Mixing
Divergent Countable Sum from Uncountable Sum
Clash Royale CLAN TAG#URR8PPP
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While pondering on the counting measure recently, I considered the following:
Let us define $sum_x in Xf(x)$ as $int_X f(x) dmu$ where $mu$ is the counting measure
Suppose $sum_x in Xf(x) = infty$ where $X$ is uncountable and $0 le f le infty$
Does there exist some countable subset $S subset X$ such that $sum_x in Sf(x) = infty$?
All the examples I tried worked out, but I'm not sure about the validity of the result in general - in fact, I remain quite skeptical. I imagine there is a simple proof either way, but haven't thought of one.
Does anyone know such a proof?
real-analysis measure-theory summation lebesgue-integral
asked Sep 8 at 4:48
Brevan Ellefsen
11.5k31549
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up vote
2
down vote
favorite
While pondering on the counting measure recently, I considered the following:
Let us define $sum_x in Xf(x)$ as $int_X f(x) dmu$ where $mu$ is the counting measure
Suppose $sum_x in Xf(x) = infty$ where $X$ is uncountable and $0 le f le infty$
Does there exist some countable subset $S subset X$ such that $sum_x in Sf(x) = infty$?
All the examples I tried worked out, but I'm not sure about the validity of the result in general - in fact, I remain quite skeptical. I imagine there is a simple proof either way, but haven't thought of one.
Does anyone know such a proof?
real-analysis measure-theory summation lebesgue-integral
asked Sep 8 at 4:48
Brevan Ellefsen
11.5k31549
Technicality, is $f ge 0$?
– copper.hat
Sep 8 at 4:50
@copper.hat yes. $0 le f le infty$. I have edited the post accordingly.
– Brevan Ellefsen
Sep 8 at 4:51
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
While pondering on the counting measure recently, I considered the following:
Let us define $sum_x in Xf(x)$ as $int_X f(x) dmu$ where $mu$ is the counting measure
Suppose $sum_x in Xf(x) = infty$ where $X$ is uncountable and $0 le f le infty$
Does there exist some countable subset $S subset X$ such that $sum_x in Sf(x) = infty$?
All the examples I tried worked out, but I'm not sure about the validity of the result in general - in fact, I remain quite skeptical. I imagine there is a simple proof either way, but haven't thought of one.
Does anyone know such a proof?
real-analysis measure-theory summation lebesgue-integral
asked Sep 8 at 4:48
Brevan Ellefsen
11.5k31549
While pondering on the counting measure recently, I considered the following:
Let us define $sum_x in Xf(x)$ as $int_X f(x) dmu$ where $mu$ is the counting measure
Suppose $sum_x in Xf(x) = infty$ where $X$ is uncountable and $0 le f le infty$
Does there exist some countable subset $S subset X$ such that $sum_x in Sf(x) = infty$?
All the examples I tried worked out, but I'm not sure about the validity of the result in general - in fact, I remain quite skeptical. I imagine there is a simple proof either way, but haven't thought of one.
Does anyone know such a proof?
real-analysis measure-theory summation lebesgue-integral
asked Sep 8 at 4:48
Brevan Ellefsen
11.5k31549
edited Sep 8 at 4:52
asked Sep 8 at 4:48
Brevan Ellefsen
11.5k31549
asked Sep 8 at 4:48
Brevan Ellefsen
11.5k31549
asked Sep 8 at 4:48
Brevan Ellefsen
11.5k31549
11.5k31549
Technicality, is $f ge 0$?
– copper.hat
Sep 8 at 4:50
@copper.hat yes. $0 le f le infty$. I have edited the post accordingly.
– Brevan Ellefsen
Sep 8 at 4:51
add a comment |Â
Technicality, is $f ge 0$?
– copper.hat
Sep 8 at 4:50
@copper.hat yes. $0 le f le infty$. I have edited the post accordingly.
– Brevan Ellefsen
Sep 8 at 4:51
Technicality, is $f ge 0$?
– copper.hat
Sep 8 at 4:50
Technicality, is $f ge 0$?
– copper.hat
Sep 8 at 4:50
@copper.hat yes. $0 le f le infty$. I have edited the post accordingly.
– Brevan Ellefsen
Sep 8 at 4:51
@copper.hat yes. $0 le f le infty$. I have edited the post accordingly.
– Brevan Ellefsen
Sep 8 at 4:51
add a comment |Â
2 Answers
2
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oldest
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up vote
4
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accepted
Since $f ge 0$ then either $f>0$ for only countably many terms, or $f>0$ for uncountably many terms. Then uncountably many must be bigger than some $1/n$ for some $n$. Hence we have the result either way.
answered Sep 8 at 4:53
David Bowman
4,1921924
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up vote
2
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Same result as David's but expressed in a more complicated way.
