Convergence of Borel measure evaluated at sequence of intervals

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Let $nu$ be a $1$-dimensional Borel measure i.e. for each $x in mathbbR$ there exists an open set $U$ containing $x$ and $nu (U) < infty$.



Does $lim_delta downarrow 0 nu(a,a+delta) = 0$ hold for all $a in mathbbR$? If so why?




real-analysis




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  • 1




    It holds for all $a$.
    – copper.hat
    Aug 27 at 18:23










  • Can you please tell me why...
    – An_876_Joke
    Aug 27 at 18:25






  • 1




    Continuity of measure from above.
    – copper.hat
    Aug 27 at 18:26















up vote
0
down vote

favorite












Let $nu$ be a $1$-dimensional Borel measure i.e. for each $x in mathbbR$ there exists an open set $U$ containing $x$ and $nu (U) < infty$.



Does $lim_delta downarrow 0 nu(a,a+delta) = 0$ hold for all $a in mathbbR$? If so why?




real-analysis




share|cite|improve this question


















  • 1




    It holds for all $a$.
    – copper.hat
    Aug 27 at 18:23










  • Can you please tell me why...
    – An_876_Joke
    Aug 27 at 18:25






  • 1




    Continuity of measure from above.
    – copper.hat
    Aug 27 at 18:26













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $nu$ be a $1$-dimensional Borel measure i.e. for each $x in mathbbR$ there exists an open set $U$ containing $x$ and $nu (U) < infty$.



Does $lim_delta downarrow 0 nu(a,a+delta) = 0$ hold for all $a in mathbbR$? If so why?




real-analysis




share|cite|improve this question














Let $nu$ be a $1$-dimensional Borel measure i.e. for each $x in mathbbR$ there exists an open set $U$ containing $x$ and $nu (U) < infty$.



Does $lim_delta downarrow 0 nu(a,a+delta) = 0$ hold for all $a in mathbbR$? If so why?





real-analysis





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 27 at 18:24

























asked Aug 27 at 18:17









An_876_Joke

935




935







  • 1




    It holds for all $a$.
    – copper.hat
    Aug 27 at 18:23










  • Can you please tell me why...
    – An_876_Joke
    Aug 27 at 18:25






  • 1




    Continuity of measure from above.
    – copper.hat
    Aug 27 at 18:26













  • 1




    It holds for all $a$.
    – copper.hat
    Aug 27 at 18:23










  • Can you please tell me why...
    – An_876_Joke
    Aug 27 at 18:25






  • 1




    Continuity of measure from above.
    – copper.hat
    Aug 27 at 18:26








1




1




It holds for all $a$.
– copper.hat
Aug 27 at 18:23




It holds for all $a$.
– copper.hat
Aug 27 at 18:23












Can you please tell me why...
– An_876_Joke
Aug 27 at 18:25




Can you please tell me why...
– An_876_Joke
Aug 27 at 18:25




1




1




Continuity of measure from above.
– copper.hat
Aug 27 at 18:26





Continuity of measure from above.
– copper.hat
Aug 27 at 18:26











1 Answer
1






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oldest

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up vote
1
down vote













As copper.hat points out in the comments, this is true. Let $(delta_n)$ be a sequence of positive real numbers such that $delta_nto 0$ as $ntoinfty$. By assumpution, there is some open neighborhood $U$ of $a$ such that $nu(U)<infty$, hence some $varepsilon>0$ such that $(a-varepsilon,a+varepsilon)subset U$.
There is some $N$ such that $delta_n<varepsilon$ for $ngeq N$. For such $n$, we have
$$nu((a,a+delta_n))leqnu((a-varepsilon,a+varepsilon))leqnu(U)<infty.$$
Continuity from above now implies that $limlimits_ntoinftynu((a,a+delta_n))=0$, and since the sequence $(delta_n)$ was arbitrary, the result follows.






share|cite|improve this answer




















  • Although I know how to show that the conditions on $nu $ imply continuity from above,) I dk whether the proposer knows why $nu$ must be continuous from above.
    – DanielWainfleet
    Aug 28 at 4:00











  • I read how to proof continuity from above.
    – An_876_Joke
    Aug 28 at 14:28






  • 1




    Although I must say that we dont have $... subset (a+delta_n+2,a) subset (a+delta_n+1,a) subset (a+delta_n,a)$.
    – An_876_Joke
    Aug 28 at 14:41







  • 1




    We can show $lim_n rightarrow infty nu((a,a+delta_phi(n))) = 0$ for some rearrangement $phi:mathbbN rightarrow mathbbN$ making the $delta_phi(n)$ monotone. Now any rearrangement of $ nu((a,a+delta_phi(n)))$ will converge to the same limit.
    – An_876_Joke
    Aug 28 at 15:17











