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$X$ a numerable set, $f$ a positive function, then $mu(A) = sum_{xin A} f(x)$ $sigma$-finite?
$begingroup$
Let $(X$, ${mathscr P}(X), mu)$ a measurable space, where $X$ is a numerable set, $f: X rightarrow [0, infty]$ a positive function.
If we define the measure as
$$mu(A) = sum_{xin A} f(x), text{for }Asubset X $$
is $mu$ $sigma$-finite? I feel like this shouldn't hold for every single positive function, any other than $text{ }f(x) = infty text{ }foralltext{ } x in X$ which is pretty ridiculous, but I can't think of any other.
measure-theory lebesgue-measure
asked Jan 6 at 9:21
sashasasha
133
$endgroup$
add a comment |
$begingroup$
Let $(X$, ${mathscr P}(X), mu)$ a measurable space, where $X$ is a numerable set, $f: X rightarrow [0, infty]$ a positive function.
If we define the measure as
$$mu(A) = sum_{xin A} f(x), text{for }Asubset X $$
is $mu$ $sigma$-finite? I feel like this shouldn't hold for every single positive function, any other than $text{ }f(x) = infty text{ }foralltext{ } x in X$ which is pretty ridiculous, but I can't think of any other.
measure-theory lebesgue-measure
asked Jan 6 at 9:21
sashasasha
133
$endgroup$
add a comment |
$begingroup$
Let $(X$, ${mathscr P}(X), mu)$ a measurable space, where $X$ is a numerable set, $f: X rightarrow [0, infty]$ a positive function.
If we define the measure as
$$mu(A) = sum_{xin A} f(x), text{for }Asubset X $$
is $mu$ $sigma$-finite? I feel like this shouldn't hold for every single positive function, any other than $text{ }f(x) = infty text{ }foralltext{ } x in X$ which is pretty ridiculous, but I can't think of any other.
measure-theory lebesgue-measure
asked Jan 6 at 9:21
sashasasha
133
$endgroup$
Let $(X$, ${mathscr P}(X), mu)$ a measurable space, where $X$ is a numerable set, $f: X rightarrow [0, infty]$ a positive function.
If we define the measure as
$$mu(A) = sum_{xin A} f(x), text{for }Asubset X $$
is $mu$ $sigma$-finite? I feel like this shouldn't hold for every single positive function, any other than $text{ }f(x) = infty text{ }foralltext{ } x in X$ which is pretty ridiculous, but I can't think of any other.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Jan 6 at 9:21
sashasasha
133
asked Jan 6 at 9:21
sashasasha
133
asked Jan 6 at 9:21
sashasasha
133
asked Jan 6 at 9:21
sashasasha
133
asked Jan 6 at 9:21
sashasasha
133
133
add a comment |
add a comment |
1 Answer
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votes
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Actually, $mu$ will be $sigma$-finite if and only if $f(x)<+infty$ for all $xin X$.
Indeed, if $f(x)<+infty$ for all $xin X$, then ${{x},xin X}$ is a countable partition of $X$ and each element as a finite measure.
If $mu$ is $sigma$-finite, there exists a sequence of subsets of $X$, say $left(A_nright)_{ngeqslant 1}$ such that $muleft(A_nright)<+infty$ and $bigcup_{ngeqslant 1}A_n=X$. Let $xin X$; there exists some $n$ such that $xin A_n$. Then
$$
f(x)=muleft({x}right)leqslant muleft(A_nright)<+infty.
$$
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
$begingroup$
Wow, you're right and thank you sm!
$endgroup$
– sasha
Jan 6 at 19:46
$begingroup$
You are welcome. But what does "sm" mean?
$endgroup$
– Davide Giraudo
Jan 6 at 20:53
$begingroup$
Oh sorry for taking so long. It means 'so much'.
$endgroup$
– sasha
Jan 20 at 17:19
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
Actually, $mu$ will be $sigma$-finite if and only if $f(x)<+infty$ for all $xin X$.
Indeed, if $f(x)<+infty$ for all $xin X$, then ${{x},xin X}$ is a countable partition of $X$ and each element as a finite measure.
If $mu$ is $sigma$-finite, there exists a sequence of subsets of $X$, say $left(A_nright)_{ngeqslant 1}$ such that $muleft(A_nright)<+infty$ and $bigcup_{ngeqslant 1}A_n=X$. Let $xin X$; there exists some $n$ such that $xin A_n$. Then
$$
f(x)=muleft({x}right)leqslant muleft(A_nright)<+infty.
$$
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
$begingroup$
Wow, you're right and thank you sm!
