The consequence of $mu$-measurability from Bogachev's book











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I am reading Bogachev's book "Measure Theory" which is in my opinion is very good book on measure theory. Let me ask you the following question:




Definition: Let $mu$ - non-negative set function with domain $mathcal{A}subset 2^X$. The set $A$ is called $mu$-measurable if
for any $varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$
such that $mu^*(Atriangle A_{varepsilon})<varepsilon,$ where
$mu^*$-outer measure.



The set of all $mu$-measurable sets is denoted by
$mathcal{A}_{mu}$.




One thing began to confuse me during reading. When author takes some set $Ain mathcal{A}_{mu}$ and then he considers $mu(A)$. This confuses me a lot.



We know that $mathcal{A}subset mathcal{A}_{mu}$ and when $Ain mathcal{A}_{mu}$ it means that $A$ is $mu$-measurable set. But $mu$ is defined only on $mathcal{A}$. Why he does not consider $mu^*(A)$?



More precisely, if $A$ is $mu$-measurable ($Ain mathcal{A}_{mu}$) then why $mu(A)$ makes sense? and does $mu(A)=mu^*(A)$?



P.S. I know the fact that $mathcal{A}_{mu}$ is $sigma$-algebra and $mu$ can be uniquely extended from algebra $mathcal{A}$ to $sigma$-algebra $mathcal{A}_{mu}$. And this extension is given by $mu^*$.



I would be very grateful if somebody can detailed answer. Because it is important two understand such things.



EDIT: I will divide my questions into 3 parts.



1) Corollary 1.5.8. (page 21). enter image description here



I was not able to understand this part. He just uses the definition of outer measure $mu^*$ but not the definition of $mu$-measurability.



My approach: Since $A$ is $mu$-measurable then $forall varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$ such that $mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using monotonicity of $mu^*$ we have $mu^*(A-A_{varepsilon})leq mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using subadditivity of $mu^*$ we have $mu^*(A)-mu^*(A_{varepsilon})leq mu^*(A-A_{varepsilon})<varepsilon.$ But since $A_{varepsilon}in mathcal{A}$ then $mu^*(A_{varepsilon})=mu(A_{varepsilon})$ so $mu^*(A)-mu(A_{varepsilon})<varepsilon$. We have that $mu^*(A)$ is finite so we can apply the definition of outer measure. Is it correct? But I am sure that it is OK.



2) enter image description here



He takes $Bin mathcal{A}_{mu}$ but why $mu(B)$ makes sense? $mu(B)$ is defined only on $malcal{A}$ but $B$ may not be in $mathcal{A}$. This also confusing me a lot.



3) enter image description here



The same question $B$ and $C$ are $mu$-measurable sets but why $mu(B)$ and $mu(C)$ makes sense?



I guess that all these questions are related to each. So would be very grateful for detailed answer and help!



EDIT 2: Also one moment which I have forgot to ask in the previous questions:



enter image description here



In the above proof note two moments which I have underlined with red line. Why Bogachev writes $mu^*(A)=mu^*(A'')$ but in he writes just $mu(B)=mu^*(X-A)$. The first is OK and I am agree with that but in the second he omits $mu^*$ and just write $mu(B)$. Could you explain it, please?










share|cite|improve this question




















  • 1




    Where exactly does he do what confuses you? Could you provide a more precise reference; this is a big book.
    – Michael Greinecker
    2 days ago










  • @MichaelGreinecker, Please take a look at the EDIT.
    – ZFR
    2 days ago















up vote
0
down vote

favorite












I am reading Bogachev's book "Measure Theory" which is in my opinion is very good book on measure theory. Let me ask you the following question:




Definition: Let $mu$ - non-negative set function with domain $mathcal{A}subset 2^X$. The set $A$ is called $mu$-measurable if
for any $varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$
such that $mu^*(Atriangle A_{varepsilon})<varepsilon,$ where
$mu^*$-outer measure.



The set of all $mu$-measurable sets is denoted by
$mathcal{A}_{mu}$.




One thing began to confuse me during reading. When author takes some set $Ain mathcal{A}_{mu}$ and then he considers $mu(A)$. This confuses me a lot.



