Exercise 3.D. in Robert G. Bartle's book
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Let us consider the problem 3.D. in the book
$textbfThe Elements of Integration and Lebesgue Measure$
of Robert G. Bartle
Let $X=mathbbN$ and $mathcalA$ be the $sigma-$algebra of all subsets of $mathbbN$. If $(a_n)$ is a sequence of nonnegative real numbers and if we define $mu$ by
$$
mu(emptyset)=0; quad mu(E)=sum_nin Ea_n, quad Eneemptyset,
$$
then $mu$ is a measure on $mathcalA$. Conversely, every measure on $mathcalA$ is obtained in this way for some sequence $(a_n)$ in $overlineR^+$.
I am stuck in proving the countably additive property of $mu$.
My attempt. I intend to use the theory of double index series to prove countably additive property.
Thank you for all solutions.
lebesgue-measure
asked Nov 2 '14 at 10:54
impartialmale
566213
add a comment |Â
up vote
1
down vote
favorite
Let us consider the problem 3.D. in the book
$textbfThe Elements of Integration and Lebesgue Measure$
of Robert G. Bartle
Let $X=mathbbN$ and $mathcalA$ be the $sigma-$algebra of all subsets of $mathbbN$. If $(a_n)$ is a sequence of nonnegative real numbers and if we define $mu$ by
$$
mu(emptyset)=0; quad mu(E)=sum_nin Ea_n, quad Eneemptyset,
$$
then $mu$ is a measure on $mathcalA$. Conversely, every measure on $mathcalA$ is obtained in this way for some sequence $(a_n)$ in $overlineR^+$.
I am stuck in proving the countably additive property of $mu$.
My attempt. I intend to use the theory of double index series to prove countably additive property.
Thank you for all solutions.
lebesgue-measure
asked Nov 2 '14 at 10:54
impartialmale
566213
can you see finite additivity?
– user87543
Nov 2 '14 at 10:57
Nope, I cannot see the finite additivity.
– impartialmale
Nov 2 '14 at 11:01
Maybe it's too obvious. If we have $A, B subseteq mathbb N$ with $A cap B = emptyset$, then $mu(A cup B) = sum_n in A cup B a_n = sum_n in A a_n + sum _n in B a_n = mu(A) + mu(B)$, since A and B are disjoint. That's finite additivity. Countable additivity is completely analogous.
– jflipp
Nov 2 '14 at 11:11
Thank you for your comments. But I do not think it is too obvious for the case finite additivity and countable additivity. We have to use some results related to theory of real series to prove two properties.
– impartialmale
Nov 2 '14 at 11:15
That's true. But luckily, we have $a_n geq 0$, so any $sum a_j$ can only diverge to $+infty$. That's the lesson to learn here.
– jflipp
Nov 2 '14 at 16:14
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let us consider the problem 3.D. in the book
$textbfThe Elements of Integration and Lebesgue Measure$
of Robert G. Bartle
Let $X=mathbbN$ and $mathcalA$ be the $sigma-$algebra of all subsets of $mathbbN$. If $(a_n)$ is a sequence of nonnegative real numbers and if we define $mu$ by
$$
mu(emptyset)=0; quad mu(E)=sum_nin Ea_n, quad Eneemptyset,
$$
then $mu$ is a measure on $mathcalA$. Conversely, every measure on $mathcalA$ is obtained in this way for some sequence $(a_n)$ in $overlineR^+$.
I am stuck in proving the countably additive property of $mu$.
My attempt. I intend to use the theory of double index series to prove countably additive property.
Thank you for all solutions.
lebesgue-measure
asked Nov 2 '14 at 10:54
impartialmale
566213
Let us consider the problem 3.D. in the book
$textbfThe Elements of Integration and Lebesgue Measure$
of Robert G. Bartle
Let $X=mathbbN$ and $mathcalA$ be the $sigma-$algebra of all subsets of $mathbbN$. If $(a_n)$ is a sequence of nonnegative real numbers and if we define $mu$ by
$$
mu(emptyset)=0; quad mu(E)=sum_nin Ea_n, quad Eneemptyset,
$$
then $mu$ is a measure on $mathcalA$. Conversely, every measure on $mathcalA$ is obtained in this way for some sequence $(a_n)$ in $overlineR^+$.
I am stuck in proving the countably additive property of $mu$.
My attempt. I intend to use the theory of double index series to prove countably additive property.
