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Showing $lim_{n to infty} L(f,P_n,[a,b]) = L(f,[a,b])$, where $P_n$ is partition of $[a,b]$ into $2^n$...
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Suppose $f : [a,b] to mathbb{R}$ is bounded. With a partition $P$ of the form $a = x_0, dots ,x_n = b$ of $[a,b]$, the lower Riemann sum is $L(f,P,[a,b]) := sum_{i=1}^{n} (x_i - x_{i-1}) inf_{[x_{i-1},x_i]} f$. Then the lower Riemann integral is $L(f,[a,b]) := sup_{P} L(f,P,[a,b])$; that is, the lower Riemann integral is the supremum over all the lower Riemann sums. Define the sequence $L(f,P_n,[a,b])$, where $P_n$ is the partition of $[a,b]$ obtained by splitting $[a,b]$ up into $2^n$ intervals of equal size. I want to prove that $lim_{n to infty} L(f,P_n,[a,b]) = L(f,[a,b])$.
Here's my approach so far: Since the list defining partition $P_{n-1}$ is a sublist of the list defining the partition $P_n$, we have $L(f, P_{n-1}, [a,b]) leq L(f,P_n, [a,b])$. That is, the sequence $L(f,P_n,[a,b])$ is monotone non-decreasing. Since it also has an upper bound of $L(f,[a,b])$, it follows from the monotone convergence theorem that $L(f,P_n,[a,b])$ is a convergent sequence, and it converges to the least upper bound of its terms. I need to prove that this limit is equal to $L(f,[a,b])$. This is where I am getting stuck.
Let the limit of $L(f,P_n,[a,b])$ be $l$. It is immediate that $l leq L(f,[a,b])$, because the supremum of a subset is at most the supremum of the original set. So I need just to show that $L(f,[a,b]) leq l$ to complete the proof. It is equivalent to show that for all $epsilon >0$, we have $L(f,[a,b]) > l - epsilon$. To this end, let $epsilon > 0$. Since $L(f,[a,b])$ is the supremum of the lower Riemann sums, there exists a partition $P$ of $[a,b]$ such that $L(f,P,[a,b]) > L(f,[a,b]) - frac{epsilon}{2}$. If I can show that there exists $N in mathbb{N}$ such that $L(f,P_N, [a,b]) geq L(f,[a,b])$ and $|L(f,P_N,[a,b]) - l| < frac{epsilon}{2}$, then I am done, since I will have as a consequence that $|l - L(f,[a,b])| < epsilon$, by the triangle inequality.
Intuitively, I want to use the triangle inequality to show a connection between four 'things'. Firstly, I have the sequence $L(f,P_n,[a,b])$, which is increasing (or at least non-decreasing) to $L(f,[a,b])$ as $n$ gets big. I know I can approximate $L(f,[a,b])$, the second thing, with a margin $epsilon$ of error, by $L(f,P,[a,b])$, the third thing, for some partition $P$. Then I just want to show that if I go far enough into the sequence $L(f,P_n,[a,b])$, the terms eventually get at least as big as $L(f,P,[a,b])$. Once they are at that threshold, the terms are within an $epsilon$ margin of error to $L(f,P,[a,b])$, and using the triangle inequality to get an upper bound on the distance between $l$, the fourth thing, and $L(f,[a,b])$, I would be finished. But how do I do this?
real-analysis riemann-integration
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
$endgroup$
add a comment |
$begingroup$
Suppose $f : [a,b] to mathbb{R}$ is bounded. With a partition $P$ of the form $a = x_0, dots ,x_n = b$ of $[a,b]$, the lower Riemann sum is $L(f,P,[a,b]) := sum_{i=1}^{n} (x_i - x_{i-1}) inf_{[x_{i-1},x_i]} f$. Then the lower Riemann integral is $L(f,[a,b]) := sup_{P} L(f,P,[a,b])$; that is, the lower Riemann integral is the supremum over all the lower Riemann sums. Define the sequence $L(f,P_n,[a,b])$, where $P_n$ is the partition of $[a,b]$ obtained by splitting $[a,b]$ up into $2^n$ intervals of equal size. I want to prove that $lim_{n to infty} L(f,P_n,[a,b]) = L(f,[a,b])$.
Here's my approach so far: Since the list defining partition $P_{n-1}$ is a sublist of the list defining the partition $P_n$, we have $L(f, P_{n-1}, [a,b]) leq L(f,P_n, [a,b])$. That is, the sequence $L(f,P_n,[a,b])$ is monotone non-decreasing. Since it also has an upper bound of $L(f,[a,b])$, it follows from the monotone convergence theorem that $L(f,P_n,[a,b])$ is a convergent sequence, and it converges to the least upper bound of its terms. I need to prove that this limit is equal to $L(f,[a,b])$. This is where I am getting stuck.
