$mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$
$begingroup$
Let $(X,mathcal{A}, mu)$ be a measure space. Prove:
$mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$
My ideas
$"Leftarrow"$
Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.
$"Rightarrow"$
I have no idea how to define this function, particularly as $f>0$, $forall x in X$
real-analysis integration measure-theory
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
Let $(X,mathcal{A}, mu)$ be a measure space. Prove:
$mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$
My ideas
$"Leftarrow"$
Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.
$"Rightarrow"$
I have no idea how to define this function, particularly as $f>0$, $forall x in X$
real-analysis integration measure-theory
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
Let $(X,mathcal{A}, mu)$ be a measure space. Prove:
$mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$
My ideas
$"Leftarrow"$
Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.
$"Rightarrow"$
I have no idea how to define this function, particularly as $f>0$, $forall x in X$
real-analysis integration measure-theory
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
$endgroup$
Let $(X,mathcal{A}, mu)$ be a measure space. Prove:
$mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$
My ideas
$"Leftarrow"$
Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.
$"Rightarrow"$
I have no idea how to define this function, particularly as $f>0$, $forall x in X$
real-analysis integration measure-theory
real-analysis integration measure-theory
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
edited Dec 4 '18 at 19:10
SABOY
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
asked Dec 4 '18 at 19:01
SABOYSABOY
656311
656311
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
$$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
$$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
$$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
$$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
$endgroup$
add a comment |
$begingroup$
Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
$$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
$$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
$endgroup$
add a comment |
$begingroup$
Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
$$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
$$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
$endgroup$
Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
$$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
$$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
answered Dec 4 '18 at 19:16
p4schp4sch
5,430317
5,430317
add a comment |
add a comment |
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