How to prove that a function similar to Thomae is integrable (using Advanced calculus only )












0












$begingroup$


$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$



Prove that $f$ is integrable [Darboux] on [0,1].



I believe that the proof will be similar to this but not exactly, so could anyone tell me the differences please?




enter image description hereenter image description here







calculus real-analysis integration analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is your "Darboux" sum deals only with partitions with equal length of subintervals?
    $endgroup$
    – xbh
    Nov 30 '18 at 9:55










  • $begingroup$
    No it does not @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 9:56






  • 2




    $begingroup$
    This one is easier. Only one interval contains infinitely many points.
    $endgroup$
    – xbh
    Nov 30 '18 at 9:57










  • $begingroup$
    How to choose the partition here? @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 10:53






  • 1




    $begingroup$
    To be clear, for each $varepsilon >0$, we only need to find one partition s.t. $U(f,P)-L(f,P)< varepsilon $, correct? [Various text may use different theorems]
    $endgroup$
    – xbh
    Nov 30 '18 at 11:05
















0












$begingroup$


$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$



Prove that $f$ is integrable [Darboux] on [0,1].



I believe that the proof will be similar to this but not exactly, so could anyone tell me the differences please?




enter image description hereenter image description here







calculus real-analysis integration analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is your "Darboux" sum deals only with partitions with equal length of subintervals?
    $endgroup$
    – xbh
    Nov 30 '18 at 9:55










  • $begingroup$
    No it does not @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 9:56






  • 2




    $begingroup$
    This one is easier. Only one interval contains infinitely many points.
    $endgroup$
    – xbh
    Nov 30 '18 at 9:57










  • $begingroup$
    How to choose the partition here? @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 10:53






  • 1




    $begingroup$
    To be clear, for each $varepsilon >0$, we only need to find one partition s.t. $U(f,P)-L(f,P)< varepsilon $, correct? [Various text may use different theorems]
    $endgroup$
    – xbh
    Nov 30 '18 at 11:05














0












0








0





$begingroup$


$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$



Prove that $f$ is integrable [Darboux] on [0,1].



I believe that the proof will be similar to this but not exactly, so could anyone tell me the differences please?




enter image description hereenter image description here







calculus real-analysis integration analysis







share|cite|improve this question











$endgroup$




$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$



Prove that $f$ is integrable [Darboux] on [0,1].



I believe that the proof will be similar to this but not exactly, so could anyone tell me the differences please?




enter image description hereenter image description here







calculus real-analysis integration analysis




calculus real-analysis integration analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 10:24









Bernard

119k639112




119k639112










asked Nov 30 '18 at 9:36









hopefullyhopefully

134112




134112












  • $begingroup$
    Is your "Darboux" sum deals only with partitions with equal length of subintervals?
    $endgroup$
    – xbh
    Nov 30 '18 at 9:55










  • $begingroup$
    No it does not @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 9:56






  • 2




    $begingroup$
    This one is easier. Only one interval contains infinitely many points.
    $endgroup$
    – xbh
    Nov 30 '18 at 9:57










  • $begingroup$
    How to choose the partition here? @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 10:53






  • 1




    $begingroup$
    To be clear, for each $varepsilon >0$, we only need to find one partition s.t. $U(f,P)-L(f,P)< varepsilon $, correct? [Various text may use different theorems]
    $endgroup$
    – xbh
    Nov 30 '18 at 11:05


















  • $begingroup$
    Is your "Darboux" sum deals only with partitions with equal length of subintervals?
    $endgroup$
    – xbh
    Nov 30 '18 at 9:55










  • $begingroup$
    No it does not @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 9:56






  • 2




    $begingroup$
    This one is easier. Only one interval contains infinitely many points.
    $endgroup$
    – xbh
    Nov 30 '18 at 9:57










  • $begingroup$
    How to choose the partition here? @xbh
    $endgroup$
    – hopefully
    Nov 30 '18 at 10:53






  • 1




    $begingroup$
    To be clear, for each $varepsilon >0$, we only need to find one partition s.t. $U(f,P)-L(f,P)< varepsilon $, correct? [Various text may use different theorems]
    $endgroup$
    – xbh
    Nov 30 '18 at 11:05
















