Mixing
Lebesgue Integral on Measurespace
Clash Royale CLAN TAG#URR8PPP
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I am reading "Real and Complex Analysis" from Rudin and could not really solve a presumably simple question I came up with.
In the chapter where the L-Integral is constructed the measure space - $(X,mathfrakM,mu)$ with set, $sigma$-algebra and measure, respectively - is considered.
Now every integral theorem (Monotone Convergence, Fatou, Dominated Convergence) is stated by integrating over the whole set $X%$. Since Rudin also proves the approximation theorem of measurable function using simple functions on all $X$, I am wondering in which sense the theorems can be transfered to problems, where I indeed have the measure space above, but only want to integrate over a subset $Ssubset X$.
I could use something like $fchi_S$ which is again measurable if $f$ and the characteristic function $chi$ are measurable, but if $f$ would be continuous on $S$ then I would destroy this property i.g. by extending to $X$.
It would also of course be necessary that $S$ lies in $mathfrakM$.
integration measure-theory
asked Aug 3 at 15:22
EpsilonDelta
4951513
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up vote
2
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I am reading "Real and Complex Analysis" from Rudin and could not really solve a presumably simple question I came up with.
In the chapter where the L-Integral is constructed the measure space - $(X,mathfrakM,mu)$ with set, $sigma$-algebra and measure, respectively - is considered.
Now every integral theorem (Monotone Convergence, Fatou, Dominated Convergence) is stated by integrating over the whole set $X%$. Since Rudin also proves the approximation theorem of measurable function using simple functions on all $X$, I am wondering in which sense the theorems can be transfered to problems, where I indeed have the measure space above, but only want to integrate over a subset $Ssubset X$.
I could use something like $fchi_S$ which is again measurable if $f$ and the characteristic function $chi$ are measurable, but if $f$ would be continuous on $S$ then I would destroy this property i.g. by extending to $X$.
It would also of course be necessary that $S$ lies in $mathfrakM$.
integration measure-theory
asked Aug 3 at 15:22
EpsilonDelta
4951513
2
You are completely right. What is the problem of $f$ not being continuous anymore? Doesn't hurt, as long as $f$ (resp. $S$) is a measurable function (resp. set)
– Jonas
Aug 3 at 15:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading "Real and Complex Analysis" from Rudin and could not really solve a presumably simple question I came up with.
In the chapter where the L-Integral is constructed the measure space - $(X,mathfrakM,mu)$ with set, $sigma$-algebra and measure, respectively - is considered.
Now every integral theorem (Monotone Convergence, Fatou, Dominated Convergence) is stated by integrating over the whole set $X%$. Since Rudin also proves the approximation theorem of measurable function using simple functions on all $X$, I am wondering in which sense the theorems can be transfered to problems, where I indeed have the measure space above, but only want to integrate over a subset $Ssubset X$.
I could use something like $fchi_S$ which is again measurable if $f$ and the characteristic function $chi$ are measurable, but if $f$ would be continuous on $S$ then I would destroy this property i.g. by extending to $X$.
It would also of course be necessary that $S$ lies in $mathfrakM$.
integration measure-theory
asked Aug 3 at 15:22
EpsilonDelta
4951513
I am reading "Real and Complex Analysis" from Rudin and could not really solve a presumably simple question I came up with.
In the chapter where the L-Integral is constructed the measure space - $(X,mathfrakM,mu)$ with set, $sigma$-algebra and measure, respectively - is considered.
Now every integral theorem (Monotone Convergence, Fatou, Dominated Convergence) is stated by integrating over the whole set $X%$. Since Rudin also proves the approximation theorem of measurable function using simple functions on all $X$, I am wondering in which sense the theorems can be transfered to problems, where I indeed have the measure space above, but only want to integrate over a subset $Ssubset X$.
I could use something like $fchi_S$ which is again measurable if $f$ and the characteristic function $chi$ are measurable, but if $f$ would be continuous on $S$ then I would destroy this property i.g. by extending to $X$.
It would also of course be necessary that $S$ lies in $mathfrakM$.
