Question on Egoroff-like theorem












4












$begingroup$


Hi all I was tackled by this question from Folland's real analysis second edition in the second chapter, it looks like a modified Egoroff theorem but I cannot really tackle it, it is question 41 of chapter 2 which reads as follows:




Let $mu$ be a $sigma$-finite measure and $ f_n rightarrow f $ a.e. Then there exist measurable sets $E_1,E_2,ldots subset X $ such that
$$muleft(left(bigcup_{i=1}^infty E_iright)^Cright)=0$$ and $f_n rightarrow f $ uniformly on each $ E_i. $




I think this might have something to do with Egoroff's theorem but that theorem mentions nothing about complement having measure zero, only as small as you would like, which is what confuses me. Can anyone point out a proof of this with an explanation?






real-analysis measure-theory uniform-convergence







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The ideas in my answer here (math.stackexchange.com/questions/1068043/…) should be helpful.
    $endgroup$
    – PhoemueX
    Jul 31 '15 at 10:10






  • 1




    $begingroup$
    @PhoemueX thanks tried your other answer but I still cannot understand it
    $endgroup$
    – Don John Prep
    Jul 31 '15 at 11:09






  • 1




    $begingroup$
    If you start with the stronger assumption that $mu$ is a finite measure, can you see how the assertion is obtained from Egorov's theorem?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:21






  • 2




    $begingroup$
    Let $F = bigcup E_n$. Then you have $Xsetminus F subset X setminus E_k$, and hence $mu(Xsetminus F) < 1/k$. Since that holds for all $k$, $mu(Xsetminus F) = 0$. Okay, the case of finite measure done. Now, for $sigma$-finite measure, write $X = bigcup A_m$, where each $A_m$ has finite measure. Perhaps it makes things easier to assume the $A_m$ disjoint, one can do that. So by the preceding, for each $m$, we have a sequence $E^{(m)}_n$ of subsets of $A_m$. How can you use that to get the desired sequence of subsets of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:48






  • 3




    $begingroup$
    @AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution.
    $endgroup$
    – Nate Eldredge
    Jul 31 '15 at 14:11
















4












$begingroup$


Hi all I was tackled by this question from Folland's real analysis second edition in the second chapter, it looks like a modified Egoroff theorem but I cannot really tackle it, it is question 41 of chapter 2 which reads as follows:




Let $mu$ be a $sigma$-finite measure and $ f_n rightarrow f $ a.e. Then there exist measurable sets $E_1,E_2,ldots subset X $ such that
$$muleft(left(bigcup_{i=1}^infty E_iright)^Cright)=0$$ and $f_n rightarrow f $ uniformly on each $ E_i. $




I think this might have something to do with Egoroff's theorem but that theorem mentions nothing about complement having measure zero, only as small as you would like, which is what confuses me. Can anyone point out a proof of this with an explanation?






real-analysis measure-theory uniform-convergence







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The ideas in my answer here (math.stackexchange.com/questions/1068043/…) should be helpful.
    $endgroup$
    – PhoemueX
    Jul 31 '15 at 10:10






  • 1




    $begingroup$
    @PhoemueX thanks tried your other answer but I still cannot understand it
    $endgroup$
    – Don John Prep
    Jul 31 '15 at 11:09






  • 1




    $begingroup$
    If you start with the stronger assumption that $mu$ is a finite measure, can you see how the assertion is obtained from Egorov's theorem?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:21






  • 2




    $begingroup$
    Let $F = bigcup E_n$. Then you have $Xsetminus F subset X setminus E_k$, and hence $mu(Xsetminus F) < 1/k$. Since that holds for all $k$, $mu(Xsetminus F) = 0$. Okay, the case of finite measure done. Now, for $sigma$-finite measure, write $X = bigcup A_m$, where each $A_m$ has finite measure. Perhaps it makes things easier to assume the $A_m$ disjoint, one can do that. So by the preceding, for each $m$, we have a sequence $E^{(m)}_n$ of subsets of $A_m$. How can you use that to get the desired sequence of subsets of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:48






  • 3




    $begingroup$
    @AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution.
    $endgroup$
    – Nate Eldredge
    Jul 31 '15 at 14:11