Let the limit of $L(f,P_n,[a,b])$ be $l$. It is immediate that $l leq L(f,[a,b])$, because the supremum of a subset is at most the supremum of the original set. So I need just to show that $L(f,[a,b]) leq l$ to complete the proof. It is equivalent to show that for all $epsilon >0$, we have $L(f,[a,b]) > l - epsilon$. To this end, let $epsilon > 0$. Since $L(f,[a,b])$ is the supremum of the lower Riemann sums, there exists a partition $P$ of $[a,b]$ such that $L(f,P,[a,b]) > L(f,[a,b]) - frac{epsilon}{2}$. If I can show that there exists $N in mathbb{N}$ such that $L(f,P_N, [a,b]) geq L(f,[a,b])$ and $|L(f,P_N,[a,b]) - l| < frac{epsilon}{2}$, then I am done, since I will have as a consequence that $|l - L(f,[a,b])| < epsilon$, by the triangle inequality.
Intuitively, I want to use the triangle inequality to show a connection between four 'things'. Firstly, I have the sequence $L(f,P_n,[a,b])$, which is increasing (or at least non-decreasing) to $L(f,[a,b])$ as $n$ gets big. I know I can approximate $L(f,[a,b])$, the second thing, with a margin $epsilon$ of error, by $L(f,P,[a,b])$, the third thing, for some partition $P$. Then I just want to show that if I go far enough into the sequence $L(f,P_n,[a,b])$, the terms eventually get at least as big as $L(f,P,[a,b])$. Once they are at that threshold, the terms are within an $epsilon$ margin of error to $L(f,P,[a,b])$, and using the triangle inequality to get an upper bound on the distance between $l$, the fourth thing, and $L(f,[a,b])$, I would be finished. But how do I do this?
real-analysis riemann-integration
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
$endgroup$
add a comment |
$begingroup$
Suppose $f : [a,b] to mathbb{R}$ is bounded. With a partition $P$ of the form $a = x_0, dots ,x_n = b$ of $[a,b]$, the lower Riemann sum is $L(f,P,[a,b]) := sum_{i=1}^{n} (x_i - x_{i-1}) inf_{[x_{i-1},x_i]} f$. Then the lower Riemann integral is $L(f,[a,b]) := sup_{P} L(f,P,[a,b])$; that is, the lower Riemann integral is the supremum over all the lower Riemann sums. Define the sequence $L(f,P_n,[a,b])$, where $P_n$ is the partition of $[a,b]$ obtained by splitting $[a,b]$ up into $2^n$ intervals of equal size. I want to prove that $lim_{n to infty} L(f,P_n,[a,b]) = L(f,[a,b])$.
Here's my approach so far: Since the list defining partition $P_{n-1}$ is a sublist of the list defining the partition $P_n$, we have $L(f, P_{n-1}, [a,b]) leq L(f,P_n, [a,b])$. That is, the sequence $L(f,P_n,[a,b])$ is monotone non-decreasing. Since it also has an upper bound of $L(f,[a,b])$, it follows from the monotone convergence theorem that $L(f,P_n,[a,b])$ is a convergent sequence, and it converges to the least upper bound of its terms. I need to prove that this limit is equal to $L(f,[a,b])$. This is where I am getting stuck.
Let the limit of $L(f,P_n,[a,b])$ be $l$. It is immediate that $l leq L(f,[a,b])$, because the supremum of a subset is at most the supremum of the original set. So I need just to show that $L(f,[a,b]) leq l$ to complete the proof. It is equivalent to show that for all $epsilon >0$, we have $L(f,[a,b]) > l - epsilon$. To this end, let $epsilon > 0$. Since $L(f,[a,b])$ is the supremum of the lower Riemann sums, there exists a partition $P$ of $[a,b]$ such that $L(f,P,[a,b]) > L(f,[a,b]) - frac{epsilon}{2}$. If I can show that there exists $N in mathbb{N}$ such that $L(f,P_N, [a,b]) geq L(f,[a,b])$ and $|L(f,P_N,[a,b]) - l| < frac{epsilon}{2}$, then I am done, since I will have as a consequence that $|l - L(f,[a,b])| < epsilon$, by the triangle inequality.