integration measure-theory
asked Aug 3 at 15:22
EpsilonDelta
4951513
asked Aug 3 at 15:22
EpsilonDelta
4951513
asked Aug 3 at 15:22
EpsilonDelta
4951513
asked Aug 3 at 15:22
EpsilonDelta
4951513
4951513
2
You are completely right. What is the problem of $f$ not being continuous anymore? Doesn't hurt, as long as $f$ (resp. $S$) is a measurable function (resp. set)
– Jonas
Aug 3 at 15:25
add a comment |Â
2
You are completely right. What is the problem of $f$ not being continuous anymore? Doesn't hurt, as long as $f$ (resp. $S$) is a measurable function (resp. set)
– Jonas
Aug 3 at 15:25
2
2
You are completely right. What is the problem of $f$ not being continuous anymore? Doesn't hurt, as long as $f$ (resp. $S$) is a measurable function (resp. set)
– Jonas
Aug 3 at 15:25
You are completely right. What is the problem of $f$ not being continuous anymore? Doesn't hurt, as long as $f$ (resp. $S$) is a measurable function (resp. set)
– Jonas
Aug 3 at 15:25
add a comment |Â
2 Answers
2
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up vote
3
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accepted
You don't need a new concept to define:
$$int_Sfoperatornamedmu.$$
In fact, if $(X,mathfrakM,mu)$ is a measure space, $SinmathfrakM$ and $f$ is measurable from $(X,mathfrakM)$ to $mathbbC$, then the following hold:
- $mathfrakM_S:=FinmathfrakM $ is a $sigma$-algebra of subset of $S;$
- $mu|_mathfrakM_S$ is a measure on $mathfrakM_S;$
- $f|_S$ is measurable from $(S,mathfrakM_S)$ to $mathbbC.$
So, if $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ you can define:
$$int_Sfoperatornamedmu:=int_Sf|_Soperatornamedmu|_mathfrakM_S.$$
Then, it is a theorem that $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ if and only if $f chi_Sin L^1(X,mathfrakM,mu)$ and in this case it holds that:
$$int_Sfoperatornamedmu=int_Xf chi_Soperatornamedmu.$$
You can prove this theorem via standard machine (for the standard machine technique, see e.g. David Williams - Probability with martingales, chapter 5).
answered Aug 3 at 15:39
Bob
1,405522
add a comment |Â
up vote
3
down vote
Yes, you are right, you can define integration of $f$ over a subset $S subset X$ to be integrating $f$ times the indicator function of $S$:
beginalign*
int_Sf dmu := int_X f cdot chi_S dmu.
endalign*
You are right, $f$ may be continuous on $S$, but may not be continuous on the whole space $X$. There are no problems though, since continuity is a stronger condition than integrability.
Also, you are right that $S$ needs to be measurable, i.e., $S in mathfrakM$.
answered Aug 3 at 15:34
amarney
978215
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You don't need a new concept to define:
$$int_Sfoperatornamedmu.$$
In fact, if $(X,mathfrakM,mu)$ is a measure space, $SinmathfrakM$ and $f$ is measurable from $(X,mathfrakM)$ to $mathbbC$, then the following hold:
- $mathfrakM_S:=FinmathfrakM $ is a $sigma$-algebra of subset of $S;$
- $mu|_mathfrakM_S$ is a measure on $mathfrakM_S;$
- $f|_S$ is measurable from $(S,mathfrakM_S)$ to $mathbbC.$
So, if $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ you can define:
$$int_Sfoperatornamedmu:=int_Sf|_Soperatornamedmu|_mathfrakM_S.$$
Then, it is a theorem that $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ if and only if $f chi_Sin L^1(X,mathfrakM,mu)$ and in this case it holds that:
$$int_Sfoperatornamedmu=int_Xf chi_Soperatornamedmu.$$
You can prove this theorem via standard machine (for the standard machine technique, see e.g. David Williams - Probability with martingales, chapter 5).
answered Aug 3 at 15:39
Bob
1,405522
add a comment |Â
up vote
3
down vote
accepted
You don't need a new concept to define:
$$int_Sfoperatornamedmu.$$
In fact, if $(X,mathfrakM,mu)$ is a measure space, $SinmathfrakM$ and $f$ is measurable from $(X,mathfrakM)$ to $mathbbC$, then the following hold:
- $mathfrakM_S:=FinmathfrakM $ is a $sigma$-algebra of subset of $S;$
- $mu|_mathfrakM_S$ is a measure on $mathfrakM_S;$
- $f|_S$ is measurable from $(S,mathfrakM_S)$ to $mathbbC.$
So, if $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ you can define:
$$int_Sfoperatornamedmu:=int_Sf|_Soperatornamedmu|_mathfrakM_S.$$
Then, it is a theorem that $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ if and only if $f chi_Sin L^1(X,mathfrakM,mu)$ and in this case it holds that:
$$int_Sfoperatornamedmu=int_Xf chi_Soperatornamedmu.$$
You can prove this theorem via standard machine (for the standard machine technique, see e.g. David Williams - Probability with martingales, chapter 5).