We know that $mathcal{A}subset mathcal{A}_{mu}$ and when $Ain mathcal{A}_{mu}$ it means that $A$ is $mu$-measurable set. But $mu$ is defined only on $mathcal{A}$. Why he does not consider $mu^*(A)$?



More precisely, if $A$ is $mu$-measurable ($Ain mathcal{A}_{mu}$) then why $mu(A)$ makes sense? and does $mu(A)=mu^*(A)$?



P.S. I know the fact that $mathcal{A}_{mu}$ is $sigma$-algebra and $mu$ can be uniquely extended from algebra $mathcal{A}$ to $sigma$-algebra $mathcal{A}_{mu}$. And this extension is given by $mu^*$.



I would be very grateful if somebody can detailed answer. Because it is important two understand such things.



EDIT: I will divide my questions into 3 parts.



1) Corollary 1.5.8. (page 21). enter image description here



I was not able to understand this part. He just uses the definition of outer measure $mu^*$ but not the definition of $mu$-measurability.



My approach: Since $A$ is $mu$-measurable then $forall varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$ such that $mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using monotonicity of $mu^*$ we have $mu^*(A-A_{varepsilon})leq mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using subadditivity of $mu^*$ we have $mu^*(A)-mu^*(A_{varepsilon})leq mu^*(A-A_{varepsilon})<varepsilon.$ But since $A_{varepsilon}in mathcal{A}$ then $mu^*(A_{varepsilon})=mu(A_{varepsilon})$ so $mu^*(A)-mu(A_{varepsilon})<varepsilon$. We have that $mu^*(A)$ is finite so we can apply the definition of outer measure. Is it correct? But I am sure that it is OK.



2) enter image description here



He takes $Bin mathcal{A}_{mu}$ but why $mu(B)$ makes sense? $mu(B)$ is defined only on $malcal{A}$ but $B$ may not be in $mathcal{A}$. This also confusing me a lot.



3) enter image description here



The same question $B$ and $C$ are $mu$-measurable sets but why $mu(B)$ and $mu(C)$ makes sense?



I guess that all these questions are related to each. So would be very grateful for detailed answer and help!



EDIT 2: Also one moment which I have forgot to ask in the previous questions:



enter image description here



In the above proof note two moments which I have underlined with red line. Why Bogachev writes $mu^*(A)=mu^*(A'')$ but in he writes just $mu(B)=mu^*(X-A)$. The first is OK and I am agree with that but in the second he omits $mu^*$ and just write $mu(B)$. Could you explain it, please?










share|cite|improve this question




















  • 1




    Where exactly does he do what confuses you? Could you provide a more precise reference; this is a big book.
    – Michael Greinecker
    2 days ago










  • @MichaelGreinecker, Please take a look at the EDIT.
    – ZFR
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am reading Bogachev's book "Measure Theory" which is in my opinion is very good book on measure theory. Let me ask you the following question:




Definition: Let $mu$ - non-negative set function with domain $mathcal{A}subset 2^X$. The set $A$ is called $mu$-measurable if
for any $varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$
such that $mu^*(Atriangle A_{varepsilon})<varepsilon,$ where
$mu^*$-outer measure.



The set of all $mu$-measurable sets is denoted by
$mathcal{A}_{mu}$.




One thing began to confuse me during reading. When author takes some set $Ain mathcal{A}_{mu}$ and then he considers $mu(A)$. This confuses me a lot.



We know that $mathcal{A}subset mathcal{A}_{mu}$ and when $Ain mathcal{A}_{mu}$ it means that $A$ is $mu$-measurable set. But $mu$ is defined only on $mathcal{A}$. Why he does not consider $mu^*(A)$?



More precisely, if $A$ is $mu$-measurable ($Ain mathcal{A}_{mu}$) then why $mu(A)$ makes sense? and does $mu(A)=mu^*(A)$?



P.S. I know the fact that $mathcal{A}_{mu}$ is $sigma$-algebra and $mu$ can be uniquely extended from algebra $mathcal{A}$ to $sigma$-algebra $mathcal{A}_{mu}$. And this extension is given by $mu^*$.



I would be very grateful if somebody can detailed answer. Because it is important two understand such things.



