Show that a given measure is equal to the Lebesgue measure on Borel subsets on $mathbb{R}$












0














Suppose we have a measure $mu$ on $left( mathbb{R},mathcal{B}(mathbb{R}) right)$ with $mu((0,1])=1$, and $mu$ invariant under translations i.e. $mu(A) = mu(A+c)$ for every $c in mathbb{R}$ and Borel set $A subseteq mathbb{R}$.



Now letting $lambda$ be the Lebesgue measure, in the first part of the question I was able to show that for any interval $A=(a,b]$ with $b-a in mathbb{Q}$ that $mu(A) = lambda(A)$.



The second part of the question asks to extend this to all Borel sets, i.e. $forall A in mathcal{B}(mathbb{R})$ we get that$mu(A) = lambda(A)$, but I am really struggling to think of a practical way of achieving this.



Clearly using the first part of the question is the correct approach, so I was attempting to come up with some ways of using this.



The first thing I thought of was maybe to approximate all of the intervals in $mathcal{B}(mathbb{R})$ by these rational intervals in the first part of the question, but I could not think of a rigorous way of doing this and I do not think this is the correct approach.



My next thought was to potentially show that the intervals from the first part form a $pi$-system of some sort and maybe generate the Borel sets and so agreeing only on these intervals is sufficient to show they agree on the whole space, but I am not too confident in arguing this so any help would be appreciated thanks.






measure-theory lebesgue-measure







share|cite|improve this question


















  • 1




    yea try the pi-lambda theorem, taking the lambda system to be the sets on which $mu$ and $lambda$ agree. en.wikipedia.org/wiki/Pi-system#The_%CF%80-%CE%BB_theorem
    – Tim kinsella
    Nov 21 at 21:45


















0














Suppose we have a measure $mu$ on $left( mathbb{R},mathcal{B}(mathbb{R}) right)$ with $mu((0,1])=1$, and $mu$ invariant under translations i.e. $mu(A) = mu(A+c)$ for every $c in mathbb{R}$ and Borel set $A subseteq mathbb{R}$.



Now letting $lambda$ be the Lebesgue measure, in the first part of the question I was able to show that for any interval $A=(a,b]$ with $b-a in mathbb{Q}$ that $mu(A) = lambda(A)$.



The second part of the question asks to extend this to all Borel sets, i.e. $forall A in mathcal{B}(mathbb{R})$ we get that$mu(A) = lambda(A)$, but I am really struggling to think of a practical way of achieving this.



Clearly using the first part of the question is the correct approach, so I was attempting to come up with some ways of using this.



The first thing I thought of was maybe to approximate all of the intervals in $mathcal{B}(mathbb{R})$ by these rational intervals in the first part of the question, but I could not think of a rigorous way of doing this and I do not think this is the correct approach.



My next thought was to potentially show that the intervals from the first part form a $pi$-system of some sort and maybe generate the Borel sets and so agreeing only on these intervals is sufficient to show they agree on the whole space, but I am not too confident in arguing this so any help would be appreciated thanks.






measure-theory lebesgue-measure







share|cite|improve this question


















  • 1




    yea try the pi-lambda theorem, taking the lambda system to be the sets on which $mu$ and $lambda$ agree. en.wikipedia.org/wiki/Pi-system#The_%CF%80-%CE%BB_theorem
    – Tim kinsella
    Nov 21 at 21:45
















0












0








0







Suppose we have a measure $mu$ on $left( mathbb{R},mathcal{B}(mathbb{R}) right)$ with $mu((0,1])=1$, and $mu$ invariant under translations i.e. $mu(A) = mu(A+c)$ for every $c in mathbb{R}$ and Borel set $A subseteq mathbb{R}$.



Now letting $lambda$ be the Lebesgue measure, in the first part of the question I was able to show that for any interval $A=(a,b]$ with $b-a in mathbb{Q}$ that $mu(A) = lambda(A)$.



The second part of the question asks to extend this to all Borel sets, i.e. $forall A in mathcal{B}(mathbb{R})$ we get that$mu(A) = lambda(A)$, but I am really struggling to think of a practical way of achieving this.