4












4








4


2



$begingroup$


Hi all I was tackled by this question from Folland's real analysis second edition in the second chapter, it looks like a modified Egoroff theorem but I cannot really tackle it, it is question 41 of chapter 2 which reads as follows:




Let $mu$ be a $sigma$-finite measure and $ f_n rightarrow f $ a.e. Then there exist measurable sets $E_1,E_2,ldots subset X $ such that
$$muleft(left(bigcup_{i=1}^infty E_iright)^Cright)=0$$ and $f_n rightarrow f $ uniformly on each $ E_i. $




I think this might have something to do with Egoroff's theorem but that theorem mentions nothing about complement having measure zero, only as small as you would like, which is what confuses me. Can anyone point out a proof of this with an explanation?






real-analysis measure-theory uniform-convergence







share|cite|improve this question











$endgroup$




Hi all I was tackled by this question from Folland's real analysis second edition in the second chapter, it looks like a modified Egoroff theorem but I cannot really tackle it, it is question 41 of chapter 2 which reads as follows:




Let $mu$ be a $sigma$-finite measure and $ f_n rightarrow f $ a.e. Then there exist measurable sets $E_1,E_2,ldots subset X $ such that
$$muleft(left(bigcup_{i=1}^infty E_iright)^Cright)=0$$ and $f_n rightarrow f $ uniformly on each $ E_i. $




I think this might have something to do with Egoroff's theorem but that theorem mentions nothing about complement having measure zero, only as small as you would like, which is what confuses me. Can anyone point out a proof of this with an explanation?






real-analysis measure-theory uniform-convergence




real-analysis measure-theory uniform-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 31 '15 at 13:25









Math1000

19.2k31745




19.2k31745










asked Jul 31 '15 at 8:41









Don John PrepDon John Prep

200111




200111








  • 3




    $begingroup$
    The ideas in my answer here (math.stackexchange.com/questions/1068043/…) should be helpful.
    $endgroup$
    – PhoemueX
    Jul 31 '15 at 10:10






  • 1




    $begingroup$
    @PhoemueX thanks tried your other answer but I still cannot understand it
    $endgroup$
    – Don John Prep
    Jul 31 '15 at 11:09






  • 1




    $begingroup$
    If you start with the stronger assumption that $mu$ is a finite measure, can you see how the assertion is obtained from Egorov's theorem?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:21






  • 2




    $begingroup$
    Let $F = bigcup E_n$. Then you have $Xsetminus F subset X setminus E_k$, and hence $mu(Xsetminus F) < 1/k$. Since that holds for all $k$, $mu(Xsetminus F) = 0$. Okay, the case of finite measure done. Now, for $sigma$-finite measure, write $X = bigcup A_m$, where each $A_m$ has finite measure. Perhaps it makes things easier to assume the $A_m$ disjoint, one can do that. So by the preceding, for each $m$, we have a sequence $E^{(m)}_n$ of subsets of $A_m$. How can you use that to get the desired sequence of subsets of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:48






  • 3




    $begingroup$
    @AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution.
    $endgroup$
    – Nate Eldredge
    Jul 31 '15 at 14:11














  • 3




    $begingroup$
    The ideas in my answer here (math.stackexchange.com/questions/1068043/…) should be helpful.
    $endgroup$
    – PhoemueX
    Jul 31 '15 at 10:10






  • 1




    $begingroup$
    @PhoemueX thanks tried your other answer but I still cannot understand it
    $endgroup$
    – Don John Prep
    Jul 31 '15 at 11:09






  • 1




    $begingroup$
    If you start with the stronger assumption that $mu$ is a finite measure, can you see how the assertion is obtained from Egorov's theorem?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:21






  • 2




    $begingroup$
    Let $F = bigcup E_n$. Then you have $Xsetminus F subset X setminus E_k$, and hence $mu(Xsetminus F) < 1/k$. Since that holds for all $k$, $mu(Xsetminus F) = 0$. Okay, the case of finite measure done. Now, for $sigma$-finite measure, write $X = bigcup A_m$, where each $A_m$ has finite measure. Perhaps it makes things easier to assume the $A_m$ disjoint, one can do that. So by the preceding, for each $m$, we have a sequence $E^{(m)}_n$ of subsets of $A_m$. How can you use that to get the desired sequence of subsets of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jul 31 '15 at 13:48