  • Yes, you are correct. It is sufficient to assume that $(delta_n)$ converges monotonically to $0$.
    – Aweygan
    Aug 28 at 16:03










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up vote
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As copper.hat points out in the comments, this is true. Let $(delta_n)$ be a sequence of positive real numbers such that $delta_nto 0$ as $ntoinfty$. By assumpution, there is some open neighborhood $U$ of $a$ such that $nu(U)<infty$, hence some $varepsilon>0$ such that $(a-varepsilon,a+varepsilon)subset U$.
There is some $N$ such that $delta_n<varepsilon$ for $ngeq N$. For such $n$, we have
$$nu((a,a+delta_n))leqnu((a-varepsilon,a+varepsilon))leqnu(U)<infty.$$
Continuity from above now implies that $limlimits_ntoinftynu((a,a+delta_n))=0$, and since the sequence $(delta_n)$ was arbitrary, the result follows.






share|cite|improve this answer




















  • Although I know how to show that the conditions on $nu $ imply continuity from above,) I dk whether the proposer knows why $nu$ must be continuous from above.
    – DanielWainfleet
    Aug 28 at 4:00











  • I read how to proof continuity from above.
    – An_876_Joke
    Aug 28 at 14:28






  • 1




    Although I must say that we dont have $... subset (a+delta_n+2,a) subset (a+delta_n+1,a) subset (a+delta_n,a)$.
    – An_876_Joke
    Aug 28 at 14:41







  • 1




    We can show $lim_n rightarrow infty nu((a,a+delta_phi(n))) = 0$ for some rearrangement $phi:mathbbN rightarrow mathbbN$ making the $delta_phi(n)$ monotone. Now any rearrangement of $ nu((a,a+delta_phi(n)))$ will converge to the same limit.
    – An_876_Joke
    Aug 28 at 15:17











  • Yes, you are correct. It is sufficient to assume that $(delta_n)$ converges monotonically to $0$.
    – Aweygan
    Aug 28 at 16:03














up vote
1
down vote













As copper.hat points out in the comments, this is true. Let $(delta_n)$ be a sequence of positive real numbers such that $delta_nto 0$ as $ntoinfty$. By assumpution, there is some open neighborhood $U$ of $a$ such that $nu(U)<infty$, hence some $varepsilon>0$ such that $(a-varepsilon,a+varepsilon)subset U$.
There is some $N$ such that $delta_n<varepsilon$ for $ngeq N$. For such $n$, we have
$$nu((a,a+delta_n))leqnu((a-varepsilon,a+varepsilon))leqnu(U)<infty.$$
Continuity from above now implies that $limlimits_ntoinftynu((a,a+delta_n))=0$, and since the sequence $(delta_n)$ was arbitrary, the result follows.






share|cite|improve this answer




















  • Although I know how to show that the conditions on $nu $ imply continuity from above,) I dk whether the proposer knows why $nu$ must be continuous from above.
    – DanielWainfleet
    Aug 28 at 4:00











  • I read how to proof continuity from above.
    – An_876_Joke
    Aug 28 at 14:28






  • 1




    Although I must say that we dont have $... subset (a+delta_n+2,a) subset (a+delta_n+1,a) subset (a+delta_n,a)$.
    – An_876_Joke
    Aug 28 at 14:41







  • 1




    We can show $lim_n rightarrow infty nu((a,a+delta_phi(n))) = 0$ for some rearrangement $phi:mathbbN rightarrow mathbbN$ making the $delta_phi(n)$ monotone. Now any rearrangement of $ nu((a,a+delta_phi(n)))$ will converge to the same limit.
    – An_876_Joke
    Aug 28 at 15:17











  • Yes, you are correct. It is sufficient to assume that $(delta_n)$ converges monotonically to $0$.
    – Aweygan
    Aug 28 at 16:03












up vote
1
down vote










up vote
1
down vote









As copper.hat points out in the comments, this is true. Let $(delta_n)$ be a sequence of positive real numbers such that $delta_nto 0$ as $ntoinfty$. By assumpution, there is some open neighborhood $U$ of $a$ such that $nu(U)<infty$, hence some $varepsilon>0$ such that $(a-varepsilon,a+varepsilon)subset U$.
There is some $N$ such that $delta_n<varepsilon$ for $ngeq N$. For such $n$, we have
$$nu((a,a+delta_n))leqnu((a-varepsilon,a+varepsilon))leqnu(U)<infty.$$
Continuity from above now implies that $limlimits_ntoinftynu((a,a+delta_n))=0$, and since the sequence $(delta_n)$ was arbitrary, the result follows.