EDIT: I will divide my questions into 3 parts.



1) Corollary 1.5.8. (page 21). enter image description here



I was not able to understand this part. He just uses the definition of outer measure $mu^*$ but not the definition of $mu$-measurability.



My approach: Since $A$ is $mu$-measurable then $forall varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$ such that $mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using monotonicity of $mu^*$ we have $mu^*(A-A_{varepsilon})leq mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using subadditivity of $mu^*$ we have $mu^*(A)-mu^*(A_{varepsilon})leq mu^*(A-A_{varepsilon})<varepsilon.$ But since $A_{varepsilon}in mathcal{A}$ then $mu^*(A_{varepsilon})=mu(A_{varepsilon})$ so $mu^*(A)-mu(A_{varepsilon})<varepsilon$. We have that $mu^*(A)$ is finite so we can apply the definition of outer measure. Is it correct? But I am sure that it is OK.



2) enter image description here



He takes $Bin mathcal{A}_{mu}$ but why $mu(B)$ makes sense? $mu(B)$ is defined only on $malcal{A}$ but $B$ may not be in $mathcal{A}$. This also confusing me a lot.



3) enter image description here



The same question $B$ and $C$ are $mu$-measurable sets but why $mu(B)$ and $mu(C)$ makes sense?



I guess that all these questions are related to each. So would be very grateful for detailed answer and help!



EDIT 2: Also one moment which I have forgot to ask in the previous questions:



enter image description here



In the above proof note two moments which I have underlined with red line. Why Bogachev writes $mu^*(A)=mu^*(A'')$ but in he writes just $mu(B)=mu^*(X-A)$. The first is OK and I am agree with that but in the second he omits $mu^*$ and just write $mu(B)$. Could you explain it, please?










share|cite|improve this question















I am reading Bogachev's book "Measure Theory" which is in my opinion is very good book on measure theory. Let me ask you the following question:




Definition: Let $mu$ - non-negative set function with domain $mathcal{A}subset 2^X$. The set $A$ is called $mu$-measurable if
for any $varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$
such that $mu^*(Atriangle A_{varepsilon})<varepsilon,$ where
$mu^*$-outer measure.



The set of all $mu$-measurable sets is denoted by
$mathcal{A}_{mu}$.




One thing began to confuse me during reading. When author takes some set $Ain mathcal{A}_{mu}$ and then he considers $mu(A)$. This confuses me a lot.



We know that $mathcal{A}subset mathcal{A}_{mu}$ and when $Ain mathcal{A}_{mu}$ it means that $A$ is $mu$-measurable set. But $mu$ is defined only on $mathcal{A}$. Why he does not consider $mu^*(A)$?



More precisely, if $A$ is $mu$-measurable ($Ain mathcal{A}_{mu}$) then why $mu(A)$ makes sense? and does $mu(A)=mu^*(A)$?



P.S. I know the fact that $mathcal{A}_{mu}$ is $sigma$-algebra and $mu$ can be uniquely extended from algebra $mathcal{A}$ to $sigma$-algebra $mathcal{A}_{mu}$. And this extension is given by $mu^*$.



I would be very grateful if somebody can detailed answer. Because it is important two understand such things.



EDIT: I will divide my questions into 3 parts.



1) Corollary 1.5.8. (page 21). enter image description here



I was not able to understand this part. He just uses the definition of outer measure $mu^*$ but not the definition of $mu$-measurability.



My approach: Since $A$ is $mu$-measurable then $forall varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$ such that $mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using monotonicity of $mu^*$ we have $mu^*(A-A_{varepsilon})leq mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using subadditivity of $mu^*$ we have $mu^*(A)-mu^*(A_{varepsilon})leq mu^*(A-A_{varepsilon})<varepsilon.$ But since $A_{varepsilon}in mathcal{A}$ then $mu^*(A_{varepsilon})=mu(A_{varepsilon})$ so $mu^*(A)-mu(A_{varepsilon})<varepsilon$. We have that $mu^*(A)$ is finite so we can apply the definition of outer measure. Is it correct? But I am sure that it is OK.



2) enter image description here



He takes $Bin mathcal{A}_{mu}$ but why $mu(B)$ makes sense? $mu(B)$ is defined only on $malcal{A}$ but $B$ may not be in $mathcal{A}$. This also confusing me a lot.