  • 3




    $begingroup$
    @AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution.
    $endgroup$
    – Nate Eldredge
    Jul 31 '15 at 14:11








3




3




$begingroup$
The ideas in my answer here (math.stackexchange.com/questions/1068043/…) should be helpful.
$endgroup$
– PhoemueX
Jul 31 '15 at 10:10




$begingroup$
The ideas in my answer here (math.stackexchange.com/questions/1068043/…) should be helpful.
$endgroup$
– PhoemueX
Jul 31 '15 at 10:10




1




1




$begingroup$
@PhoemueX thanks tried your other answer but I still cannot understand it
$endgroup$
– Don John Prep
Jul 31 '15 at 11:09




$begingroup$
@PhoemueX thanks tried your other answer but I still cannot understand it
$endgroup$
– Don John Prep
Jul 31 '15 at 11:09




1




1




$begingroup$
If you start with the stronger assumption that $mu$ is a finite measure, can you see how the assertion is obtained from Egorov's theorem?
$endgroup$
– Daniel Fischer♦
Jul 31 '15 at 13:21




$begingroup$
If you start with the stronger assumption that $mu$ is a finite measure, can you see how the assertion is obtained from Egorov's theorem?
$endgroup$
– Daniel Fischer♦
Jul 31 '15 at 13:21




2




2




$begingroup$
Let $F = bigcup E_n$. Then you have $Xsetminus F subset X setminus E_k$, and hence $mu(Xsetminus F) < 1/k$. Since that holds for all $k$, $mu(Xsetminus F) = 0$. Okay, the case of finite measure done. Now, for $sigma$-finite measure, write $X = bigcup A_m$, where each $A_m$ has finite measure. Perhaps it makes things easier to assume the $A_m$ disjoint, one can do that. So by the preceding, for each $m$, we have a sequence $E^{(m)}_n$ of subsets of $A_m$. How can you use that to get the desired sequence of subsets of $X$?
$endgroup$
– Daniel Fischer♦
Jul 31 '15 at 13:48




$begingroup$
Let $F = bigcup E_n$. Then you have $Xsetminus F subset X setminus E_k$, and hence $mu(Xsetminus F) < 1/k$. Since that holds for all $k$, $mu(Xsetminus F) = 0$. Okay, the case of finite measure done. Now, for $sigma$-finite measure, write $X = bigcup A_m$, where each $A_m$ has finite measure. Perhaps it makes things easier to assume the $A_m$ disjoint, one can do that. So by the preceding, for each $m$, we have a sequence $E^{(m)}_n$ of subsets of $A_m$. How can you use that to get the desired sequence of subsets of $X$?
$endgroup$
– Daniel Fischer♦
Jul 31 '15 at 13:48




3




3




$begingroup$
@AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution.
$endgroup$
– Nate Eldredge
Jul 31 '15 at 14:11




$begingroup$
@AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution.
$endgroup$
– Nate Eldredge
Jul 31 '15 at 14:11










1 Answer
1






active

oldest

votes


















1












$begingroup$

Since $mu$ is a $sigma-$finite measure, there exists ${A_n}_{nin mathbb{N}}$ such that $bigcup_{n}A_n=X$ and $mu(A_n)<infty$ $forall n$.



Let $epsilon>0$ be given. Note that $f_nrightarrow f$ a.e on each $A_n$. Therefore, by Egoroff's theorem, there exists $E_nsubset A_n$ such that $mu(A_nsetminus E_n)<frac{epsilon}{2^n}$ and $f_n rightarrow f$ uniformly on $E_n$ for each $nin mathbb{N}$.



Now, observe that



begin{align*}
mu(left(bigcup_{n}E_n right)^c)&= muleft( bigcup_n A_n setminus bigcup_n E_n right)\ &= muleft(bigcup_{n}(A_nsetminus E_n)right)\&leq sum_n mu(A_nsetminus E_n)\&<sum_nfrac{epsilon}{2^n}=epsilon hspace{1 in}forall nin mathbb{N}.
end{align*}



Since $epsilon >0$ is arbitrary, $mu(left(bigcup_{n}E_n right)^c)=0$.