Intuitively, I want to use the triangle inequality to show a connection between four 'things'. Firstly, I have the sequence $L(f,P_n,[a,b])$, which is increasing (or at least non-decreasing) to $L(f,[a,b])$ as $n$ gets big. I know I can approximate $L(f,[a,b])$, the second thing, with a margin $epsilon$ of error, by $L(f,P,[a,b])$, the third thing, for some partition $P$. Then I just want to show that if I go far enough into the sequence $L(f,P_n,[a,b])$, the terms eventually get at least as big as $L(f,P,[a,b])$. Once they are at that threshold, the terms are within an $epsilon$ margin of error to $L(f,P,[a,b])$, and using the triangle inequality to get an upper bound on the distance between $l$, the fourth thing, and $L(f,[a,b])$, I would be finished. But how do I do this?
real-analysis riemann-integration
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
$endgroup$
Suppose $f : [a,b] to mathbb{R}$ is bounded. With a partition $P$ of the form $a = x_0, dots ,x_n = b$ of $[a,b]$, the lower Riemann sum is $L(f,P,[a,b]) := sum_{i=1}^{n} (x_i - x_{i-1}) inf_{[x_{i-1},x_i]} f$. Then the lower Riemann integral is $L(f,[a,b]) := sup_{P} L(f,P,[a,b])$; that is, the lower Riemann integral is the supremum over all the lower Riemann sums. Define the sequence $L(f,P_n,[a,b])$, where $P_n$ is the partition of $[a,b]$ obtained by splitting $[a,b]$ up into $2^n$ intervals of equal size. I want to prove that $lim_{n to infty} L(f,P_n,[a,b]) = L(f,[a,b])$.
Here's my approach so far: Since the list defining partition $P_{n-1}$ is a sublist of the list defining the partition $P_n$, we have $L(f, P_{n-1}, [a,b]) leq L(f,P_n, [a,b])$. That is, the sequence $L(f,P_n,[a,b])$ is monotone non-decreasing. Since it also has an upper bound of $L(f,[a,b])$, it follows from the monotone convergence theorem that $L(f,P_n,[a,b])$ is a convergent sequence, and it converges to the least upper bound of its terms. I need to prove that this limit is equal to $L(f,[a,b])$. This is where I am getting stuck.
Let the limit of $L(f,P_n,[a,b])$ be $l$. It is immediate that $l leq L(f,[a,b])$, because the supremum of a subset is at most the supremum of the original set. So I need just to show that $L(f,[a,b]) leq l$ to complete the proof. It is equivalent to show that for all $epsilon >0$, we have $L(f,[a,b]) > l - epsilon$. To this end, let $epsilon > 0$. Since $L(f,[a,b])$ is the supremum of the lower Riemann sums, there exists a partition $P$ of $[a,b]$ such that $L(f,P,[a,b]) > L(f,[a,b]) - frac{epsilon}{2}$. If I can show that there exists $N in mathbb{N}$ such that $L(f,P_N, [a,b]) geq L(f,[a,b])$ and $|L(f,P_N,[a,b]) - l| < frac{epsilon}{2}$, then I am done, since I will have as a consequence that $|l - L(f,[a,b])| < epsilon$, by the triangle inequality.
Intuitively, I want to use the triangle inequality to show a connection between four 'things'. Firstly, I have the sequence $L(f,P_n,[a,b])$, which is increasing (or at least non-decreasing) to $L(f,[a,b])$ as $n$ gets big. I know I can approximate $L(f,[a,b])$, the second thing, with a margin $epsilon$ of error, by $L(f,P,[a,b])$, the third thing, for some partition $P$. Then I just want to show that if I go far enough into the sequence $L(f,P_n,[a,b])$, the terms eventually get at least as big as $L(f,P,[a,b])$. Once they are at that threshold, the terms are within an $epsilon$ margin of error to $L(f,P,[a,b])$, and using the triangle inequality to get an upper bound on the distance between $l$, the fourth thing, and $L(f,[a,b])$, I would be finished. But how do I do this?
real-analysis riemann-integration
real-analysis riemann-integration
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
edited Nov 29 '18 at 14:25
Brahadeesh
6,35442363
6,35442363
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
asked Nov 29 '18 at 10:33
FChaosiFChaosi
263
263
add a comment |
add a comment |
3 Answers
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The essential thing here is that $|P_n|to 0$.
Use the standard inequality $$L(f,P)-L(f,Q) le 2,M,p, |Q|$$ where $M$ is an upper bound of $|f|$ and $p$ is the number of partition points of $P$.
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
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Following up on your idea (which is good btw), I think you are only missing one small link in your chain of thought.