answered Aug 3 at 15:39
Bob
1,405522
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You don't need a new concept to define:
$$int_Sfoperatornamedmu.$$
In fact, if $(X,mathfrakM,mu)$ is a measure space, $SinmathfrakM$ and $f$ is measurable from $(X,mathfrakM)$ to $mathbbC$, then the following hold:
- $mathfrakM_S:=FinmathfrakM $ is a $sigma$-algebra of subset of $S;$
- $mu|_mathfrakM_S$ is a measure on $mathfrakM_S;$
- $f|_S$ is measurable from $(S,mathfrakM_S)$ to $mathbbC.$
So, if $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ you can define:
$$int_Sfoperatornamedmu:=int_Sf|_Soperatornamedmu|_mathfrakM_S.$$
Then, it is a theorem that $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ if and only if $f chi_Sin L^1(X,mathfrakM,mu)$ and in this case it holds that:
$$int_Sfoperatornamedmu=int_Xf chi_Soperatornamedmu.$$
You can prove this theorem via standard machine (for the standard machine technique, see e.g. David Williams - Probability with martingales, chapter 5).
answered Aug 3 at 15:39
Bob
1,405522
You don't need a new concept to define:
$$int_Sfoperatornamedmu.$$
In fact, if $(X,mathfrakM,mu)$ is a measure space, $SinmathfrakM$ and $f$ is measurable from $(X,mathfrakM)$ to $mathbbC$, then the following hold:
- $mathfrakM_S:=FinmathfrakM $ is a $sigma$-algebra of subset of $S;$
- $mu|_mathfrakM_S$ is a measure on $mathfrakM_S;$
- $f|_S$ is measurable from $(S,mathfrakM_S)$ to $mathbbC.$
So, if $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ you can define:
$$int_Sfoperatornamedmu:=int_Sf|_Soperatornamedmu|_mathfrakM_S.$$
Then, it is a theorem that $f|_Sin L^1(S,mathfrakM_S,mu|_mathfrakM_S)$ if and only if $f chi_Sin L^1(X,mathfrakM,mu)$ and in this case it holds that:
$$int_Sfoperatornamedmu=int_Xf chi_Soperatornamedmu.$$
You can prove this theorem via standard machine (for the standard machine technique, see e.g. David Williams - Probability with martingales, chapter 5).
answered Aug 3 at 15:39
Bob
1,405522
edited Aug 3 at 16:50
answered Aug 3 at 15:39
Bob
1,405522
answered Aug 3 at 15:39
Bob
1,405522
answered Aug 3 at 15:39
Bob
1,405522
1,405522
add a comment |Â
add a comment |Â
up vote
3
down vote
Yes, you are right, you can define integration of $f$ over a subset $S subset X$ to be integrating $f$ times the indicator function of $S$:
beginalign*
int_Sf dmu := int_X f cdot chi_S dmu.
endalign*
You are right, $f$ may be continuous on $S$, but may not be continuous on the whole space $X$. There are no problems though, since continuity is a stronger condition than integrability.
Also, you are right that $S$ needs to be measurable, i.e., $S in mathfrakM$.
answered Aug 3 at 15:34
amarney
978215
add a comment |Â
up vote
3
down vote
Yes, you are right, you can define integration of $f$ over a subset $S subset X$ to be integrating $f$ times the indicator function of $S$:
beginalign*
int_Sf dmu := int_X f cdot chi_S dmu.
endalign*
You are right, $f$ may be continuous on $S$, but may not be continuous on the whole space $X$. There are no problems though, since continuity is a stronger condition than integrability.
Also, you are right that $S$ needs to be measurable, i.e., $S in mathfrakM$.
answered Aug 3 at 15:34
amarney
978215
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes, you are right, you can define integration of $f$ over a subset $S subset X$ to be integrating $f$ times the indicator function of $S$:
beginalign*
int_Sf dmu := int_X f cdot chi_S dmu.
endalign*
You are right, $f$ may be continuous on $S$, but may not be continuous on the whole space $X$. There are no problems though, since continuity is a stronger condition than integrability.
Also, you are right that $S$ needs to be measurable, i.e., $S in mathfrakM$.
answered Aug 3 at 15:34
amarney
978215
Yes, you are right, you can define integration of $f$ over a subset $S subset X$ to be integrating $f$ times the indicator function of $S$:
beginalign*
int_Sf dmu := int_X f cdot chi_S dmu.
endalign*
You are right, $f$ may be continuous on $S$, but may not be continuous on the whole space $X$. There are no problems though, since continuity is a stronger condition than integrability.
Also, you are right that $S$ needs to be measurable, i.e., $S in mathfrakM$.
answered Aug 3 at 15:34
amarney
978215
answered Aug 3 at 15:34
amarney
978215
answered Aug 3 at 15:34
amarney
978215
answered Aug 3 at 15:34
amarney
978215
978215
add a comment |Â
add a comment |Â
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2
You are completely right. What is the problem of $f$ not being continuous anymore? Doesn't hurt, as long as $f$ (resp. $S$) is a measurable function (resp. set)
– Jonas
Aug 3 at 15:25