Since $f_nrightarrow f$ uniformly on each $E_n$ by construction, we are done.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Hi, when you write $mu(E_n^c) < frac{1}{n}$ you actually mean $mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $sigma- $ finite case by using your construction and using a standard $frac{epsilon}{2^n}$ argument.
    $endgroup$
    – Soham
    Dec 11 '18 at 5:36










  • $begingroup$
    @LucyferZedd I did not realize that! I fixed it :) Thank you!
    $endgroup$
    – Lev Ban
    Dec 14 '18 at 16:06











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1 Answer
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1 Answer
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oldest

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active

oldest

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1












$begingroup$

Since $mu$ is a $sigma-$finite measure, there exists ${A_n}_{nin mathbb{N}}$ such that $bigcup_{n}A_n=X$ and $mu(A_n)<infty$ $forall n$.



Let $epsilon>0$ be given. Note that $f_nrightarrow f$ a.e on each $A_n$. Therefore, by Egoroff's theorem, there exists $E_nsubset A_n$ such that $mu(A_nsetminus E_n)<frac{epsilon}{2^n}$ and $f_n rightarrow f$ uniformly on $E_n$ for each $nin mathbb{N}$.



Now, observe that



begin{align*}
mu(left(bigcup_{n}E_n right)^c)&= muleft( bigcup_n A_n setminus bigcup_n E_n right)\ &= muleft(bigcup_{n}(A_nsetminus E_n)right)\&leq sum_n mu(A_nsetminus E_n)\&<sum_nfrac{epsilon}{2^n}=epsilon hspace{1 in}forall nin mathbb{N}.
end{align*}



Since $epsilon >0$ is arbitrary, $mu(left(bigcup_{n}E_n right)^c)=0$.



Since $f_nrightarrow f$ uniformly on each $E_n$ by construction, we are done.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Hi, when you write $mu(E_n^c) < frac{1}{n}$ you actually mean $mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $sigma- $ finite case by using your construction and using a standard $frac{epsilon}{2^n}$ argument.
    $endgroup$
    – Soham
    Dec 11 '18 at 5:36










  • $begingroup$
    @LucyferZedd I did not realize that! I fixed it :) Thank you!
    $endgroup$
    – Lev Ban
    Dec 14 '18 at 16:06
















1












$begingroup$

Since $mu$ is a $sigma-$finite measure, there exists ${A_n}_{nin mathbb{N}}$ such that $bigcup_{n}A_n=X$ and $mu(A_n)<infty$ $forall n$.



Let $epsilon>0$ be given. Note that $f_nrightarrow f$ a.e on each $A_n$. Therefore, by Egoroff's theorem, there exists $E_nsubset A_n$ such that $mu(A_nsetminus E_n)<frac{epsilon}{2^n}$ and $f_n rightarrow f$ uniformly on $E_n$ for each $nin mathbb{N}$.



Now, observe that



begin{align*}
mu(left(bigcup_{n}E_n right)^c)&= muleft( bigcup_n A_n setminus bigcup_n E_n right)\ &= muleft(bigcup_{n}(A_nsetminus E_n)right)\&leq sum_n mu(A_nsetminus E_n)\&<sum_nfrac{epsilon}{2^n}=epsilon hspace{1 in}forall nin mathbb{N}.
end{align*}



Since $epsilon >0$ is arbitrary, $mu(left(bigcup_{n}E_n right)^c)=0$.



Since $f_nrightarrow f$ uniformly on each $E_n$ by construction, we are done.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Hi, when you write $mu(E_n^c) < frac{1}{n}$ you actually mean $mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $sigma- $ finite case by using your construction and using a standard $frac{epsilon}{2^n}$ argument.
    $endgroup$
    – Soham
    Dec 11 '18 at 5:36










  • $begingroup$
    @LucyferZedd I did not realize that! I fixed it :) Thank you!
    $endgroup$
    – Lev Ban
    Dec 14 '18 at 16:06














1












1








1





$begingroup$

Since $mu$ is a $sigma-$finite measure, there exists ${A_n}_{nin mathbb{N}}$ such that $bigcup_{n}A_n=X$ and $mu(A_n)<infty$ $forall n$.