$begingroup$
Is your "Darboux" sum deals only with partitions with equal length of subintervals?
$endgroup$
– xbh
Nov 30 '18 at 9:55




$begingroup$
Is your "Darboux" sum deals only with partitions with equal length of subintervals?
$endgroup$
– xbh
Nov 30 '18 at 9:55












$begingroup$
No it does not @xbh
$endgroup$
– hopefully
Nov 30 '18 at 9:56




$begingroup$
No it does not @xbh
$endgroup$
– hopefully
Nov 30 '18 at 9:56




2




2




$begingroup$
This one is easier. Only one interval contains infinitely many points.
$endgroup$
– xbh
Nov 30 '18 at 9:57




$begingroup$
This one is easier. Only one interval contains infinitely many points.
$endgroup$
– xbh
Nov 30 '18 at 9:57












$begingroup$
How to choose the partition here? @xbh
$endgroup$
– hopefully
Nov 30 '18 at 10:53




$begingroup$
How to choose the partition here? @xbh
$endgroup$
– hopefully
Nov 30 '18 at 10:53




1




1




$begingroup$
To be clear, for each $varepsilon >0$, we only need to find one partition s.t. $U(f,P)-L(f,P)< varepsilon $, correct? [Various text may use different theorems]
$endgroup$
– xbh
Nov 30 '18 at 11:05




$begingroup$
To be clear, for each $varepsilon >0$, we only need to find one partition s.t. $U(f,P)-L(f,P)< varepsilon $, correct? [Various text may use different theorems]
$endgroup$
– xbh
Nov 30 '18 at 11:05










1 Answer
1






active

oldest

votes


















2












$begingroup$

For each $varepsilon >0$, there is some $Nin mathbb N^* colon N < 2/varepsilon$. Let $1/(N+1) leqslant x_1 < 1/N$, then $[x_1, 1]$ contains $S = {1/N, 1/(N-1), dots, 1/2, 1}$. Now use closed intervals with length $leqslant 1/N^2$ to cover each single point $1/j$ foe $j =1, dots, N$, and add other points if you like, then
$$
U(f,P) leqslant 1 (x_1 - 0) + 1 cdot frac 1{N^2} cdot N + 0cdot sum_{S cap [x_{j-1},x_j] =varnothing} (x_j - x_{j-1})leqslant frac 1{N+1} + frac 1N < frac 2N < varepsilon,
$$
where $P={0, x_1, cdots, 1}$ that contains the endpoints of the aforementioned intervals. Clearly $L(f,P) =0$, since every interval contains points not in $S$, so $U(f,P) - L(f,P) < varepsilon$.



Remark



We first notice that ${1/n}_1^infty$ converges to $0$, so we could choose $x_1$ s.t. $[0,x_1]$ covers infinitely many points of them. For the rest of them, each are "isolated", so we could use intervals with length as small as possible to cover them. Thus the upper sum w.r.t. such partition would be small enough as well.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    math.stackexchange.com/questions/3019732/… Can you look at this question for me please?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:21










  • $begingroup$
    this is the partition of the new problem not an explanation for the example correct?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:25






  • 1




    $begingroup$
    @hopefully Yes.
    $endgroup$
    – xbh
    Nov 30 '18 at 11:26










  • $begingroup$
    what do you mean by the statement "the rest of them are isolated"?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:41










  • $begingroup$
    $x_{1}$ is so strange for me, what values it can take?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:43











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

For each $varepsilon >0$, there is some $Nin mathbb N^* colon N < 2/varepsilon$. Let $1/(N+1) leqslant x_1 < 1/N$, then $[x_1, 1]$ contains $S = {1/N, 1/(N-1), dots, 1/2, 1}$. Now use closed intervals with length $leqslant 1/N^2$ to cover each single point $1/j$ foe $j =1, dots, N$, and add other points if you like, then
$$
U(f,P) leqslant 1 (x_1 - 0) + 1 cdot frac 1{N^2} cdot N + 0cdot sum_{S cap [x_{j-1},x_j] =varnothing} (x_j - x_{j-1})leqslant frac 1{N+1} + frac 1N < frac 2N < varepsilon,
$$
where $P={0, x_1, cdots, 1}$ that contains the endpoints of the aforementioned intervals. Clearly $L(f,P) =0$, since every interval contains points not in $S$, so $U(f,P) - L(f,P) < varepsilon$.