3) enter image description here



The same question $B$ and $C$ are $mu$-measurable sets but why $mu(B)$ and $mu(C)$ makes sense?



I guess that all these questions are related to each. So would be very grateful for detailed answer and help!



EDIT 2: Also one moment which I have forgot to ask in the previous questions:



enter image description here



In the above proof note two moments which I have underlined with red line. Why Bogachev writes $mu^*(A)=mu^*(A'')$ but in he writes just $mu(B)=mu^*(X-A)$. The first is OK and I am agree with that but in the second he omits $mu^*$ and just write $mu(B)$. Could you explain it, please?







measure-theory






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share|cite|improve this question








edited 2 days ago

























asked 2 days ago









ZFR

4,86631337




4,86631337








  • 1




    Where exactly does he do what confuses you? Could you provide a more precise reference; this is a big book.
    – Michael Greinecker
    2 days ago










  • @MichaelGreinecker, Please take a look at the EDIT.
    – ZFR
    2 days ago














  • 1




    Where exactly does he do what confuses you? Could you provide a more precise reference; this is a big book.
    – Michael Greinecker
    2 days ago










  • @MichaelGreinecker, Please take a look at the EDIT.
    – ZFR
    2 days ago








1




1




Where exactly does he do what confuses you? Could you provide a more precise reference; this is a big book.
– Michael Greinecker
2 days ago




Where exactly does he do what confuses you? Could you provide a more precise reference; this is a big book.
– Michael Greinecker
2 days ago












@MichaelGreinecker, Please take a look at the EDIT.
– ZFR
2 days ago




@MichaelGreinecker, Please take a look at the EDIT.
– ZFR
2 days ago










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










In 1) he does indeed not use the definition of $mu$-measurability, he refers to the definition of the outer measure.



2) He just started to denote the unique nonnegative countable additive extension $mu^*$ of $mu$ by $mu$ itself. Indeed, he does so already before definition 1.5.10 when he calls $(X,mathcal{A}_mu,mu)$ the Lebesgue completion of $(X,mathcal{A},mu)$. Using the same name for a function and an extension of it is a widespread harmless abuse of notation in much of mathematics.



3) Same as 2)






share|cite|improve this answer























  • Could you take a look at my approach in question 1?
    – ZFR
    2 days ago










  • I am beginner in measure theory and that's why I am asking quite stupid questions. My approach explained to me why we can use the definition of outer measure. If the set $A$ is $mu$-measurable then it is outer measure, i.e. $mu^*(A)<infty$ and we can use the definition of outer measure (that what Bogachev did.)
    – ZFR
    2 days ago










  • Well, Bogachev starts with set functions whose values are real numbers. In particular, a nonnegative measure on algebra must be bounded above by the value of the total set $X$. He only starts to allow for infinite measures in 1.6.
    – Michael Greinecker
    2 days ago










  • And evry set has an outer measure.
    – Michael Greinecker
    2 days ago






  • 1




    No, he just mentions that the definition makes sense for set functions that can have the value $infty$, but that does not mean the results include such set functions. Indeed, not all results of this section hold true. The extension from a measure on an algebra to the generated $sigma$-algebra need for example not be unique. However, outer measure on the generated $sigma$-algebra is one such extension and if you want to allow for infinite measures, you need a proof that works for sets with infinite measure.
    – Michael Greinecker
    2 days ago











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










In 1) he does indeed not use the definition of $mu$-measurability, he refers to the definition of the outer measure.



2) He just started to denote the unique nonnegative countable additive extension $mu^*$ of $mu$ by $mu$ itself. Indeed, he does so already before definition 1.5.10 when he calls $(X,mathcal{A}_mu,mu)$ the Lebesgue completion of $(X,mathcal{A},mu)$. Using the same name for a function and an extension of it is a widespread harmless abuse of notation in much of mathematics.