$endgroup$
– sasha
Jan 6 at 19:46
$begingroup$
You are welcome. But what does "sm" mean?
$endgroup$
– Davide Giraudo
Jan 6 at 20:53
$begingroup$
Oh sorry for taking so long. It means 'so much'.
$endgroup$
– sasha
Jan 20 at 17:19
add a comment |
$begingroup$
Actually, $mu$ will be $sigma$-finite if and only if $f(x)<+infty$ for all $xin X$.
Indeed, if $f(x)<+infty$ for all $xin X$, then ${{x},xin X}$ is a countable partition of $X$ and each element as a finite measure.
If $mu$ is $sigma$-finite, there exists a sequence of subsets of $X$, say $left(A_nright)_{ngeqslant 1}$ such that $muleft(A_nright)<+infty$ and $bigcup_{ngeqslant 1}A_n=X$. Let $xin X$; there exists some $n$ such that $xin A_n$. Then
$$
f(x)=muleft({x}right)leqslant muleft(A_nright)<+infty.
$$
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
$begingroup$
Wow, you're right and thank you sm!
$endgroup$
– sasha
Jan 6 at 19:46
$begingroup$
You are welcome. But what does "sm" mean?
$endgroup$
– Davide Giraudo
Jan 6 at 20:53
$begingroup$
Oh sorry for taking so long. It means 'so much'.
$endgroup$
– sasha
Jan 20 at 17:19
add a comment |
$begingroup$
Actually, $mu$ will be $sigma$-finite if and only if $f(x)<+infty$ for all $xin X$.
Indeed, if $f(x)<+infty$ for all $xin X$, then ${{x},xin X}$ is a countable partition of $X$ and each element as a finite measure.
If $mu$ is $sigma$-finite, there exists a sequence of subsets of $X$, say $left(A_nright)_{ngeqslant 1}$ such that $muleft(A_nright)<+infty$ and $bigcup_{ngeqslant 1}A_n=X$. Let $xin X$; there exists some $n$ such that $xin A_n$. Then
$$
f(x)=muleft({x}right)leqslant muleft(A_nright)<+infty.
$$
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
Actually, $mu$ will be $sigma$-finite if and only if $f(x)<+infty$ for all $xin X$.
Indeed, if $f(x)<+infty$ for all $xin X$, then ${{x},xin X}$ is a countable partition of $X$ and each element as a finite measure.
If $mu$ is $sigma$-finite, there exists a sequence of subsets of $X$, say $left(A_nright)_{ngeqslant 1}$ such that $muleft(A_nright)<+infty$ and $bigcup_{ngeqslant 1}A_n=X$. Let $xin X$; there exists some $n$ such that $xin A_n$. Then
$$
f(x)=muleft({x}right)leqslant muleft(A_nright)<+infty.
$$
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 6 at 10:28
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
$begingroup$
Wow, you're right and thank you sm!
$endgroup$
– sasha
Jan 6 at 19:46
$begingroup$
You are welcome. But what does "sm" mean?
$endgroup$
– Davide Giraudo
Jan 6 at 20:53
$begingroup$
Oh sorry for taking so long. It means 'so much'.
$endgroup$
– sasha
Jan 20 at 17:19
add a comment |
$begingroup$
Wow, you're right and thank you sm!
$endgroup$
– sasha
Jan 6 at 19:46
$begingroup$
You are welcome. But what does "sm" mean?
$endgroup$
– Davide Giraudo
Jan 6 at 20:53
$begingroup$
Oh sorry for taking so long. It means 'so much'.
$endgroup$
– sasha
Jan 20 at 17:19
$begingroup$
Wow, you're right and thank you sm!
$endgroup$
– sasha
Jan 6 at 19:46
$begingroup$
Wow, you're right and thank you sm!
$endgroup$
– sasha
Jan 6 at 19:46
$begingroup$
You are welcome. But what does "sm" mean?
$endgroup$
– Davide Giraudo
Jan 6 at 20:53
$begingroup$
You are welcome. But what does "sm" mean?
$endgroup$
– Davide Giraudo
Jan 6 at 20:53
$begingroup$
Oh sorry for taking so long. It means 'so much'.
$endgroup$
– sasha
Jan 20 at 17:19
$begingroup$
Oh sorry for taking so long. It means 'so much'.
$endgroup$
– sasha
Jan 20 at 17:19
add a comment |
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