Remark



We first notice that ${1/n}_1^infty$ converges to $0$, so we could choose $x_1$ s.t. $[0,x_1]$ covers infinitely many points of them. For the rest of them, each are "isolated", so we could use intervals with length as small as possible to cover them. Thus the upper sum w.r.t. such partition would be small enough as well.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    math.stackexchange.com/questions/3019732/… Can you look at this question for me please?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:21










  • $begingroup$
    this is the partition of the new problem not an explanation for the example correct?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:25






  • 1




    $begingroup$
    @hopefully Yes.
    $endgroup$
    – xbh
    Nov 30 '18 at 11:26










  • $begingroup$
    what do you mean by the statement "the rest of them are isolated"?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:41










  • $begingroup$
    $x_{1}$ is so strange for me, what values it can take?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:43
















2












$begingroup$

For each $varepsilon >0$, there is some $Nin mathbb N^* colon N < 2/varepsilon$. Let $1/(N+1) leqslant x_1 < 1/N$, then $[x_1, 1]$ contains $S = {1/N, 1/(N-1), dots, 1/2, 1}$. Now use closed intervals with length $leqslant 1/N^2$ to cover each single point $1/j$ foe $j =1, dots, N$, and add other points if you like, then
$$
U(f,P) leqslant 1 (x_1 - 0) + 1 cdot frac 1{N^2} cdot N + 0cdot sum_{S cap [x_{j-1},x_j] =varnothing} (x_j - x_{j-1})leqslant frac 1{N+1} + frac 1N < frac 2N < varepsilon,
$$
where $P={0, x_1, cdots, 1}$ that contains the endpoints of the aforementioned intervals. Clearly $L(f,P) =0$, since every interval contains points not in $S$, so $U(f,P) - L(f,P) < varepsilon$.



Remark



We first notice that ${1/n}_1^infty$ converges to $0$, so we could choose $x_1$ s.t. $[0,x_1]$ covers infinitely many points of them. For the rest of them, each are "isolated", so we could use intervals with length as small as possible to cover them. Thus the upper sum w.r.t. such partition would be small enough as well.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    math.stackexchange.com/questions/3019732/… Can you look at this question for me please?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:21










  • $begingroup$
    this is the partition of the new problem not an explanation for the example correct?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:25






  • 1




    $begingroup$
    @hopefully Yes.
    $endgroup$
    – xbh
    Nov 30 '18 at 11:26










  • $begingroup$
    what do you mean by the statement "the rest of them are isolated"?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:41










  • $begingroup$
    $x_{1}$ is so strange for me, what values it can take?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:43














2












2








2





$begingroup$

For each $varepsilon >0$, there is some $Nin mathbb N^* colon N < 2/varepsilon$. Let $1/(N+1) leqslant x_1 < 1/N$, then $[x_1, 1]$ contains $S = {1/N, 1/(N-1), dots, 1/2, 1}$. Now use closed intervals with length $leqslant 1/N^2$ to cover each single point $1/j$ foe $j =1, dots, N$, and add other points if you like, then
$$
U(f,P) leqslant 1 (x_1 - 0) + 1 cdot frac 1{N^2} cdot N + 0cdot sum_{S cap [x_{j-1},x_j] =varnothing} (x_j - x_{j-1})leqslant frac 1{N+1} + frac 1N < frac 2N < varepsilon,
$$
where $P={0, x_1, cdots, 1}$ that contains the endpoints of the aforementioned intervals. Clearly $L(f,P) =0$, since every interval contains points not in $S$, so $U(f,P) - L(f,P) < varepsilon$.