3) Same as 2)






share|cite|improve this answer























  • Could you take a look at my approach in question 1?
    – ZFR
    2 days ago










  • I am beginner in measure theory and that's why I am asking quite stupid questions. My approach explained to me why we can use the definition of outer measure. If the set $A$ is $mu$-measurable then it is outer measure, i.e. $mu^*(A)<infty$ and we can use the definition of outer measure (that what Bogachev did.)
    – ZFR
    2 days ago










  • Well, Bogachev starts with set functions whose values are real numbers. In particular, a nonnegative measure on algebra must be bounded above by the value of the total set $X$. He only starts to allow for infinite measures in 1.6.
    – Michael Greinecker
    2 days ago










  • And evry set has an outer measure.
    – Michael Greinecker
    2 days ago






  • 1




    No, he just mentions that the definition makes sense for set functions that can have the value $infty$, but that does not mean the results include such set functions. Indeed, not all results of this section hold true. The extension from a measure on an algebra to the generated $sigma$-algebra need for example not be unique. However, outer measure on the generated $sigma$-algebra is one such extension and if you want to allow for infinite measures, you need a proof that works for sets with infinite measure.
    – Michael Greinecker
    2 days ago















up vote
1
down vote



accepted










In 1) he does indeed not use the definition of $mu$-measurability, he refers to the definition of the outer measure.



2) He just started to denote the unique nonnegative countable additive extension $mu^*$ of $mu$ by $mu$ itself. Indeed, he does so already before definition 1.5.10 when he calls $(X,mathcal{A}_mu,mu)$ the Lebesgue completion of $(X,mathcal{A},mu)$. Using the same name for a function and an extension of it is a widespread harmless abuse of notation in much of mathematics.



3) Same as 2)






share|cite|improve this answer























  • Could you take a look at my approach in question 1?
    – ZFR
    2 days ago










  • I am beginner in measure theory and that's why I am asking quite stupid questions. My approach explained to me why we can use the definition of outer measure. If the set $A$ is $mu$-measurable then it is outer measure, i.e. $mu^*(A)<infty$ and we can use the definition of outer measure (that what Bogachev did.)
    – ZFR
    2 days ago










  • Well, Bogachev starts with set functions whose values are real numbers. In particular, a nonnegative measure on algebra must be bounded above by the value of the total set $X$. He only starts to allow for infinite measures in 1.6.
    – Michael Greinecker
    2 days ago










  • And evry set has an outer measure.
    – Michael Greinecker
    2 days ago






  • 1




    No, he just mentions that the definition makes sense for set functions that can have the value $infty$, but that does not mean the results include such set functions. Indeed, not all results of this section hold true. The extension from a measure on an algebra to the generated $sigma$-algebra need for example not be unique. However, outer measure on the generated $sigma$-algebra is one such extension and if you want to allow for infinite measures, you need a proof that works for sets with infinite measure.
    – Michael Greinecker
    2 days ago













up vote
1
down vote



accepted







up vote
1
down vote



accepted






In 1) he does indeed not use the definition of $mu$-measurability, he refers to the definition of the outer measure.



2) He just started to denote the unique nonnegative countable additive extension $mu^*$ of $mu$ by $mu$ itself. Indeed, he does so already before definition 1.5.10 when he calls $(X,mathcal{A}_mu,mu)$ the Lebesgue completion of $(X,mathcal{A},mu)$. Using the same name for a function and an extension of it is a widespread harmless abuse of notation in much of mathematics.



3) Same as 2)






share|cite|improve this answer














In 1) he does indeed not use the definition of $mu$-measurability, he refers to the definition of the outer measure.



2) He just started to denote the unique nonnegative countable additive extension $mu^*$ of $mu$ by $mu$ itself. Indeed, he does so already before definition 1.5.10 when he calls $(X,mathcal{A}_mu,mu)$ the Lebesgue completion of $(X,mathcal{A},mu)$. Using the same name for a function and an extension of it is a widespread harmless abuse of notation in much of mathematics.