You are approximating $L(f, [a,b])$ by $L(f,P ,[a,b])$ and you'd like to approximate $L(f,P, [a,b])$ by $L(f, P_n, [a,b])$. To do this, assume first that your partition $P$ looks like $P : x_0=a < x_1 <...<x_k=b$ for some fixed k. Then note that if n is sufficiently large, the intervals in the partition $P_n$ are either contained in the intervals $[x_i, x_{i+1}]$ from P, or they overlap with two consecutive intervals of the form $[x_i, x_{i+1}]$, $[x_{i+1}, x_{i+2}]$ from P. But(here is the key!) note that the overlaps are negligible(they can be made arbitrarily small) because f is bounded and because the lengths of these overlaps go to 0. Also note that if an interval from the $P_n$ partition is contained in an interval from $P$ its infimum is larger.
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
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Indeed, I was struggling to express concisely why the overlaps must be negligible. A clear answer. +1
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– Brahadeesh
Nov 29 '18 at 14:22
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Let $P = { t_0,dots,t_n }$ be a partition of $[a,b]$, that is, $a = t_0 < t_1 < dots < t_n = b$. Let $N$ be a sufficiently large integer; in particular, let $2^{-(N+1)} > d$, where $d = max { t_i - t_{i-1} : 1 leq i leq n }$. Let $P_N = { s_0,dots,s_{2^N} }$.
If for each $i$ we had that $t_i = s_{k}$ for some $k = k(i)$, then $P_N$ would be a refinement of $P$ and we could easily compare the the summations of $L(f,P,[a,b])$ and $L(f,P_N,[a,b])$ to conclude that $L(f,P_N,[a,b]) > L(f,P,[a,b])$.
The difficulty arises when $t_i in (s_{k(i) - 1}, s_{k(i)})$ for some $k(i)$. In this case, what we can show is that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$, where $mathcal{E}_N to 0$ as $N to infty$. This will imply that $L(f,P,[a,b]) leq l$, so if we had chosen the partition $P$ so that $L(f,P,[a,b]) > L(f,[a,b]) - epsilon$, then we have $L(f,[a,b]) < l + epsilon$ as desired.
The way to show that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$ is to compare the partition in $P_N$ lying between $t_i$ and $t_{i+1}$ and then dealing with the end points separately. The complete solution is given below.
Let $m_j = inf { f(x) : s_{j-1} leq x leq s_{j} }$ and $tilde{m}_j = inf { f(x) : t_{j-1} leq x leq t_j }$. Let us start by comparing $$(t_i - t_{i-1}) tilde{m}_i$$ with $$sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j.$$
Then, we have
$$
(t_i - t_{i-1}) = (s_{k(i-1)} - t_{i-1}) + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1}) right) + (t_i - s_{k(i)-1}).
$$
Then,
$$
begin{align*}
(t_i - t_{i-1})tilde{m}_i &= (s_{k(i-1)} - t_{i-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})tilde{m}_i right) + (t_i - s_{k(i)-1})tilde{m}_i\
&leq (s_{k(i-1)} - s_{k(i-1)-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})m_j right) + (s_{k(i)} - s_{k(i)-1})tilde{m}_i\
end{align*}
$$
Now, in the first terms on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i-1)} + m_{k(i-1)})$. Similarly, in the third term on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i)} + m_{k(i)})$. Expand the brackets appropriately, collect the terms together appropriately, and use $s_j - s_{j-1} = 2^{-N}$ to get
$$
(t_i - t_{i-1})tilde{m}_i leq sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j
+ frac{tilde{m}_i - m_{k(i-1)}}{2^N} + frac{tilde{m}_i - m_{k(i)}}{2^N}.
$$
So, by summing over all $0 leq i leq n$, we get
$$
L(f,P,[a,b]) leq L(f,P_N,[a,b]) + 2 sum_{i = 0}^n frac{|m_{k(i)} - tilde{m}_i|}{2^N}.
$$
Since $f$ is bounded, $f(x) leq B$ for all $x in [a,b]$ for some $B > 0$. Thus, the error is no greater than $4nB/2^N$, which tends to $0$ as $N$ tends to infinity.
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
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3 Answers
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3 Answers
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$begingroup$
The essential thing here is that $|P_n|to 0$.
Use the standard inequality $$L(f,P)-L(f,Q) le 2,M,p, |Q|$$ where $M$ is an upper bound of $|f|$ and $p$ is the number of partition points of $P$.
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
3,2152719
$endgroup$
add a comment |
$begingroup$
The essential thing here is that $|P_n|to 0$.
Use the standard inequality $$L(f,P)-L(f,Q) le 2,M,p, |Q|$$ where $M$ is an upper bound of $|f|$ and $p$ is the number of partition points of $P$.
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
3,2152719
$endgroup$
add a comment |
$begingroup$
The essential thing here is that $|P_n|to 0$.
Use the standard inequality $$L(f,P)-L(f,Q) le 2,M,p, |Q|$$ where $M$ is an upper bound of $|f|$ and $p$ is the number of partition points of $P$.