Thank you for all solutions.
lebesgue-measure
asked Nov 2 '14 at 10:54
impartialmale
566213
asked Nov 2 '14 at 10:54
impartialmale
566213
asked Nov 2 '14 at 10:54
impartialmale
566213
asked Nov 2 '14 at 10:54
impartialmale
566213
566213
can you see finite additivity?
– user87543
Nov 2 '14 at 10:57
Nope, I cannot see the finite additivity.
– impartialmale
Nov 2 '14 at 11:01
Maybe it's too obvious. If we have $A, B subseteq mathbb N$ with $A cap B = emptyset$, then $mu(A cup B) = sum_n in A cup B a_n = sum_n in A a_n + sum _n in B a_n = mu(A) + mu(B)$, since A and B are disjoint. That's finite additivity. Countable additivity is completely analogous.
– jflipp
Nov 2 '14 at 11:11
Thank you for your comments. But I do not think it is too obvious for the case finite additivity and countable additivity. We have to use some results related to theory of real series to prove two properties.
– impartialmale
Nov 2 '14 at 11:15
That's true. But luckily, we have $a_n geq 0$, so any $sum a_j$ can only diverge to $+infty$. That's the lesson to learn here.
– jflipp
Nov 2 '14 at 16:14
add a comment |Â
can you see finite additivity?
– user87543
Nov 2 '14 at 10:57
Nope, I cannot see the finite additivity.
– impartialmale
Nov 2 '14 at 11:01
Maybe it's too obvious. If we have $A, B subseteq mathbb N$ with $A cap B = emptyset$, then $mu(A cup B) = sum_n in A cup B a_n = sum_n in A a_n + sum _n in B a_n = mu(A) + mu(B)$, since A and B are disjoint. That's finite additivity. Countable additivity is completely analogous.
– jflipp
Nov 2 '14 at 11:11
Thank you for your comments. But I do not think it is too obvious for the case finite additivity and countable additivity. We have to use some results related to theory of real series to prove two properties.
– impartialmale
Nov 2 '14 at 11:15
That's true. But luckily, we have $a_n geq 0$, so any $sum a_j$ can only diverge to $+infty$. That's the lesson to learn here.
– jflipp
Nov 2 '14 at 16:14
can you see finite additivity?
– user87543
Nov 2 '14 at 10:57
can you see finite additivity?
– user87543
Nov 2 '14 at 10:57
Nope, I cannot see the finite additivity.
– impartialmale
Nov 2 '14 at 11:01
Nope, I cannot see the finite additivity.
– impartialmale
Nov 2 '14 at 11:01
Maybe it's too obvious. If we have $A, B subseteq mathbb N$ with $A cap B = emptyset$, then $mu(A cup B) = sum_n in A cup B a_n = sum_n in A a_n + sum _n in B a_n = mu(A) + mu(B)$, since A and B are disjoint. That's finite additivity. Countable additivity is completely analogous.
– jflipp
Nov 2 '14 at 11:11
Maybe it's too obvious. If we have $A, B subseteq mathbb N$ with $A cap B = emptyset$, then $mu(A cup B) = sum_n in A cup B a_n = sum_n in A a_n + sum _n in B a_n = mu(A) + mu(B)$, since A and B are disjoint. That's finite additivity. Countable additivity is completely analogous.
– jflipp
Nov 2 '14 at 11:11
Thank you for your comments. But I do not think it is too obvious for the case finite additivity and countable additivity. We have to use some results related to theory of real series to prove two properties.
– impartialmale
Nov 2 '14 at 11:15
Thank you for your comments. But I do not think it is too obvious for the case finite additivity and countable additivity. We have to use some results related to theory of real series to prove two properties.
– impartialmale
Nov 2 '14 at 11:15
That's true. But luckily, we have $a_n geq 0$, so any $sum a_j$ can only diverge to $+infty$. That's the lesson to learn here.
– jflipp
Nov 2 '14 at 16:14
That's true. But luckily, we have $a_n geq 0$, so any $sum a_j$ can only diverge to $+infty$. That's the lesson to learn here.
– jflipp
Nov 2 '14 at 16:14
add a comment |Â
1 Answer
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We have to check the following
$mu(emptyset)=0$, by definition
$mu(E)=sum_nin Ea_ngeq 0$, since $a_ngeq 0$, for all $n$.
