If $f$ be non-negative, bounded and Riemann Integrable on $[a, b]$, then $sqrt{f}$ is R- integrable.
The easiest way to solve this is to use the theorem: If $g$ be a continuous function in $[M, m] subset [j, k]$ and $f$ be bounded and Riemann integrable, then $g circ f$ is Riemann integrable too. Using $g(x) = x^{1/2}$ solves the problem, but I want to prove it more directly.
Here is my approach:
From the necessary and sufficient condition of integrability, we have
$forall epsilon>0$, $exists$ a $delta >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r=U(P,f)-L(P,f) < epsilon $ $ forall ||P||< delta$, $P$ being any partition of $[a,b]$.
We now create a set ( more precisely, a subset of the index set $I$ ), $Z$, defined as $Z={r in I: sup_{[x_{r-1},x_r]}f=sup_{[x_{r-1},x_r]}sqrt{f}=0}$. Since the function is non negative over the entire interval, therefore, in every interval indexed in $Z$ must have $0$ as the infimum.
By the property of non negative real numbers, $m_r leq f leq M_r implies sqrt{m_r} leq sqrt{f} leq sqrt{M_r}$
Now taking $W(P^*,sqrt{f})= sum_{rin IZ}(sqrt{M_r}-sqrt{m_r})delta_r =sum_{rin IZ}frac{({M_r}-{m_r})}{(sqrt{M_r}+sqrt{m_r})}delta_r leq musum_{rin IZ}({M_r}-{m_r})delta_r $
choosing $min {sqrt{M_r}+sqrt{m_r}: r in IZ}= K = 1/mu >0 (neq0$, by our choice of $r$) .
We further have $sqrt{M_r}-sqrt{m_r} = 0 =M_r -m_r forall rin Z.$ Hence $W(P^*,sqrt{f})= W(P,sqrt{f}) $
Now, $forall epsilon >0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r< epsilon/mu $ [$mu$ is a fixed positive real number] $ forall ||P||< delta_mu$, $P$ being any partition of $[a,b]$.
Thus, we finally have the following:
$forall epsilon>0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,sqrt{f}) leq mu W(P,f) < epsilon $, $forall ||P||< delta_mu.$
P.S : I am aware of the duplicates. I just want my solution verified.
real-analysis proof-verification alternative-proof riemann-integration
asked Dec 26 at 16:05
Subhasis Biswas
413211
add a comment |
The easiest way to solve this is to use the theorem: If $g$ be a continuous function in $[M, m] subset [j, k]$ and $f$ be bounded and Riemann integrable, then $g circ f$ is Riemann integrable too. Using $g(x) = x^{1/2}$ solves the problem, but I want to prove it more directly.
Here is my approach:
From the necessary and sufficient condition of integrability, we have
$forall epsilon>0$, $exists$ a $delta >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r=U(P,f)-L(P,f) < epsilon $ $ forall ||P||< delta$, $P$ being any partition of $[a,b]$.
We now create a set ( more precisely, a subset of the index set $I$ ), $Z$, defined as $Z={r in I: sup_{[x_{r-1},x_r]}f=sup_{[x_{r-1},x_r]}sqrt{f}=0}$. Since the function is non negative over the entire interval, therefore, in every interval indexed in $Z$ must have $0$ as the infimum.
By the property of non negative real numbers, $m_r leq f leq M_r implies sqrt{m_r} leq sqrt{f} leq sqrt{M_r}$
Now taking $W(P^*,sqrt{f})= sum_{rin IZ}(sqrt{M_r}-sqrt{m_r})delta_r =sum_{rin IZ}frac{({M_r}-{m_r})}{(sqrt{M_r}+sqrt{m_r})}delta_r leq musum_{rin IZ}({M_r}-{m_r})delta_r $
choosing $min {sqrt{M_r}+sqrt{m_r}: r in IZ}= K = 1/mu >0 (neq0$, by our choice of $r$) .