Remark



We first notice that ${1/n}_1^infty$ converges to $0$, so we could choose $x_1$ s.t. $[0,x_1]$ covers infinitely many points of them. For the rest of them, each are "isolated", so we could use intervals with length as small as possible to cover them. Thus the upper sum w.r.t. such partition would be small enough as well.






share|cite|improve this answer











$endgroup$



For each $varepsilon >0$, there is some $Nin mathbb N^* colon N < 2/varepsilon$. Let $1/(N+1) leqslant x_1 < 1/N$, then $[x_1, 1]$ contains $S = {1/N, 1/(N-1), dots, 1/2, 1}$. Now use closed intervals with length $leqslant 1/N^2$ to cover each single point $1/j$ foe $j =1, dots, N$, and add other points if you like, then
$$
U(f,P) leqslant 1 (x_1 - 0) + 1 cdot frac 1{N^2} cdot N + 0cdot sum_{S cap [x_{j-1},x_j] =varnothing} (x_j - x_{j-1})leqslant frac 1{N+1} + frac 1N < frac 2N < varepsilon,
$$
where $P={0, x_1, cdots, 1}$ that contains the endpoints of the aforementioned intervals. Clearly $L(f,P) =0$, since every interval contains points not in $S$, so $U(f,P) - L(f,P) < varepsilon$.



Remark



We first notice that ${1/n}_1^infty$ converges to $0$, so we could choose $x_1$ s.t. $[0,x_1]$ covers infinitely many points of them. For the rest of them, each are "isolated", so we could use intervals with length as small as possible to cover them. Thus the upper sum w.r.t. such partition would be small enough as well.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 12:23

























answered Nov 30 '18 at 11:14









xbhxbh

5,9031522




5,9031522












  • $begingroup$
    math.stackexchange.com/questions/3019732/… Can you look at this question for me please?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:21










  • $begingroup$
    this is the partition of the new problem not an explanation for the example correct?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:25






  • 1




    $begingroup$
    @hopefully Yes.
    $endgroup$
    – xbh
    Nov 30 '18 at 11:26










  • $begingroup$
    what do you mean by the statement "the rest of them are isolated"?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:41










  • $begingroup$
    $x_{1}$ is so strange for me, what values it can take?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:43


















  • $begingroup$
    math.stackexchange.com/questions/3019732/… Can you look at this question for me please?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:21










  • $begingroup$
    this is the partition of the new problem not an explanation for the example correct?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:25






  • 1




    $begingroup$
    @hopefully Yes.
    $endgroup$
    – xbh
    Nov 30 '18 at 11:26










  • $begingroup$
    what do you mean by the statement "the rest of them are isolated"?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:41










  • $begingroup$
    $x_{1}$ is so strange for me, what values it can take?
    $endgroup$
    – hopefully
    Nov 30 '18 at 11:43
















$begingroup$
math.stackexchange.com/questions/3019732/… Can you look at this question for me please?
$endgroup$
– hopefully
Nov 30 '18 at 11:21




$begingroup$
math.stackexchange.com/questions/3019732/… Can you look at this question for me please?
$endgroup$
– hopefully
Nov 30 '18 at 11:21












$begingroup$
this is the partition of the new problem not an explanation for the example correct?
$endgroup$
– hopefully
Nov 30 '18 at 11:25




$begingroup$
this is the partition of the new problem not an explanation for the example correct?
$endgroup$
– hopefully
Nov 30 '18 at 11:25




1




1




$begingroup$
@hopefully Yes.
$endgroup$
– xbh
Nov 30 '18 at 11:26




$begingroup$
@hopefully Yes.
$endgroup$
– xbh
Nov 30 '18 at 11:26












$begingroup$
what do you mean by the statement "the rest of them are isolated"?
$endgroup$
– hopefully
Nov 30 '18 at 11:41




$begingroup$
what do you mean by the statement "the rest of them are isolated"?
$endgroup$
– hopefully
Nov 30 '18 at 11:41












$begingroup$
$x_{1}$ is so strange for me, what values it can take?
$endgroup$
– hopefully
Nov 30 '18 at 11:43




$begingroup$
$x_{1}$ is so strange for me, what values it can take?
$endgroup$
– hopefully
Nov 30 '18 at 11:43


















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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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