If $fcdot 1_X$ is $|cdot|$ measurable there is a sequence of simple functions $s_n$ such that $s_n le f cdot 1_X$ and $int s_n to infty$.
If the support of all the $s_n$ is countable, then the union of the supports is countable and the integral over this set is $infty$.
If the support of any of the $s_n$ is uncountable, then we have
$f ge s_n ge alpha cdot1_A$ for some uncountable $A$ with $alpha >0$. Now choose any countable subset of $A$ to get the desired result.
answered Sep 8 at 5:24
copper.hat
123k557156
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Since $f ge 0$ then either $f>0$ for only countably many terms, or $f>0$ for uncountably many terms. Then uncountably many must be bigger than some $1/n$ for some $n$. Hence we have the result either way.
answered Sep 8 at 4:53
David Bowman
4,1921924
add a comment |Â
up vote
4
down vote
accepted
Since $f ge 0$ then either $f>0$ for only countably many terms, or $f>0$ for uncountably many terms. Then uncountably many must be bigger than some $1/n$ for some $n$. Hence we have the result either way.
answered Sep 8 at 4:53
David Bowman
4,1921924
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Since $f ge 0$ then either $f>0$ for only countably many terms, or $f>0$ for uncountably many terms. Then uncountably many must be bigger than some $1/n$ for some $n$. Hence we have the result either way.
answered Sep 8 at 4:53
David Bowman
4,1921924
Since $f ge 0$ then either $f>0$ for only countably many terms, or $f>0$ for uncountably many terms. Then uncountably many must be bigger than some $1/n$ for some $n$. Hence we have the result either way.
answered Sep 8 at 4:53
David Bowman
4,1921924
answered Sep 8 at 4:53
David Bowman
4,1921924
answered Sep 8 at 4:53
David Bowman
4,1921924
answered Sep 8 at 4:53
David Bowman
4,1921924
4,1921924
add a comment |Â
add a comment |Â
up vote
2
down vote
Same result as David's but expressed in a more complicated way.
If $fcdot 1_X$ is $|cdot|$ measurable there is a sequence of simple functions $s_n$ such that $s_n le f cdot 1_X$ and $int s_n to infty$.
If the support of all the $s_n$ is countable, then the union of the supports is countable and the integral over this set is $infty$.
If the support of any of the $s_n$ is uncountable, then we have
$f ge s_n ge alpha cdot1_A$ for some uncountable $A$ with $alpha >0$. Now choose any countable subset of $A$ to get the desired result.
answered Sep 8 at 5:24
copper.hat
123k557156
add a comment |Â
up vote
2
down vote
Same result as David's but expressed in a more complicated way.
If $fcdot 1_X$ is $|cdot|$ measurable there is a sequence of simple functions $s_n$ such that $s_n le f cdot 1_X$ and $int s_n to infty$.
If the support of all the $s_n$ is countable, then the union of the supports is countable and the integral over this set is $infty$.
If the support of any of the $s_n$ is uncountable, then we have
$f ge s_n ge alpha cdot1_A$ for some uncountable $A$ with $alpha >0$. Now choose any countable subset of $A$ to get the desired result.
answered Sep 8 at 5:24
copper.hat
123k557156
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Same result as David's but expressed in a more complicated way.
If $fcdot 1_X$ is $|cdot|$ measurable there is a sequence of simple functions $s_n$ such that $s_n le f cdot 1_X$ and $int s_n to infty$.
If the support of all the $s_n$ is countable, then the union of the supports is countable and the integral over this set is $infty$.
If the support of any of the $s_n$ is uncountable, then we have
$f ge s_n ge alpha cdot1_A$ for some uncountable $A$ with $alpha >0$. Now choose any countable subset of $A$ to get the desired result.
answered Sep 8 at 5:24
copper.hat
123k557156
Same result as David's but expressed in a more complicated way.
If $fcdot 1_X$ is $|cdot|$ measurable there is a sequence of simple functions $s_n$ such that $s_n le f cdot 1_X$ and $int s_n to infty$.
If the support of all the $s_n$ is countable, then the union of the supports is countable and the integral over this set is $infty$.
If the support of any of the $s_n$ is uncountable, then we have
$f ge s_n ge alpha cdot1_A$ for some uncountable $A$ with $alpha >0$. Now choose any countable subset of $A$ to get the desired result.
answered Sep 8 at 5:24
copper.hat
123k557156
answered Sep 8 at 5:24
copper.hat
123k557156
answered Sep 8 at 5:24
copper.hat
123k557156
answered Sep 8 at 5:24
copper.hat
123k557156
123k557156
add a comment |Â
add a comment |Â
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Technicality, is $f ge 0$?
– copper.hat
Sep 8 at 4:50
@copper.hat yes. $0 le f le infty$. I have edited the post accordingly.
– Brevan Ellefsen
Sep 8 at 4:51