We further have $sqrt{M_r}-sqrt{m_r} = 0 =M_r -m_r forall rin Z.$ Hence $W(P^*,sqrt{f})= W(P,sqrt{f}) $
Now, $forall epsilon >0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r< epsilon/mu $ [$mu$ is a fixed positive real number] $ forall ||P||< delta_mu$, $P$ being any partition of $[a,b]$.
Thus, we finally have the following:
$forall epsilon>0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,sqrt{f}) leq mu W(P,f) < epsilon $, $forall ||P||< delta_mu.$
P.S : I am aware of the duplicates. I just want my solution verified.
real-analysis proof-verification alternative-proof riemann-integration
asked Dec 26 at 16:05
Subhasis Biswas
413211
It would have been pretty neat and clean if $f$ were $>0$. The problem arises when $0$ is in the range of $f$. This is when for some $i$, $epsilon / (sqrt{M_i}+sqrt{m_i})$ might become $infty$. I just have modified the partition index to remove the trouble. Otherwise, it's just the same thing.
– Subhasis Biswas
Dec 26 at 16:41
add a comment |
The easiest way to solve this is to use the theorem: If $g$ be a continuous function in $[M, m] subset [j, k]$ and $f$ be bounded and Riemann integrable, then $g circ f$ is Riemann integrable too. Using $g(x) = x^{1/2}$ solves the problem, but I want to prove it more directly.
Here is my approach:
From the necessary and sufficient condition of integrability, we have
$forall epsilon>0$, $exists$ a $delta >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r=U(P,f)-L(P,f) < epsilon $ $ forall ||P||< delta$, $P$ being any partition of $[a,b]$.
We now create a set ( more precisely, a subset of the index set $I$ ), $Z$, defined as $Z={r in I: sup_{[x_{r-1},x_r]}f=sup_{[x_{r-1},x_r]}sqrt{f}=0}$. Since the function is non negative over the entire interval, therefore, in every interval indexed in $Z$ must have $0$ as the infimum.
By the property of non negative real numbers, $m_r leq f leq M_r implies sqrt{m_r} leq sqrt{f} leq sqrt{M_r}$
Now taking $W(P^*,sqrt{f})= sum_{rin IZ}(sqrt{M_r}-sqrt{m_r})delta_r =sum_{rin IZ}frac{({M_r}-{m_r})}{(sqrt{M_r}+sqrt{m_r})}delta_r leq musum_{rin IZ}({M_r}-{m_r})delta_r $
choosing $min {sqrt{M_r}+sqrt{m_r}: r in IZ}= K = 1/mu >0 (neq0$, by our choice of $r$) .
We further have $sqrt{M_r}-sqrt{m_r} = 0 =M_r -m_r forall rin Z.$ Hence $W(P^*,sqrt{f})= W(P,sqrt{f}) $
Now, $forall epsilon >0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r< epsilon/mu $ [$mu$ is a fixed positive real number] $ forall ||P||< delta_mu$, $P$ being any partition of $[a,b]$.
Thus, we finally have the following:
$forall epsilon>0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,sqrt{f}) leq mu W(P,f) < epsilon $, $forall ||P||< delta_mu.$
P.S : I am aware of the duplicates. I just want my solution verified.
real-analysis proof-verification alternative-proof riemann-integration
asked Dec 26 at 16:05
Subhasis Biswas
413211
The easiest way to solve this is to use the theorem: If $g$ be a continuous function in $[M, m] subset [j, k]$ and $f$ be bounded and Riemann integrable, then $g circ f$ is Riemann integrable too. Using $g(x) = x^{1/2}$ solves the problem, but I want to prove it more directly.
Here is my approach:
From the necessary and sufficient condition of integrability, we have
$forall epsilon>0$, $exists$ a $delta >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r=U(P,f)-L(P,f) < epsilon $ $ forall ||P||< delta$, $P$ being any partition of $[a,b]$.
We now create a set ( more precisely, a subset of the index set $I$ ), $Z$, defined as $Z={r in I: sup_{[x_{r-1},x_r]}f=sup_{[x_{r-1},x_r]}sqrt{f}=0}$. Since the function is non negative over the entire interval, therefore, in every interval indexed in $Z$ must have $0$ as the infimum.