Clearly using the first part of the question is the correct approach, so I was attempting to come up with some ways of using this.



The first thing I thought of was maybe to approximate all of the intervals in $mathcal{B}(mathbb{R})$ by these rational intervals in the first part of the question, but I could not think of a rigorous way of doing this and I do not think this is the correct approach.



My next thought was to potentially show that the intervals from the first part form a $pi$-system of some sort and maybe generate the Borel sets and so agreeing only on these intervals is sufficient to show they agree on the whole space, but I am not too confident in arguing this so any help would be appreciated thanks.






measure-theory lebesgue-measure







share|cite|improve this question













Suppose we have a measure $mu$ on $left( mathbb{R},mathcal{B}(mathbb{R}) right)$ with $mu((0,1])=1$, and $mu$ invariant under translations i.e. $mu(A) = mu(A+c)$ for every $c in mathbb{R}$ and Borel set $A subseteq mathbb{R}$.



Now letting $lambda$ be the Lebesgue measure, in the first part of the question I was able to show that for any interval $A=(a,b]$ with $b-a in mathbb{Q}$ that $mu(A) = lambda(A)$.



The second part of the question asks to extend this to all Borel sets, i.e. $forall A in mathcal{B}(mathbb{R})$ we get that$mu(A) = lambda(A)$, but I am really struggling to think of a practical way of achieving this.



Clearly using the first part of the question is the correct approach, so I was attempting to come up with some ways of using this.



The first thing I thought of was maybe to approximate all of the intervals in $mathcal{B}(mathbb{R})$ by these rational intervals in the first part of the question, but I could not think of a rigorous way of doing this and I do not think this is the correct approach.



My next thought was to potentially show that the intervals from the first part form a $pi$-system of some sort and maybe generate the Borel sets and so agreeing only on these intervals is sufficient to show they agree on the whole space, but I am not too confident in arguing this so any help would be appreciated thanks.






measure-theory lebesgue-measure




measure-theory lebesgue-measure






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asked Nov 21 at 21:40









UsernameInvalid

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  • 1




    yea try the pi-lambda theorem, taking the lambda system to be the sets on which $mu$ and $lambda$ agree. en.wikipedia.org/wiki/Pi-system#The_%CF%80-%CE%BB_theorem
    – Tim kinsella
    Nov 21 at 21:45
















  • 1




    yea try the pi-lambda theorem, taking the lambda system to be the sets on which $mu$ and $lambda$ agree. en.wikipedia.org/wiki/Pi-system#The_%CF%80-%CE%BB_theorem
    – Tim kinsella
    Nov 21 at 21:45










1




1




yea try the pi-lambda theorem, taking the lambda system to be the sets on which $mu$ and $lambda$ agree. en.wikipedia.org/wiki/Pi-system#The_%CF%80-%CE%BB_theorem
– Tim kinsella
Nov 21 at 21:45






yea try the pi-lambda theorem, taking the lambda system to be the sets on which $mu$ and $lambda$ agree. en.wikipedia.org/wiki/Pi-system#The_%CF%80-%CE%BB_theorem
– Tim kinsella
Nov 21 at 21:45












1 Answer
1






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oldest

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Having $mu (a,b]=b-a$ for rational $a,b$ gives $mu (a,b)=b-a$ for all real $a<b.$ Proof: Write $(a,b)$ as the increasing union of $(a_n,b_n]$ for appropritate rational $a_n,b_n.$ Standard measure theory with your result for rationals then gives the result.



Since every open set in $mathbb R $ is the disjoint union of open intervals, we see $mu(U) = lambda (U)$ for all open $Usubset mathbb R.$



I'll stop here for now. Ask questions if you like.






share|cite|improve this answer





















  • Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work?
    – UsernameInvalid
    Nov 21 at 22:09






  • 1




    But if both $a_n,b_n$ are rational, you have it covered.
    – zhw.
    Nov 21 at 22:11












  • Ah yes of course, I suppose I was looking into it too much, thanks for the help :)
    – UsernameInvalid
    Nov 21 at 22:14











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Having $mu (a,b]=b-a$ for rational $a,b$ gives $mu (a,b)=b-a$ for all real $a<b.$ Proof: Write $(a,b)$ as the increasing union of $(a_n,b_n]$ for appropritate rational $a_n,b_n.$ Standard measure theory with your result for rationals then gives the result.