share|cite|improve this answer












As copper.hat points out in the comments, this is true. Let $(delta_n)$ be a sequence of positive real numbers such that $delta_nto 0$ as $ntoinfty$. By assumpution, there is some open neighborhood $U$ of $a$ such that $nu(U)<infty$, hence some $varepsilon>0$ such that $(a-varepsilon,a+varepsilon)subset U$.
There is some $N$ such that $delta_n<varepsilon$ for $ngeq N$. For such $n$, we have
$$nu((a,a+delta_n))leqnu((a-varepsilon,a+varepsilon))leqnu(U)<infty.$$
Continuity from above now implies that $limlimits_ntoinftynu((a,a+delta_n))=0$, and since the sequence $(delta_n)$ was arbitrary, the result follows.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 27 at 19:09









Aweygan

12k21438




12k21438











  • Although I know how to show that the conditions on $nu $ imply continuity from above,) I dk whether the proposer knows why $nu$ must be continuous from above.
    – DanielWainfleet
    Aug 28 at 4:00











  • I read how to proof continuity from above.
    – An_876_Joke
    Aug 28 at 14:28






  • 1




    Although I must say that we dont have $... subset (a+delta_n+2,a) subset (a+delta_n+1,a) subset (a+delta_n,a)$.
    – An_876_Joke
    Aug 28 at 14:41







  • 1




    We can show $lim_n rightarrow infty nu((a,a+delta_phi(n))) = 0$ for some rearrangement $phi:mathbbN rightarrow mathbbN$ making the $delta_phi(n)$ monotone. Now any rearrangement of $ nu((a,a+delta_phi(n)))$ will converge to the same limit.
    – An_876_Joke
    Aug 28 at 15:17











  • Yes, you are correct. It is sufficient to assume that $(delta_n)$ converges monotonically to $0$.
    – Aweygan
    Aug 28 at 16:03
















  • Although I know how to show that the conditions on $nu $ imply continuity from above,) I dk whether the proposer knows why $nu$ must be continuous from above.
    – DanielWainfleet
    Aug 28 at 4:00











  • I read how to proof continuity from above.
    – An_876_Joke
    Aug 28 at 14:28






  • 1




    Although I must say that we dont have $... subset (a+delta_n+2,a) subset (a+delta_n+1,a) subset (a+delta_n,a)$.
    – An_876_Joke
    Aug 28 at 14:41







  • 1




    We can show $lim_n rightarrow infty nu((a,a+delta_phi(n))) = 0$ for some rearrangement $phi:mathbbN rightarrow mathbbN$ making the $delta_phi(n)$ monotone. Now any rearrangement of $ nu((a,a+delta_phi(n)))$ will converge to the same limit.
    – An_876_Joke
    Aug 28 at 15:17











  • Yes, you are correct. It is sufficient to assume that $(delta_n)$ converges monotonically to $0$.
    – Aweygan
    Aug 28 at 16:03















Although I know how to show that the conditions on $nu $ imply continuity from above,) I dk whether the proposer knows why $nu$ must be continuous from above.
– DanielWainfleet
Aug 28 at 4:00





Although I know how to show that the conditions on $nu $ imply continuity from above,) I dk whether the proposer knows why $nu$ must be continuous from above.
– DanielWainfleet
Aug 28 at 4:00













I read how to proof continuity from above.
– An_876_Joke
Aug 28 at 14:28




I read how to proof continuity from above.
– An_876_Joke
Aug 28 at 14:28




1




1




Although I must say that we dont have $... subset (a+delta_n+2,a) subset (a+delta_n+1,a) subset (a+delta_n,a)$.
– An_876_Joke
Aug 28 at 14:41





Although I must say that we dont have $... subset (a+delta_n+2,a) subset (a+delta_n+1,a) subset (a+delta_n,a)$.
– An_876_Joke
Aug 28 at 14:41





1




1




We can show $lim_n rightarrow infty nu((a,a+delta_phi(n))) = 0$ for some rearrangement $phi:mathbbN rightarrow mathbbN$ making the $delta_phi(n)$ monotone. Now any rearrangement of $ nu((a,a+delta_phi(n)))$ will converge to the same limit.
– An_876_Joke
Aug 28 at 15:17





We can show $lim_n rightarrow infty nu((a,a+delta_phi(n))) = 0$ for some rearrangement $phi:mathbbN rightarrow mathbbN$ making the $delta_phi(n)$ monotone. Now any rearrangement of $ nu((a,a+delta_phi(n)))$ will converge to the same limit.
– An_876_Joke
Aug 28 at 15:17













Yes, you are correct. It is sufficient to assume that $(delta_n)$ converges monotonically to $0$.
– Aweygan
Aug 28 at 16:03




Yes, you are correct. It is sufficient to assume that $(delta_n)$ converges monotonically to $0$.
– Aweygan
Aug 28 at 16:03

















 

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