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
3,2152719
$endgroup$
The essential thing here is that $|P_n|to 0$.
Use the standard inequality $$L(f,P)-L(f,Q) le 2,M,p, |Q|$$ where $M$ is an upper bound of $|f|$ and $p$ is the number of partition points of $P$.
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
3,2152719
edited Jan 12 at 23:40
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
3,2152719
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
3,2152719
answered Dec 16 '18 at 18:45
Tony PiccoloTony Piccolo
3,2152719
3,2152719
add a comment |
add a comment |
$begingroup$
Following up on your idea (which is good btw), I think you are only missing one small link in your chain of thought.
You are approximating $L(f, [a,b])$ by $L(f,P ,[a,b])$ and you'd like to approximate $L(f,P, [a,b])$ by $L(f, P_n, [a,b])$. To do this, assume first that your partition $P$ looks like $P : x_0=a < x_1 <...<x_k=b$ for some fixed k. Then note that if n is sufficiently large, the intervals in the partition $P_n$ are either contained in the intervals $[x_i, x_{i+1}]$ from P, or they overlap with two consecutive intervals of the form $[x_i, x_{i+1}]$, $[x_{i+1}, x_{i+2}]$ from P. But(here is the key!) note that the overlaps are negligible(they can be made arbitrarily small) because f is bounded and because the lengths of these overlaps go to 0. Also note that if an interval from the $P_n$ partition is contained in an interval from $P$ its infimum is larger.
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
$endgroup$
$begingroup$
Indeed, I was struggling to express concisely why the overlaps must be negligible. A clear answer. +1
$endgroup$
– Brahadeesh
Nov 29 '18 at 14:22
add a comment |
$begingroup$
Following up on your idea (which is good btw), I think you are only missing one small link in your chain of thought.
You are approximating $L(f, [a,b])$ by $L(f,P ,[a,b])$ and you'd like to approximate $L(f,P, [a,b])$ by $L(f, P_n, [a,b])$. To do this, assume first that your partition $P$ looks like $P : x_0=a < x_1 <...<x_k=b$ for some fixed k. Then note that if n is sufficiently large, the intervals in the partition $P_n$ are either contained in the intervals $[x_i, x_{i+1}]$ from P, or they overlap with two consecutive intervals of the form $[x_i, x_{i+1}]$, $[x_{i+1}, x_{i+2}]$ from P. But(here is the key!) note that the overlaps are negligible(they can be made arbitrarily small) because f is bounded and because the lengths of these overlaps go to 0. Also note that if an interval from the $P_n$ partition is contained in an interval from $P$ its infimum is larger.
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
$endgroup$
$begingroup$
Indeed, I was struggling to express concisely why the overlaps must be negligible. A clear answer. +1
$endgroup$
– Brahadeesh
Nov 29 '18 at 14:22
add a comment |
$begingroup$
Following up on your idea (which is good btw), I think you are only missing one small link in your chain of thought.
You are approximating $L(f, [a,b])$ by $L(f,P ,[a,b])$ and you'd like to approximate $L(f,P, [a,b])$ by $L(f, P_n, [a,b])$. To do this, assume first that your partition $P$ looks like $P : x_0=a < x_1 <...<x_k=b$ for some fixed k. Then note that if n is sufficiently large, the intervals in the partition $P_n$ are either contained in the intervals $[x_i, x_{i+1}]$ from P, or they overlap with two consecutive intervals of the form $[x_i, x_{i+1}]$, $[x_{i+1}, x_{i+2}]$ from P. But(here is the key!) note that the overlaps are negligible(they can be made arbitrarily small) because f is bounded and because the lengths of these overlaps go to 0. Also note that if an interval from the $P_n$ partition is contained in an interval from $P$ its infimum is larger.
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
$endgroup$
Following up on your idea (which is good btw), I think you are only missing one small link in your chain of thought.
You are approximating $L(f, [a,b])$ by $L(f,P ,[a,b])$ and you'd like to approximate $L(f,P, [a,b])$ by $L(f, P_n, [a,b])$. To do this, assume first that your partition $P$ looks like $P : x_0=a < x_1 <...<x_k=b$ for some fixed k. Then note that if n is sufficiently large, the intervals in the partition $P_n$ are either contained in the intervals $[x_i, x_{i+1}]$ from P, or they overlap with two consecutive intervals of the form $[x_i, x_{i+1}]$, $[x_{i+1}, x_{i+2}]$ from P. But(here is the key!) note that the overlaps are negligible(they can be made arbitrarily small) because f is bounded and because the lengths of these overlaps go to 0. Also note that if an interval from the $P_n$ partition is contained in an interval from $P$ its infimum is larger.
