How to prove that a function similar to Thomae is integrable (using Advanced calculus only )
$begingroup$
$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$
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?
calculus real-analysis integration analysis
edited Nov 30 '18 at 10:24
Bernard
119k639112
asked Nov 30 '18 at 9:36
hopefullyhopefully
134112
$endgroup$
|
show 2 more comments
$begingroup$
$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$
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?
calculus real-analysis integration analysis
edited Nov 30 '18 at 10:24
Bernard
119k639112
asked Nov 30 '18 at 9:36
hopefullyhopefully
134112
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$
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?
calculus real-analysis integration analysis
edited Nov 30 '18 at 10:24
Bernard
119k639112
asked Nov 30 '18 at 9:36
hopefullyhopefully
134112
$endgroup$
$$ f(x) = left{ begin{array}{ll}
1 & { x= 1/n}, n in mathbb{N}, \
0 &text{otherwise}
end{array} right. $$
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?
calculus real-analysis integration analysis
calculus real-analysis integration analysis
edited Nov 30 '18 at 10:24
Bernard
119k639112
asked Nov 30 '18 at 9:36
hopefullyhopefully
134112
edited Nov 30 '18 at 10:24
Bernard
119k639112
asked Nov 30 '18 at 9:36
hopefullyhopefully
134112
edited Nov 30 '18 at 10:24
Bernard
119k639112
edited Nov 30 '18 at 10:24
Bernard
119k639112
edited Nov 30 '18 at 10:24
Bernard
119k639112
119k639112
asked Nov 30 '18 at 9:36
hopefullyhopefully
134112
asked Nov 30 '18 at 9:36
hopefullyhopefully
134112
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Nov 30 '18 at 11:14
xbhxbh
5,9031522
$endgroup$
$begingroup$
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
|
show 5 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Nov 30 '18 at 11:14
xbhxbh
5,9031522
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
answered Nov 30 '18 at 11:14
xbhxbh
5,9031522
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
answered Nov 30 '18 at 11:14
xbhxbh
5,9031522
$endgroup$
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.
answered Nov 30 '18 at 11:14
xbhxbh
5,9031522
edited Nov 30 '18 at 12:23
answered Nov 30 '18 at 11:14
xbhxbh
5,9031522
answered Nov 30 '18 at 11:14
xbhxbh
5,9031522
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
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$begingroup$
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