By the property of non negative real numbers, $m_r leq f leq M_r implies sqrt{m_r} leq sqrt{f} leq sqrt{M_r}$
Now taking $W(P^*,sqrt{f})= sum_{rin IZ}(sqrt{M_r}-sqrt{m_r})delta_r =sum_{rin IZ}frac{({M_r}-{m_r})}{(sqrt{M_r}+sqrt{m_r})}delta_r leq musum_{rin IZ}({M_r}-{m_r})delta_r $
choosing $min {sqrt{M_r}+sqrt{m_r}: r in IZ}= K = 1/mu >0 (neq0$, by our choice of $r$) .
We further have $sqrt{M_r}-sqrt{m_r} = 0 =M_r -m_r forall rin Z.$ Hence $W(P^*,sqrt{f})= W(P,sqrt{f}) $
Now, $forall epsilon >0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,f) = sum_{r=1}^{n} (M_r-m_r)delta_r< epsilon/mu $ [$mu$ is a fixed positive real number] $ forall ||P||< delta_mu$, $P$ being any partition of $[a,b]$.
Thus, we finally have the following:
$forall epsilon>0$, $exists$ a $delta_mu >0$ such that, $0 leq W(P,sqrt{f}) leq mu W(P,f) < epsilon $, $forall ||P||< delta_mu.$
P.S : I am aware of the duplicates. I just want my solution verified.
real-analysis proof-verification alternative-proof riemann-integration
real-analysis proof-verification alternative-proof riemann-integration
asked Dec 26 at 16:05
Subhasis Biswas
413211
asked Dec 26 at 16:05
Subhasis Biswas
413211
asked Dec 26 at 16:05
Subhasis Biswas
413211
asked Dec 26 at 16:05
Subhasis Biswas
413211
asked Dec 26 at 16:05
Subhasis Biswas
413211
413211
It would have been pretty neat and clean if $f$ were $>0$. The problem arises when $0$ is in the range of $f$. This is when for some $i$, $epsilon / (sqrt{M_i}+sqrt{m_i})$ might become $infty$. I just have modified the partition index to remove the trouble. Otherwise, it's just the same thing.
– Subhasis Biswas
Dec 26 at 16:41
add a comment |
It would have been pretty neat and clean if $f$ were $>0$. The problem arises when $0$ is in the range of $f$. This is when for some $i$, $epsilon / (sqrt{M_i}+sqrt{m_i})$ might become $infty$. I just have modified the partition index to remove the trouble. Otherwise, it's just the same thing.
– Subhasis Biswas
Dec 26 at 16:41
It would have been pretty neat and clean if $f$ were $>0$. The problem arises when $0$ is in the range of $f$. This is when for some $i$, $epsilon / (sqrt{M_i}+sqrt{m_i})$ might become $infty$. I just have modified the partition index to remove the trouble. Otherwise, it's just the same thing.
– Subhasis Biswas
Dec 26 at 16:41
It would have been pretty neat and clean if $f$ were $>0$. The problem arises when $0$ is in the range of $f$. This is when for some $i$, $epsilon / (sqrt{M_i}+sqrt{m_i})$ might become $infty$. I just have modified the partition index to remove the trouble. Otherwise, it's just the same thing.
– Subhasis Biswas
Dec 26 at 16:41
add a comment |
1 Answer
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Following your notation, note that we have
$$sqrt{M_i} - sqrt{m_i} leq sqrt {M_i-m_i}$$
since $sqrt{(M_i-m_i) + m_i} leq sqrt {M_i-m_i} + sqrt{m_i}$.
To bound the Riemann sum,
$$sum_i (sqrt{M_i} - sqrt{m_i}) delta_i leq sum_i sqrt {(M_i-m_i)delta_i} sqrt delta_i leq left( sum_i (M_i-m_i)delta_iright)^{1/2} left( sum_i delta_iright)^{1/2} $$
because $f$ is Riemann integrable, $sum_i (M_i-m_i)delta_i$ can be made arbritrily small, and $sum_i delta_i = b-a$ is bounded, we are done.
answered Dec 26 at 17:11
Xiao
4,72711434
Good one. But is mine alright?