Let $epsilon>0$ be given. Note that $f_nrightarrow f$ a.e on each $A_n$. Therefore, by Egoroff's theorem, there exists $E_nsubset A_n$ such that $mu(A_nsetminus E_n)<frac{epsilon}{2^n}$ and $f_n rightarrow f$ uniformly on $E_n$ for each $nin mathbb{N}$.



Now, observe that



begin{align*}
mu(left(bigcup_{n}E_n right)^c)&= muleft( bigcup_n A_n setminus bigcup_n E_n right)\ &= muleft(bigcup_{n}(A_nsetminus E_n)right)\&leq sum_n mu(A_nsetminus E_n)\&<sum_nfrac{epsilon}{2^n}=epsilon hspace{1 in}forall nin mathbb{N}.
end{align*}



Since $epsilon >0$ is arbitrary, $mu(left(bigcup_{n}E_n right)^c)=0$.



Since $f_nrightarrow f$ uniformly on each $E_n$ by construction, we are done.






share|cite|improve this answer











$endgroup$



Since $mu$ is a $sigma-$finite measure, there exists ${A_n}_{nin mathbb{N}}$ such that $bigcup_{n}A_n=X$ and $mu(A_n)<infty$ $forall n$.



Let $epsilon>0$ be given. Note that $f_nrightarrow f$ a.e on each $A_n$. Therefore, by Egoroff's theorem, there exists $E_nsubset A_n$ such that $mu(A_nsetminus E_n)<frac{epsilon}{2^n}$ and $f_n rightarrow f$ uniformly on $E_n$ for each $nin mathbb{N}$.



Now, observe that



begin{align*}
mu(left(bigcup_{n}E_n right)^c)&= muleft( bigcup_n A_n setminus bigcup_n E_n right)\ &= muleft(bigcup_{n}(A_nsetminus E_n)right)\&leq sum_n mu(A_nsetminus E_n)\&<sum_nfrac{epsilon}{2^n}=epsilon hspace{1 in}forall nin mathbb{N}.
end{align*}



Since $epsilon >0$ is arbitrary, $mu(left(bigcup_{n}E_n right)^c)=0$.



Since $f_nrightarrow f$ uniformly on each $E_n$ by construction, we are done.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 16:05

























answered Aug 6 '18 at 14:31









Lev BanLev Ban

1,0671317




1,0671317








  • 1




    $begingroup$
    Hi, when you write $mu(E_n^c) < frac{1}{n}$ you actually mean $mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $sigma- $ finite case by using your construction and using a standard $frac{epsilon}{2^n}$ argument.
    $endgroup$
    – Soham
    Dec 11 '18 at 5:36










  • $begingroup$
    @LucyferZedd I did not realize that! I fixed it :) Thank you!
    $endgroup$
    – Lev Ban
    Dec 14 '18 at 16:06














  • 1




    $begingroup$
    Hi, when you write $mu(E_n^c) < frac{1}{n}$ you actually mean $mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $sigma- $ finite case by using your construction and using a standard $frac{epsilon}{2^n}$ argument.
    $endgroup$
    – Soham
    Dec 11 '18 at 5:36










  • $begingroup$
    @LucyferZedd I did not realize that! I fixed it :) Thank you!
    $endgroup$
    – Lev Ban
    Dec 14 '18 at 16:06








1




1




$begingroup$
Hi, when you write $mu(E_n^c) < frac{1}{n}$ you actually mean $mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $sigma- $ finite case by using your construction and using a standard $frac{epsilon}{2^n}$ argument.
$endgroup$
– Soham
Dec 11 '18 at 5:36




$begingroup$
Hi, when you write $mu(E_n^c) < frac{1}{n}$ you actually mean $mu(A_n - E_n)$. Thus your argument doesn't work as stated. But you can actually prove that Egoroff works for $sigma- $ finite case by using your construction and using a standard $frac{epsilon}{2^n}$ argument.
$endgroup$
– Soham
Dec 11 '18 at 5:36












$begingroup$
@LucyferZedd I did not realize that! I fixed it :) Thank you!
$endgroup$
– Lev Ban
Dec 14 '18 at 16:06




$begingroup$
@LucyferZedd I did not realize that! I fixed it :) Thank you!
$endgroup$
– Lev Ban
Dec 14 '18 at 16:06


















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Bundesstraße 106

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Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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