroup$
– Junglemath
Mar 20 at 19:12
$begingroup$
Because connected components are disjoint, by definition, so their interiors certainly are, too!
$endgroup$
– tomasz
Mar 20 at 19:34
$begingroup$
I still don't see why it's a countable collection. Plus, what if two intervals overlap? How are their interiors disjoint?
$endgroup$
– Junglemath
Mar 20 at 20:59
$begingroup$
If they overlap, they are in the same connected components. A disjoint family of open subsets of reals cannot be uncountable.
$endgroup$
– tomasz
Mar 21 at 11:17
add a comment |
$begingroup$
I have seen similar questions before but the replies given were less than satisfactory.
Let $mathcalF = F_alpha_alpha in J$ be an arbitrary family of closed intervals of real numbers. Let $E = displaystyle bigcup_alpha in J F_alpha$, and let $E_n = E cap [-n, n]$. Let $V$ be a Vitali covering of $E$; then $V$ is a Vitali covering of $E_n$. By the Vitali Lemma, we can find a finite, pairwise disjoint collection $V_i_i = 1^m$ of $V$ such that $m^ast(E_n smallsetminus displaystyle bigcup_i = 1^m V_i) < varepsilon$. Since each $V_i$ is an interval of real numbers, and intervals are Lebesgue measurable, it follows that $displaystyle bigcup_i = 1^m V_i$ is measurable. Thus
$$bigcup_n = 1^ infty E_n$$
is a countable collection of measurable sets and is therefore measurable. Moreover, the Lebesgue measure of this union is equal to the Lebasgue measure of $E$, which is what we sought to prove.
Does this make any sense?
real-analysis measure-theory proof-verification
asked Mar 20 at 18:21
JunglemathJunglemath
184
$endgroup$
I have seen similar questions before but the replies given were less than satisfactory.
Let $mathcalF = F_alpha_alpha in J$ be an arbitrary family of closed intervals of real numbers. Let $E = displaystyle bigcup_alpha in J F_alpha$, and let $E_n = E cap [-n, n]$. Let $V$ be a Vitali covering of $E$; then $V$ is a Vitali covering of $E_n$. By the Vitali Lemma, we can find a finite, pairwise disjoint collection $V_i_i = 1^m$ of $V$ such that $m^ast(E_n smallsetminus displaystyle bigcup_i = 1^m V_i) < varepsilon$. Since each $V_i$ is an interval of real numbers, and intervals are Lebesgue measurable, it follows that $displaystyle bigcup_i = 1^m V_i$ is measurable. Thus
$$bigcup_n = 1^ infty E_n$$
is a countable collection of measurable sets and is therefore measurable. Moreover, the Lebesgue measure of this union is equal to the Lebasgue measure of $E$, which is what we sought to prove.
Does this make any sense?
real-analysis measure-theory proof-verification
real-analysis measure-theory proof-verification
asked Mar 20 at 18:21
JunglemathJunglemath
184
asked Mar 20 at 18:21
JunglemathJunglemath
184
asked Mar 20 at 18:21
JunglemathJunglemath
184
asked Mar 20 at 18:21
JunglemathJunglemath
184
asked Mar 20 at 18:21
JunglemathJunglemath
184
184
$begingroup$
You don't really need Vitali's covering lemma for this: just note that an arbitrary union of (nontrivial) closed intervals has countably many connected components (because they have nonempty disjoint interiors), and every connected subset of reals is an interval, and thus measurable.
$endgroup$
– tomasz
Mar 20 at 18:37
$begingroup$
@tomasz If my collection of closed intervals is uncountable, why are there only countably many connected components in the union? Why must the interiors be disjoint?
$endgroup$
– Junglemath
Mar 20 at 19:12
$begingroup$
Because connected components are disjoint, by definition, so their interiors certainly are, too!
$endgroup$
– tomasz
Mar 20 at 19:34
$begingroup$
I still don't see why it's a countable collection. Plus, what if two intervals overlap? How are their interiors disjoint?
$endgroup$
– Junglemath
Mar 20 at 20:59
$begingroup$
If they overlap, they are in the same connected components. A disjoint family of open subsets of reals cannot be uncountable.
$endgroup$
– tomasz
Mar 21 at 11:17
add a comment |
$begingroup$
You don't really need Vitali's covering lemma for this: just note that an arbitrary union of (nontrivial) closed intervals has countably many connected components (because they have nonempty disjoint interiors), and every connected subset of reals is an interval, and thus measurable.
$endgroup$
– tomasz
Mar 20 at 18:37
$begingroup$
@tomasz If my collection of closed intervals is uncountable, why are there only countably many connected components in the union? Why must the interiors be disjoint?
$endgroup$
– Junglemath
Mar 20 at 19:12
$begingroup$
Because connected components are disjoint, by definition, so their interiors certainly are, too!