Let $E_k$ be disjoint, thenbeginalign*
muleft( bigcup_k=1^infty E_kright)&=sum_ninbigcup E_ka_n=sum_k=1^infty sum_nin E_ka_n=sum_k=1^inftymu (E_k).
endalign*
Thus, $mu$ is a measure on X. Conversely, for $(a_n)$ in $overlinemathbbR^+$, we can obtain any measure on X using the enunciated definition of $mu$.
answered Oct 7 '17 at 23:24
f. wolfe
842321
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We have to check the following
$mu(emptyset)=0$, by definition
$mu(E)=sum_nin Ea_ngeq 0$, since $a_ngeq 0$, for all $n$.
Let $E_k$ be disjoint, thenbeginalign*
muleft( bigcup_k=1^infty E_kright)&=sum_ninbigcup E_ka_n=sum_k=1^infty sum_nin E_ka_n=sum_k=1^inftymu (E_k).
endalign*
Thus, $mu$ is a measure on X. Conversely, for $(a_n)$ in $overlinemathbbR^+$, we can obtain any measure on X using the enunciated definition of $mu$.
answered Oct 7 '17 at 23:24
f. wolfe
842321
add a comment |Â
up vote
0
down vote
We have to check the following
$mu(emptyset)=0$, by definition
$mu(E)=sum_nin Ea_ngeq 0$, since $a_ngeq 0$, for all $n$.
Let $E_k$ be disjoint, thenbeginalign*
muleft( bigcup_k=1^infty E_kright)&=sum_ninbigcup E_ka_n=sum_k=1^infty sum_nin E_ka_n=sum_k=1^inftymu (E_k).
endalign*
Thus, $mu$ is a measure on X. Conversely, for $(a_n)$ in $overlinemathbbR^+$, we can obtain any measure on X using the enunciated definition of $mu$.
answered Oct 7 '17 at 23:24
f. wolfe
842321
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have to check the following
$mu(emptyset)=0$, by definition
$mu(E)=sum_nin Ea_ngeq 0$, since $a_ngeq 0$, for all $n$.
Let $E_k$ be disjoint, thenbeginalign*
muleft( bigcup_k=1^infty E_kright)&=sum_ninbigcup E_ka_n=sum_k=1^infty sum_nin E_ka_n=sum_k=1^inftymu (E_k).
endalign*
Thus, $mu$ is a measure on X. Conversely, for $(a_n)$ in $overlinemathbbR^+$, we can obtain any measure on X using the enunciated definition of $mu$.
answered Oct 7 '17 at 23:24
f. wolfe
842321
We have to check the following
$mu(emptyset)=0$, by definition
$mu(E)=sum_nin Ea_ngeq 0$, since $a_ngeq 0$, for all $n$.
Let $E_k$ be disjoint, thenbeginalign*
muleft( bigcup_k=1^infty E_kright)&=sum_ninbigcup E_ka_n=sum_k=1^infty sum_nin E_ka_n=sum_k=1^inftymu (E_k).
endalign*
Thus, $mu$ is a measure on X. Conversely, for $(a_n)$ in $overlinemathbbR^+$, we can obtain any measure on X using the enunciated definition of $mu$.
answered Oct 7 '17 at 23:24
f. wolfe
842321
answered Oct 7 '17 at 23:24
f. wolfe
842321
answered Oct 7 '17 at 23:24
f. wolfe
842321
answered Oct 7 '17 at 23:24
f. wolfe
842321
842321
add a comment |Â
add a comment |Â
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can you see finite additivity?
– user87543
Nov 2 '14 at 10:57
Nope, I cannot see the finite additivity.
– impartialmale
Nov 2 '14 at 11:01
Maybe it's too obvious. If we have $A, B subseteq mathbb N$ with $A cap B = emptyset$, then $mu(A cup B) = sum_n in A cup B a_n = sum_n in A a_n + sum _n in B a_n = mu(A) + mu(B)$, since A and B are disjoint. That's finite additivity. Countable additivity is completely analogous.
– jflipp
Nov 2 '14 at 11:11
Thank you for your comments. But I do not think it is too obvious for the case finite additivity and countable additivity. We have to use some results related to theory of real series to prove two properties.
– impartialmale
Nov 2 '14 at 11:15
That's true. But luckily, we have $a_n geq 0$, so any $sum a_j$ can only diverge to $+infty$. That's the lesson to learn here.
– jflipp
Nov 2 '14 at 16:14