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
answered Nov 29 '18 at 13:55
Sorin TircSorin Tirc
1,755213
1,755213
$begingroup$
Indeed, I was struggling to express concisely why the overlaps must be negligible. A clear answer. +1
$endgroup$
– Brahadeesh
Nov 29 '18 at 14:22
add a comment |
$begingroup$
Indeed, I was struggling to express concisely why the overlaps must be negligible. A clear answer. +1
$endgroup$
– Brahadeesh
Nov 29 '18 at 14:22
$begingroup$
Indeed, I was struggling to express concisely why the overlaps must be negligible. A clear answer. +1
$endgroup$
– Brahadeesh
Nov 29 '18 at 14:22
$begingroup$
Indeed, I was struggling to express concisely why the overlaps must be negligible. A clear answer. +1
$endgroup$
– Brahadeesh
Nov 29 '18 at 14:22
add a comment |
$begingroup$
Let $P = { t_0,dots,t_n }$ be a partition of $[a,b]$, that is, $a = t_0 < t_1 < dots < t_n = b$. Let $N$ be a sufficiently large integer; in particular, let $2^{-(N+1)} > d$, where $d = max { t_i - t_{i-1} : 1 leq i leq n }$. Let $P_N = { s_0,dots,s_{2^N} }$.
If for each $i$ we had that $t_i = s_{k}$ for some $k = k(i)$, then $P_N$ would be a refinement of $P$ and we could easily compare the the summations of $L(f,P,[a,b])$ and $L(f,P_N,[a,b])$ to conclude that $L(f,P_N,[a,b]) > L(f,P,[a,b])$.
The difficulty arises when $t_i in (s_{k(i) - 1}, s_{k(i)})$ for some $k(i)$. In this case, what we can show is that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$, where $mathcal{E}_N to 0$ as $N to infty$. This will imply that $L(f,P,[a,b]) leq l$, so if we had chosen the partition $P$ so that $L(f,P,[a,b]) > L(f,[a,b]) - epsilon$, then we have $L(f,[a,b]) < l + epsilon$ as desired.
The way to show that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$ is to compare the partition in $P_N$ lying between $t_i$ and $t_{i+1}$ and then dealing with the end points separately. The complete solution is given below.
Let $m_j = inf { f(x) : s_{j-1} leq x leq s_{j} }$ and $tilde{m}_j = inf { f(x) : t_{j-1} leq x leq t_j }$. Let us start by comparing $$(t_i - t_{i-1}) tilde{m}_i$$ with $$sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j.$$
Then, we have
$$
(t_i - t_{i-1}) = (s_{k(i-1)} - t_{i-1}) + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1}) right) + (t_i - s_{k(i)-1}).
$$
Then,
$$
begin{align*}
(t_i - t_{i-1})tilde{m}_i &= (s_{k(i-1)} - t_{i-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})tilde{m}_i right) + (t_i - s_{k(i)-1})tilde{m}_i\
&leq (s_{k(i-1)} - s_{k(i-1)-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})m_j right) + (s_{k(i)} - s_{k(i)-1})tilde{m}_i\
end{align*}
$$
Now, in the first terms on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i-1)} + m_{k(i-1)})$. Similarly, in the third term on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i)} + m_{k(i)})$. Expand the brackets appropriately, collect the terms together appropriately, and use $s_j - s_{j-1} = 2^{-N}$ to get
$$
(t_i - t_{i-1})tilde{m}_i leq sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j
+ frac{tilde{m}_i - m_{k(i-1)}}{2^N} + frac{tilde{m}_i - m_{k(i)}}{2^N}.
$$
So, by summing over all $0 leq i leq n$, we get
$$
L(f,P,[a,b]) leq L(f,P_N,[a,b]) + 2 sum_{i = 0}^n frac{|m_{k(i)} - tilde{m}_i|}{2^N}.
$$
Since $f$ is bounded, $f(x) leq B$ for all $x in [a,b]$ for some $B > 0$. Thus, the error is no greater than $4nB/2^N$, which tends to $0$ as $N$ tends to infinity.
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
$endgroup$
add a comment |
$begingroup$
Let $P = { t_0,dots,t_n }$ be a partition of $[a,b]$, that is, $a = t_0 < t_1 < dots < t_n = b$. Let $N$ be a sufficiently large integer; in particular, let $2^{-(N+1)} > d$, where $d = max { t_i - t_{i-1} : 1 leq i leq n }$. Let $P_N = { s_0,dots,s_{2^N} }$.
If for each $i$ we had that $t_i = s_{k}$ for some $k = k(i)$, then $P_N$ would be a refinement of $P$ and we could easily compare the the summations of $L(f,P,[a,b])$ and $L(f,P_N,[a,b])$ to conclude that $L(f,P_N,[a,b]) > L(f,P,[a,b])$.