– Subhasis Biswas
Dec 26 at 17:45
@SubhasisBiswas In your argument, does $mu$ (or the index set $Z$ and $I$) depend on the partition?
– Xiao
Dec 26 at 17:57
Everything you mentioned changes according to the mode of Partition. But ultimately, we can show that it can be made less than any predetermined positive quantity.
– Subhasis Biswas
Dec 26 at 18:00
1
@SubhasisBiswas In the second last line, you mentioned [$mu$ is a fixed constant], how did you pick this fixed constant? Also it would be hard to estimate "$mu sum_{iin Isetminus Z} (M_i - m_i) delta i$". Take $f(x) = x$ on $(0,1]$ and $f(0) = 1$, in this case $Z$ is empty. The quantity $mu sum_{iin I} (M_i - m_i) delta_i$ is hard to control as the partition size goes to zero, since $mu$ goes to infinity and $sum_{iin I} (M_i - m_i) delta i$ goes to zero.
– Xiao
Dec 26 at 18:38
unfortunately, I have to agree.
– Subhasis Biswas
Dec 26 at 19:25
|
show 1 more comment
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Following your notation, note that we have
$$sqrt{M_i} - sqrt{m_i} leq sqrt {M_i-m_i}$$
since $sqrt{(M_i-m_i) + m_i} leq sqrt {M_i-m_i} + sqrt{m_i}$.
To bound the Riemann sum,
$$sum_i (sqrt{M_i} - sqrt{m_i}) delta_i leq sum_i sqrt {(M_i-m_i)delta_i} sqrt delta_i leq left( sum_i (M_i-m_i)delta_iright)^{1/2} left( sum_i delta_iright)^{1/2} $$
because $f$ is Riemann integrable, $sum_i (M_i-m_i)delta_i$ can be made arbritrily small, and $sum_i delta_i = b-a$ is bounded, we are done.
answered Dec 26 at 17:11
Xiao
4,72711434
Good one. But is mine alright?
– Subhasis Biswas
Dec 26 at 17:45
@SubhasisBiswas In your argument, does $mu$ (or the index set $Z$ and $I$) depend on the partition?
– Xiao
Dec 26 at 17:57
Everything you mentioned changes according to the mode of Partition. But ultimately, we can show that it can be made less than any predetermined positive quantity.
– Subhasis Biswas
Dec 26 at 18:00
1
@SubhasisBiswas In the second last line, you mentioned [$mu$ is a fixed constant], how did you pick this fixed constant? Also it would be hard to estimate "$mu sum_{iin Isetminus Z} (M_i - m_i) delta i$". Take $f(x) = x$ on $(0,1]$ and $f(0) = 1$, in this case $Z$ is empty. The quantity $mu sum_{iin I} (M_i - m_i) delta_i$ is hard to control as the partition size goes to zero, since $mu$ goes to infinity and $sum_{iin I} (M_i - m_i) delta i$ goes to zero.
– Xiao
Dec 26 at 18:38
unfortunately, I have to agree.
– Subhasis Biswas
Dec 26 at 19:25
|
show 1 more comment
Following your notation, note that we have
$$sqrt{M_i} - sqrt{m_i} leq sqrt {M_i-m_i}$$
since $sqrt{(M_i-m_i) + m_i} leq sqrt {M_i-m_i} + sqrt{m_i}$.
To bound the Riemann sum,
$$sum_i (sqrt{M_i} - sqrt{m_i}) delta_i leq sum_i sqrt {(M_i-m_i)delta_i} sqrt delta_i leq left( sum_i (M_i-m_i)delta_iright)^{1/2} left( sum_i delta_iright)^{1/2} $$
because $f$ is Riemann integrable, $sum_i (M_i-m_i)delta_i$ can be made arbritrily small, and $sum_i delta_i = b-a$ is bounded, we are done.
answered Dec 26 at 17:11
Xiao
4,72711434
Good one. But is mine alright?