3) Same as 2)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered 2 days ago


























community wiki





Michael Greinecker













  • Could you take a look at my approach in question 1?
    – ZFR
    2 days ago










  • I am beginner in measure theory and that's why I am asking quite stupid questions. My approach explained to me why we can use the definition of outer measure. If the set $A$ is $mu$-measurable then it is outer measure, i.e. $mu^*(A)<infty$ and we can use the definition of outer measure (that what Bogachev did.)
    – ZFR
    2 days ago










  • Well, Bogachev starts with set functions whose values are real numbers. In particular, a nonnegative measure on algebra must be bounded above by the value of the total set $X$. He only starts to allow for infinite measures in 1.6.
    – Michael Greinecker
    2 days ago










  • And evry set has an outer measure.
    – Michael Greinecker
    2 days ago






  • 1




    No, he just mentions that the definition makes sense for set functions that can have the value $infty$, but that does not mean the results include such set functions. Indeed, not all results of this section hold true. The extension from a measure on an algebra to the generated $sigma$-algebra need for example not be unique. However, outer measure on the generated $sigma$-algebra is one such extension and if you want to allow for infinite measures, you need a proof that works for sets with infinite measure.
    – Michael Greinecker
    2 days ago


















  • Could you take a look at my approach in question 1?
    – ZFR
    2 days ago










  • I am beginner in measure theory and that's why I am asking quite stupid questions. My approach explained to me why we can use the definition of outer measure. If the set $A$ is $mu$-measurable then it is outer measure, i.e. $mu^*(A)<infty$ and we can use the definition of outer measure (that what Bogachev did.)
    – ZFR
    2 days ago










  • Well, Bogachev starts with set functions whose values are real numbers. In particular, a nonnegative measure on algebra must be bounded above by the value of the total set $X$. He only starts to allow for infinite measures in 1.6.
    – Michael Greinecker
    2 days ago










  • And evry set has an outer measure.
    – Michael Greinecker
    2 days ago






  • 1




    No, he just mentions that the definition makes sense for set functions that can have the value $infty$, but that does not mean the results include such set functions. Indeed, not all results of this section hold true. The extension from a measure on an algebra to the generated $sigma$-algebra need for example not be unique. However, outer measure on the generated $sigma$-algebra is one such extension and if you want to allow for infinite measures, you need a proof that works for sets with infinite measure.
    – Michael Greinecker
    2 days ago
















Could you take a look at my approach in question 1?
– ZFR
2 days ago




Could you take a look at my approach in question 1?
– ZFR
2 days ago












I am beginner in measure theory and that's why I am asking quite stupid questions. My approach explained to me why we can use the definition of outer measure. If the set $A$ is $mu$-measurable then it is outer measure, i.e. $mu^*(A)<infty$ and we can use the definition of outer measure (that what Bogachev did.)
– ZFR
2 days ago




I am beginner in measure theory and that's why I am asking quite stupid questions. My approach explained to me why we can use the definition of outer measure. If the set $A$ is $mu$-measurable then it is outer measure, i.e. $mu^*(A)<infty$ and we can use the definition of outer measure (that what Bogachev did.)
– ZFR
2 days ago












Well, Bogachev starts with set functions whose values are real numbers. In particular, a nonnegative measure on algebra must be bounded above by the value of the total set $X$. He only starts to allow for infinite measures in 1.6.
– Michael Greinecker
2 days ago




Well, Bogachev starts with set functions whose values are real numbers. In particular, a nonnegative measure on algebra must be bounded above by the value of the total set $X$. He only starts to allow for infinite measures in 1.6.
– Michael Greinecker
2 days ago












And evry set has an outer measure.
– Michael Greinecker
2 days ago




And evry set has an outer measure.
– Michael Greinecker
2 days ago




1




1




No, he just mentions that the definition makes sense for set functions that can have the value $infty$, but that does not mean the results include such set functions. Indeed, not all results of this section hold true. The extension from a measure on an algebra to the generated $sigma$-algebra need for example not be unique. However, outer measure on the generated $sigma$-algebra is one such extension and if you want to allow for infinite measures, you need a proof that works for sets with infinite measure.
– Michael Greinecker
2 days ago




No, he just mentions that the definition makes sense for set functions that can have the value $infty$, but that does not mean the results include such set functions. Indeed, not all results of this section hold true. The extension from a measure on an algebra to the generated $sigma$-algebra need for example not be unique. However, outer measure on the generated $sigma$-algebra is one such extension and if you want to allow for infinite measures, you need a proof that works for sets with infinite measure.
– Michael Greinecker
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