The difficulty arises when $t_i in (s_{k(i) - 1}, s_{k(i)})$ for some $k(i)$. In this case, what we can show is that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$, where $mathcal{E}_N to 0$ as $N to infty$. This will imply that $L(f,P,[a,b]) leq l$, so if we had chosen the partition $P$ so that $L(f,P,[a,b]) > L(f,[a,b]) - epsilon$, then we have $L(f,[a,b]) < l + epsilon$ as desired.
The way to show that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$ is to compare the partition in $P_N$ lying between $t_i$ and $t_{i+1}$ and then dealing with the end points separately. The complete solution is given below.
Let $m_j = inf { f(x) : s_{j-1} leq x leq s_{j} }$ and $tilde{m}_j = inf { f(x) : t_{j-1} leq x leq t_j }$. Let us start by comparing $$(t_i - t_{i-1}) tilde{m}_i$$ with $$sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j.$$
Then, we have
$$
(t_i - t_{i-1}) = (s_{k(i-1)} - t_{i-1}) + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1}) right) + (t_i - s_{k(i)-1}).
$$
Then,
$$
begin{align*}
(t_i - t_{i-1})tilde{m}_i &= (s_{k(i-1)} - t_{i-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})tilde{m}_i right) + (t_i - s_{k(i)-1})tilde{m}_i\
&leq (s_{k(i-1)} - s_{k(i-1)-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})m_j right) + (s_{k(i)} - s_{k(i)-1})tilde{m}_i\
end{align*}
$$
Now, in the first terms on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i-1)} + m_{k(i-1)})$. Similarly, in the third term on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i)} + m_{k(i)})$. Expand the brackets appropriately, collect the terms together appropriately, and use $s_j - s_{j-1} = 2^{-N}$ to get
$$
(t_i - t_{i-1})tilde{m}_i leq sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j
+ frac{tilde{m}_i - m_{k(i-1)}}{2^N} + frac{tilde{m}_i - m_{k(i)}}{2^N}.
$$
So, by summing over all $0 leq i leq n$, we get
$$
L(f,P,[a,b]) leq L(f,P_N,[a,b]) + 2 sum_{i = 0}^n frac{|m_{k(i)} - tilde{m}_i|}{2^N}.
$$
Since $f$ is bounded, $f(x) leq B$ for all $x in [a,b]$ for some $B > 0$. Thus, the error is no greater than $4nB/2^N$, which tends to $0$ as $N$ tends to infinity.
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
$endgroup$
add a comment |
$begingroup$
Let $P = { t_0,dots,t_n }$ be a partition of $[a,b]$, that is, $a = t_0 < t_1 < dots < t_n = b$. Let $N$ be a sufficiently large integer; in particular, let $2^{-(N+1)} > d$, where $d = max { t_i - t_{i-1} : 1 leq i leq n }$. Let $P_N = { s_0,dots,s_{2^N} }$.
If for each $i$ we had that $t_i = s_{k}$ for some $k = k(i)$, then $P_N$ would be a refinement of $P$ and we could easily compare the the summations of $L(f,P,[a,b])$ and $L(f,P_N,[a,b])$ to conclude that $L(f,P_N,[a,b]) > L(f,P,[a,b])$.
The difficulty arises when $t_i in (s_{k(i) - 1}, s_{k(i)})$ for some $k(i)$. In this case, what we can show is that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$, where $mathcal{E}_N to 0$ as $N to infty$. This will imply that $L(f,P,[a,b]) leq l$, so if we had chosen the partition $P$ so that $L(f,P,[a,b]) > L(f,[a,b]) - epsilon$, then we have $L(f,[a,b]) < l + epsilon$ as desired.
The way to show that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$ is to compare the partition in $P_N$ lying between $t_i$ and $t_{i+1}$ and then dealing with the end points separately. The complete solution is given below.
Let $m_j = inf { f(x) : s_{j-1} leq x leq s_{j} }$ and $tilde{m}_j = inf { f(x) : t_{j-1} leq x leq t_j }$. Let us start by comparing $$(t_i - t_{i-1}) tilde{m}_i$$ with $$sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j.$$
Then, we have
$$
(t_i - t_{i-1}) = (s_{k(i-1)} - t_{i-1}) + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1}) right) + (t_i - s_{k(i)-1}).