– Subhasis Biswas
Dec 26 at 17:45
@SubhasisBiswas In your argument, does $mu$ (or the index set $Z$ and $I$) depend on the partition?
– Xiao
Dec 26 at 17:57
Everything you mentioned changes according to the mode of Partition. But ultimately, we can show that it can be made less than any predetermined positive quantity.
– Subhasis Biswas
Dec 26 at 18:00
1
@SubhasisBiswas In the second last line, you mentioned [$mu$ is a fixed constant], how did you pick this fixed constant? Also it would be hard to estimate "$mu sum_{iin Isetminus Z} (M_i - m_i) delta i$". Take $f(x) = x$ on $(0,1]$ and $f(0) = 1$, in this case $Z$ is empty. The quantity $mu sum_{iin I} (M_i - m_i) delta_i$ is hard to control as the partition size goes to zero, since $mu$ goes to infinity and $sum_{iin I} (M_i - m_i) delta i$ goes to zero.
– Xiao
Dec 26 at 18:38
unfortunately, I have to agree.
– Subhasis Biswas
Dec 26 at 19:25
|
show 1 more comment
Following your notation, note that we have
$$sqrt{M_i} - sqrt{m_i} leq sqrt {M_i-m_i}$$
since $sqrt{(M_i-m_i) + m_i} leq sqrt {M_i-m_i} + sqrt{m_i}$.
To bound the Riemann sum,
$$sum_i (sqrt{M_i} - sqrt{m_i}) delta_i leq sum_i sqrt {(M_i-m_i)delta_i} sqrt delta_i leq left( sum_i (M_i-m_i)delta_iright)^{1/2} left( sum_i delta_iright)^{1/2} $$
because $f$ is Riemann integrable, $sum_i (M_i-m_i)delta_i$ can be made arbritrily small, and $sum_i delta_i = b-a$ is bounded, we are done.
answered Dec 26 at 17:11
Xiao
4,72711434
Following your notation, note that we have
$$sqrt{M_i} - sqrt{m_i} leq sqrt {M_i-m_i}$$
since $sqrt{(M_i-m_i) + m_i} leq sqrt {M_i-m_i} + sqrt{m_i}$.
To bound the Riemann sum,
$$sum_i (sqrt{M_i} - sqrt{m_i}) delta_i leq sum_i sqrt {(M_i-m_i)delta_i} sqrt delta_i leq left( sum_i (M_i-m_i)delta_iright)^{1/2} left( sum_i delta_iright)^{1/2} $$
because $f$ is Riemann integrable, $sum_i (M_i-m_i)delta_i$ can be made arbritrily small, and $sum_i delta_i = b-a$ is bounded, we are done.
answered Dec 26 at 17:11
Xiao
4,72711434
answered Dec 26 at 17:11
Xiao
4,72711434
answered Dec 26 at 17:11
Xiao
4,72711434
answered Dec 26 at 17:11
Xiao
4,72711434
4,72711434
Good one. But is mine alright?
– Subhasis Biswas
Dec 26 at 17:45
@SubhasisBiswas In your argument, does $mu$ (or the index set $Z$ and $I$) depend on the partition?
– Xiao
Dec 26 at 17:57
Everything you mentioned changes according to the mode of Partition. But ultimately, we can show that it can be made less than any predetermined positive quantity.
– Subhasis Biswas
Dec 26 at 18:00
1
@SubhasisBiswas In the second last line, you mentioned [$mu$ is a fixed constant], how did you pick this fixed constant? Also it would be hard to estimate "$mu sum_{iin Isetminus Z} (M_i - m_i) delta i$". Take $f(x) = x$ on $(0,1]$ and $f(0) = 1$, in this case $Z$ is empty. The quantity $mu sum_{iin I} (M_i - m_i) delta_i$ is hard to control as the partition size goes to zero, since $mu$ goes to infinity and $sum_{iin I} (M_i - m_i) delta i$ goes to zero.