Since every open set in $mathbb R $ is the disjoint union of open intervals, we see $mu(U) = lambda (U)$ for all open $Usubset mathbb R.$



I'll stop here for now. Ask questions if you like.






share|cite|improve this answer





















  • Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work?
    – UsernameInvalid
    Nov 21 at 22:09






  • 1




    But if both $a_n,b_n$ are rational, you have it covered.
    – zhw.
    Nov 21 at 22:11












  • Ah yes of course, I suppose I was looking into it too much, thanks for the help :)
    – UsernameInvalid
    Nov 21 at 22:14
















1














Having $mu (a,b]=b-a$ for rational $a,b$ gives $mu (a,b)=b-a$ for all real $a<b.$ Proof: Write $(a,b)$ as the increasing union of $(a_n,b_n]$ for appropritate rational $a_n,b_n.$ Standard measure theory with your result for rationals then gives the result.



Since every open set in $mathbb R $ is the disjoint union of open intervals, we see $mu(U) = lambda (U)$ for all open $Usubset mathbb R.$



I'll stop here for now. Ask questions if you like.






share|cite|improve this answer





















  • Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work?
    – UsernameInvalid
    Nov 21 at 22:09






  • 1




    But if both $a_n,b_n$ are rational, you have it covered.
    – zhw.
    Nov 21 at 22:11












  • Ah yes of course, I suppose I was looking into it too much, thanks for the help :)
    – UsernameInvalid
    Nov 21 at 22:14














1












1








1






Having $mu (a,b]=b-a$ for rational $a,b$ gives $mu (a,b)=b-a$ for all real $a<b.$ Proof: Write $(a,b)$ as the increasing union of $(a_n,b_n]$ for appropritate rational $a_n,b_n.$ Standard measure theory with your result for rationals then gives the result.



Since every open set in $mathbb R $ is the disjoint union of open intervals, we see $mu(U) = lambda (U)$ for all open $Usubset mathbb R.$



I'll stop here for now. Ask questions if you like.






share|cite|improve this answer












Having $mu (a,b]=b-a$ for rational $a,b$ gives $mu (a,b)=b-a$ for all real $a<b.$ Proof: Write $(a,b)$ as the increasing union of $(a_n,b_n]$ for appropritate rational $a_n,b_n.$ Standard measure theory with your result for rationals then gives the result.



Since every open set in $mathbb R $ is the disjoint union of open intervals, we see $mu(U) = lambda (U)$ for all open $Usubset mathbb R.$



I'll stop here for now. Ask questions if you like.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 22:03









zhw.

71.4k43075




71.4k43075












  • Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work?
    – UsernameInvalid
    Nov 21 at 22:09






  • 1




    But if both $a_n,b_n$ are rational, you have it covered.
    – zhw.
    Nov 21 at 22:11












  • Ah yes of course, I suppose I was looking into it too much, thanks for the help :)
    – UsernameInvalid
    Nov 21 at 22:14


















  • Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work?
    – UsernameInvalid
    Nov 21 at 22:09






  • 1




    But if both $a_n,b_n$ are rational, you have it covered.
    – zhw.
    Nov 21 at 22:11












  • Ah yes of course, I suppose I was looking into it too much, thanks for the help :)
    – UsernameInvalid
    Nov 21 at 22:14
















Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work?
– UsernameInvalid
Nov 21 at 22:09




Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work?
– UsernameInvalid
Nov 21 at 22:09




1




1




But if both $a_n,b_n$ are rational, you have it covered.
– zhw.
Nov 21 at 22:11






But if both $a_n,b_n$ are rational, you have it covered.
– zhw.
Nov 21 at 22:11














Ah yes of course, I suppose I was looking into it too much, thanks for the help :)
– UsernameInvalid
Nov 21 at 22:14




Ah yes of course, I suppose I was looking into it too much, thanks for the help :)
– UsernameInvalid
Nov 21 at 22:14


















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