$endgroup$
– tomasz
Mar 20 at 19:34
$begingroup$
I still don't see why it's a countable collection. Plus, what if two intervals overlap? How are their interiors disjoint?
$endgroup$
– Junglemath
Mar 20 at 20:59
$begingroup$
If they overlap, they are in the same connected components. A disjoint family of open subsets of reals cannot be uncountable.
$endgroup$
– tomasz
Mar 21 at 11:17
$begingroup$
You don't really need Vitali's covering lemma for this: just note that an arbitrary union of (nontrivial) closed intervals has countably many connected components (because they have nonempty disjoint interiors), and every connected subset of reals is an interval, and thus measurable.
$endgroup$
– tomasz
Mar 20 at 18:37
$begingroup$
You don't really need Vitali's covering lemma for this: just note that an arbitrary union of (nontrivial) closed intervals has countably many connected components (because they have nonempty disjoint interiors), and every connected subset of reals is an interval, and thus measurable.
$endgroup$
– tomasz
Mar 20 at 18:37
$begingroup$
@tomasz If my collection of closed intervals is uncountable, why are there only countably many connected components in the union? Why must the interiors be disjoint?
$endgroup$
– Junglemath
Mar 20 at 19:12
$begingroup$
@tomasz If my collection of closed intervals is uncountable, why are there only countably many connected components in the union? Why must the interiors be disjoint?
$endgroup$
– Junglemath
Mar 20 at 19:12
$begingroup$
Because connected components are disjoint, by definition, so their interiors certainly are, too!
$endgroup$
– tomasz
Mar 20 at 19:34
$begingroup$
Because connected components are disjoint, by definition, so their interiors certainly are, too!
$endgroup$
– tomasz
Mar 20 at 19:34
$begingroup$
I still don't see why it's a countable collection. Plus, what if two intervals overlap? How are their interiors disjoint?
$endgroup$
– Junglemath
Mar 20 at 20:59
$begingroup$
I still don't see why it's a countable collection. Plus, what if two intervals overlap? How are their interiors disjoint?
$endgroup$
– Junglemath
Mar 20 at 20:59
$begingroup$
If they overlap, they are in the same connected components. A disjoint family of open subsets of reals cannot be uncountable.
$endgroup$
– tomasz
Mar 21 at 11:17
$begingroup$
If they overlap, they are in the same connected components. A disjoint family of open subsets of reals cannot be uncountable.
$endgroup$
– tomasz
Mar 21 at 11:17
add a comment |
1 Answer
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I cannot write a comment because I have no reputation...
Yes, that is correct. $E_n$ is measurable iff $exists mathcalO$ so that $m*(E_n/mathcalO)<epsilon$ for any $epsilon>0$. So your proof is complete.
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
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add a comment |
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I cannot write a comment because I have no reputation...
Yes, that is correct. $E_n$ is measurable iff $exists mathcalO$ so that $m*(E_n/mathcalO)<epsilon$ for any $epsilon>0$. So your proof is complete.
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
$endgroup$
add a comment |
$begingroup$
I cannot write a comment because I have no reputation...
Yes, that is correct. $E_n$ is measurable iff $exists mathcalO$ so that $m*(E_n/mathcalO)<epsilon$ for any $epsilon>0$. So your proof is complete.
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
$endgroup$
add a comment |
$begingroup$
I cannot write a comment because I have no reputation...
Yes, that is correct. $E_n$ is measurable iff $exists mathcalO$ so that $m*(E_n/mathcalO)<epsilon$ for any $epsilon>0$. So your proof is complete.
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
$endgroup$
I cannot write a comment because I have no reputation...
Yes, that is correct. $E_n$ is measurable iff $exists mathcalO$ so that $m*(E_n/mathcalO)<epsilon$ for any $epsilon>0$. So your proof is complete.
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
answered Mar 20 at 18:30
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
213
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You don't really need Vitali's covering lemma for this: just note that an arbitrary union of (nontrivial) closed intervals has countably many connected components (because they have nonempty disjoint interiors), and every connected subset of reals is an interval, and thus measurable.
$endgroup$
– tomasz
Mar 20 at 18:37
$begingroup$
@tomasz If my collection of closed intervals is uncountable, why are there only countably many connected components in the union? Why must the interiors be disjoint?
$endgroup$
– Junglemath
Mar 20 at 19:12
$begingroup$
Because connected components are disjoint, by definition, so their interiors certainly are, too!
$endgroup$
– tomasz
Mar 20 at 19:34
$begingroup$
I still don't see why it's a countable collection. Plus, what if two intervals overlap? How are their interiors disjoint?
$endgroup$
– Junglemath
Mar 20 at 20:59
$begingroup$
If they overlap, they are in the same connected components. A disjoint family of open subsets of reals cannot be uncountable.
$endgroup$
– tomasz
Mar 21 at 11:17