– Xiao
Dec 26 at 18:38
unfortunately, I have to agree.
– Subhasis Biswas
Dec 26 at 19:25
|
show 1 more comment
Good one. But is mine alright?
– Subhasis Biswas
Dec 26 at 17:45
@SubhasisBiswas In your argument, does $mu$ (or the index set $Z$ and $I$) depend on the partition?
– Xiao
Dec 26 at 17:57
Everything you mentioned changes according to the mode of Partition. But ultimately, we can show that it can be made less than any predetermined positive quantity.
– Subhasis Biswas
Dec 26 at 18:00
1
@SubhasisBiswas In the second last line, you mentioned [$mu$ is a fixed constant], how did you pick this fixed constant? Also it would be hard to estimate "$mu sum_{iin Isetminus Z} (M_i - m_i) delta i$". Take $f(x) = x$ on $(0,1]$ and $f(0) = 1$, in this case $Z$ is empty. The quantity $mu sum_{iin I} (M_i - m_i) delta_i$ is hard to control as the partition size goes to zero, since $mu$ goes to infinity and $sum_{iin I} (M_i - m_i) delta i$ goes to zero.
– Xiao
Dec 26 at 18:38
unfortunately, I have to agree.
– Subhasis Biswas
Dec 26 at 19:25
Good one. But is mine alright?
– Subhasis Biswas
Dec 26 at 17:45
Good one. But is mine alright?
– Subhasis Biswas
Dec 26 at 17:45
@SubhasisBiswas In your argument, does $mu$ (or the index set $Z$ and $I$) depend on the partition?
– Xiao
Dec 26 at 17:57
@SubhasisBiswas In your argument, does $mu$ (or the index set $Z$ and $I$) depend on the partition?
– Xiao
Dec 26 at 17:57
Everything you mentioned changes according to the mode of Partition. But ultimately, we can show that it can be made less than any predetermined positive quantity.
– Subhasis Biswas
Dec 26 at 18:00
Everything you mentioned changes according to the mode of Partition. But ultimately, we can show that it can be made less than any predetermined positive quantity.
– Subhasis Biswas
Dec 26 at 18:00
1
1
@SubhasisBiswas In the second last line, you mentioned [$mu$ is a fixed constant], how did you pick this fixed constant? Also it would be hard to estimate "$mu sum_{iin Isetminus Z} (M_i - m_i) delta i$". Take $f(x) = x$ on $(0,1]$ and $f(0) = 1$, in this case $Z$ is empty. The quantity $mu sum_{iin I} (M_i - m_i) delta_i$ is hard to control as the partition size goes to zero, since $mu$ goes to infinity and $sum_{iin I} (M_i - m_i) delta i$ goes to zero.
– Xiao
Dec 26 at 18:38
@SubhasisBiswas In the second last line, you mentioned [$mu$ is a fixed constant], how did you pick this fixed constant? Also it would be hard to estimate "$mu sum_{iin Isetminus Z} (M_i - m_i) delta i$". Take $f(x) = x$ on $(0,1]$ and $f(0) = 1$, in this case $Z$ is empty. The quantity $mu sum_{iin I} (M_i - m_i) delta_i$ is hard to control as the partition size goes to zero, since $mu$ goes to infinity and $sum_{iin I} (M_i - m_i) delta i$ goes to zero.
– Xiao
Dec 26 at 18:38
unfortunately, I have to agree.
– Subhasis Biswas
Dec 26 at 19:25
unfortunately, I have to agree.
– Subhasis Biswas
Dec 26 at 19:25
|
show 1 more comment
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It would have been pretty neat and clean if $f$ were $>0$. The problem arises when $0$ is in the range of $f$. This is when for some $i$, $epsilon / (sqrt{M_i}+sqrt{m_i})$ might become $infty$. I just have modified the partition index to remove the trouble. Otherwise, it's just the same thing.
– Subhasis Biswas
Dec 26 at 16:41