Sigma-finite measure (countable sequence)
Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.
How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that
$A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?
I got:
Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$
So $X=A dotcup (X$ $g^{-1}(0))^c$
By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$
By non-negativity: $mu(A)<infty$
I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.
measure-theory
asked Nov 24 at 11:53
Tartulop
656
add a comment |
Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.
How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that
$A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?
I got:
Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$
So $X=A dotcup (X$ $g^{-1}(0))^c$
By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$
By non-negativity: $mu(A)<infty$
I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.
measure-theory
asked Nov 24 at 11:53
Tartulop
656
add a comment |
Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.
How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that
$A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?
I got:
Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$
So $X=A dotcup (X$ $g^{-1}(0))^c$
By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$
By non-negativity: $mu(A)<infty$
I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.
measure-theory
asked Nov 24 at 11:53
Tartulop
656
Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.
How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that
$A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?
I got:
Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$
So $X=A dotcup (X$ $g^{-1}(0))^c$
By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$
By non-negativity: $mu(A)<infty$
I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.
measure-theory
measure-theory
asked Nov 24 at 11:53
Tartulop
656
asked Nov 24 at 11:53
Tartulop
656
asked Nov 24 at 11:53
Tartulop
656
asked Nov 24 at 11:53
Tartulop
656
asked Nov 24 at 11:53
Tartulop
656
656
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.
Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.
Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.
It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.
answered Nov 24 at 12:08
drhab
96.5k544127
Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
– sharpe
Nov 24 at 12:11
It is not demanded that they cover $X$, but that their union is $A$.
– drhab
Nov 24 at 12:13
Sorry. I understood.
– sharpe
Nov 24 at 12:14
add a comment |
The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.
Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.
answered Nov 24 at 12:07
Masacroso
12.8k41746
No. $X$ is just a measure space.
– sharpe
Nov 24 at 12:08
@sharpe it is the same.
– Masacroso
Nov 24 at 12:10
Probably, your argument is mistaken.
– sharpe
Nov 24 at 12:22
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.
Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.
Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.
It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.
answered Nov 24 at 12:08
drhab
96.5k544127
Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
– sharpe
Nov 24 at 12:11
It is not demanded that they cover $X$, but that their union is $A$.
– drhab
Nov 24 at 12:13
Sorry. I understood.
– sharpe
Nov 24 at 12:14
add a comment |
You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.
Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.
Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.
It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.
answered Nov 24 at 12:08
drhab
96.5k544127
Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
– sharpe
Nov 24 at 12:11
It is not demanded that they cover $X$, but that their union is $A$.
– drhab
Nov 24 at 12:13
Sorry. I understood.
– sharpe
Nov 24 at 12:14
add a comment |
You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.
Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.
Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.
It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.
answered Nov 24 at 12:08
drhab
96.5k544127
You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.
Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.
Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.
It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.
answered Nov 24 at 12:08
drhab
96.5k544127
edited Nov 24 at 12:39
answered Nov 24 at 12:08
drhab
96.5k544127
answered Nov 24 at 12:08
drhab
96.5k544127
answered Nov 24 at 12:08
drhab
96.5k544127
96.5k544127
Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
– sharpe
Nov 24 at 12:11
It is not demanded that they cover $X$, but that their union is $A$.
– drhab
Nov 24 at 12:13
Sorry. I understood.
– sharpe
Nov 24 at 12:14
add a comment |
Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
– sharpe
Nov 24 at 12:11
It is not demanded that they cover $X$, but that their union is $A$.
– drhab
Nov 24 at 12:13
Sorry. I understood.
– sharpe
Nov 24 at 12:14
Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
– sharpe
Nov 24 at 12:11
Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
– sharpe
Nov 24 at 12:11
It is not demanded that they cover $X$, but that their union is $A$.
– drhab
Nov 24 at 12:13
It is not demanded that they cover $X$, but that their union is $A$.
– drhab
Nov 24 at 12:13
Sorry. I understood.
– sharpe
Nov 24 at 12:14
Sorry. I understood.
– sharpe
Nov 24 at 12:14
add a comment |
The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.
Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.
answered Nov 24 at 12:07
Masacroso
12.8k41746
No. $X$ is just a measure space.
– sharpe
Nov 24 at 12:08
@sharpe it is the same.
– Masacroso
Nov 24 at 12:10
Probably, your argument is mistaken.
– sharpe
Nov 24 at 12:22
add a comment |
The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.
Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.
answered Nov 24 at 12:07
Masacroso
12.8k41746
No. $X$ is just a measure space.
– sharpe
Nov 24 at 12:08
@sharpe it is the same.
– Masacroso
Nov 24 at 12:10
Probably, your argument is mistaken.
– sharpe
Nov 24 at 12:22
add a comment |
The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.
Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.
answered Nov 24 at 12:07
Masacroso
12.8k41746
The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.
Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.
answered Nov 24 at 12:07
Masacroso
12.8k41746
edited Nov 24 at 12:08
answered Nov 24 at 12:07
Masacroso
12.8k41746
answered Nov 24 at 12:07
Masacroso
12.8k41746
answered Nov 24 at 12:07
Masacroso
12.8k41746
12.8k41746
No. $X$ is just a measure space.
– sharpe
Nov 24 at 12:08
@sharpe it is the same.
– Masacroso
Nov 24 at 12:10
Probably, your argument is mistaken.
– sharpe
Nov 24 at 12:22
add a comment |
No. $X$ is just a measure space.
– sharpe
Nov 24 at 12:08
@sharpe it is the same.
– Masacroso
Nov 24 at 12:10
Probably, your argument is mistaken.
– sharpe
Nov 24 at 12:22
No. $X$ is just a measure space.
– sharpe
Nov 24 at 12:08
No. $X$ is just a measure space.
– sharpe
Nov 24 at 12:08
@sharpe it is the same.
– Masacroso
Nov 24 at 12:10
@sharpe it is the same.
– Masacroso
Nov 24 at 12:10
Probably, your argument is mistaken.
– sharpe
Nov 24 at 12:22
Probably, your argument is mistaken.
– sharpe
Nov 24 at 12:22
add a comment |
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