$$
Then,
$$
begin{align*}
(t_i - t_{i-1})tilde{m}_i &= (s_{k(i-1)} - t_{i-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})tilde{m}_i right) + (t_i - s_{k(i)-1})tilde{m}_i\
&leq (s_{k(i-1)} - s_{k(i-1)-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})m_j right) + (s_{k(i)} - s_{k(i)-1})tilde{m}_i\
end{align*}
$$
Now, in the first terms on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i-1)} + m_{k(i-1)})$. Similarly, in the third term on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i)} + m_{k(i)})$. Expand the brackets appropriately, collect the terms together appropriately, and use $s_j - s_{j-1} = 2^{-N}$ to get
$$
(t_i - t_{i-1})tilde{m}_i leq sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j
+ frac{tilde{m}_i - m_{k(i-1)}}{2^N} + frac{tilde{m}_i - m_{k(i)}}{2^N}.
$$
So, by summing over all $0 leq i leq n$, we get
$$
L(f,P,[a,b]) leq L(f,P_N,[a,b]) + 2 sum_{i = 0}^n frac{|m_{k(i)} - tilde{m}_i|}{2^N}.
$$
Since $f$ is bounded, $f(x) leq B$ for all $x in [a,b]$ for some $B > 0$. Thus, the error is no greater than $4nB/2^N$, which tends to $0$ as $N$ tends to infinity.
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
$endgroup$
Let $P = { t_0,dots,t_n }$ be a partition of $[a,b]$, that is, $a = t_0 < t_1 < dots < t_n = b$. Let $N$ be a sufficiently large integer; in particular, let $2^{-(N+1)} > d$, where $d = max { t_i - t_{i-1} : 1 leq i leq n }$. Let $P_N = { s_0,dots,s_{2^N} }$.
If for each $i$ we had that $t_i = s_{k}$ for some $k = k(i)$, then $P_N$ would be a refinement of $P$ and we could easily compare the the summations of $L(f,P,[a,b])$ and $L(f,P_N,[a,b])$ to conclude that $L(f,P_N,[a,b]) > L(f,P,[a,b])$.
The difficulty arises when $t_i in (s_{k(i) - 1}, s_{k(i)})$ for some $k(i)$. In this case, what we can show is that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$, where $mathcal{E}_N to 0$ as $N to infty$. This will imply that $L(f,P,[a,b]) leq l$, so if we had chosen the partition $P$ so that $L(f,P,[a,b]) > L(f,[a,b]) - epsilon$, then we have $L(f,[a,b]) < l + epsilon$ as desired.
The way to show that $L(f,P,[a,b]) leq L(f,P_N,[a,b]) + mathcal{E}_N$ is to compare the partition in $P_N$ lying between $t_i$ and $t_{i+1}$ and then dealing with the end points separately. The complete solution is given below.
Let $m_j = inf { f(x) : s_{j-1} leq x leq s_{j} }$ and $tilde{m}_j = inf { f(x) : t_{j-1} leq x leq t_j }$. Let us start by comparing $$(t_i - t_{i-1}) tilde{m}_i$$ with $$sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j.$$
Then, we have
$$
(t_i - t_{i-1}) = (s_{k(i-1)} - t_{i-1}) + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1}) right) + (t_i - s_{k(i)-1}).
$$
Then,
$$
begin{align*}
(t_i - t_{i-1})tilde{m}_i &= (s_{k(i-1)} - t_{i-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})tilde{m}_i right) + (t_i - s_{k(i)-1})tilde{m}_i\
&leq (s_{k(i-1)} - s_{k(i-1)-1})tilde{m}_i + left( sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})m_j right) + (s_{k(i)} - s_{k(i)-1})tilde{m}_i\
end{align*}
$$
Now, in the first terms on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i-1)} + m_{k(i-1)})$. Similarly, in the third term on the RHS replace $tilde{m}_i$ with $(tilde{m}_i - m_{k(i)} + m_{k(i)})$. Expand the brackets appropriately, collect the terms together appropriately, and use $s_j - s_{j-1} = 2^{-N}$ to get
$$
(t_i - t_{i-1})tilde{m}_i leq sum_{k(i-1) leq j leq k(i)} (s_j - s_{j-1}) m_j
+ frac{tilde{m}_i - m_{k(i-1)}}{2^N} + frac{tilde{m}_i - m_{k(i)}}{2^N}.
$$
So, by summing over all $0 leq i leq n$, we get
$$
L(f,P,[a,b]) leq L(f,P_N,[a,b]) + 2 sum_{i = 0}^n frac{|m_{k(i)} - tilde{m}_i|}{2^N}.
$$
Since $f$ is bounded, $f(x) leq B$ for all $x in [a,b]$ for some $B > 0$. Thus, the error is no greater than $4nB/2^N$, which tends to $0$ as $N$ tends to infinity.
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
answered Nov 29 '18 at 14:19
BrahadeeshBrahadeesh
6,35442363
6,35442363
add a comment |
add a comment |
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