Mixing
(Baby Rudin) ch6 Theorem 6.11 (The Riemann-Stieltjes Integral)
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The theorem and the proof above is from Rudin.
I have a question about the last part of the proof.
For the inequality $sum_i in A(M_i^*-m^*_i)Deltaalpha_i+sum_i in B(M_i^*-m^*_i)Deltaalpha_i le epsilon[alpha(b) - alpha(a)]+2Kdelta$, I understand that the second term in the right side is made so because $M_i^*-m^*_ile2K$ for $iin B$ and $sum_i in BDeltaalpha_i<delta$ as proved in the proof. However, as for the first term in the right side of the inequality, How do we know that $alpha(b)$ and $-alpha(a)$ terms are in $sum_i in A Delta alpha_i$? $A$ might not contain $1$ and $n$ so that $Deltaalpha(x_n)=alpha(b) - alpha(x_n-1)$ and $Deltaalpha(x_1)=alpha(x_1)-alpha(a)$ are not a part of $sum_i in A Delta alpha_i$.
Thank you in advance!
real-analysis general-topology analysis riemann-integration stieltjes-integral
edited 47 mins ago
Henno Brandsma
99.5k344107
asked 5 hours ago
Hunnam
615
add a comment |
up vote
2
down vote
favorite
The theorem and the proof above is from Rudin.
I have a question about the last part of the proof.
For the inequality $sum_i in A(M_i^*-m^*_i)Deltaalpha_i+sum_i in B(M_i^*-m^*_i)Deltaalpha_i le epsilon[alpha(b) - alpha(a)]+2Kdelta$, I understand that the second term in the right side is made so because $M_i^*-m^*_ile2K$ for $iin B$ and $sum_i in BDeltaalpha_i<delta$ as proved in the proof. However, as for the first term in the right side of the inequality, How do we know that $alpha(b)$ and $-alpha(a)$ terms are in $sum_i in A Delta alpha_i$? $A$ might not contain $1$ and $n$ so that $Deltaalpha(x_n)=alpha(b) - alpha(x_n-1)$ and $Deltaalpha(x_1)=alpha(x_1)-alpha(a)$ are not a part of $sum_i in A Delta alpha_i$.
Thank you in advance!
real-analysis general-topology analysis riemann-integration stieltjes-integral
edited 47 mins ago
Henno Brandsma
99.5k344107
asked 5 hours ago
Hunnam
615
What is $alpha$? Is it an increasing function?
– Lord Shark the Unknown
4 hours ago
Yes. According to Rudin, $alpha$ is a monotonically increasing function on $[a, b]$ which is bounded.
– Hunnam
4 hours ago
Note that $A subseteq 1,2,dots,n$.
– xbh
4 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The theorem and the proof above is from Rudin.
I have a question about the last part of the proof.
For the inequality $sum_i in A(M_i^*-m^*_i)Deltaalpha_i+sum_i in B(M_i^*-m^*_i)Deltaalpha_i le epsilon[alpha(b) - alpha(a)]+2Kdelta$, I understand that the second term in the right side is made so because $M_i^*-m^*_ile2K$ for $iin B$ and $sum_i in BDeltaalpha_i<delta$ as proved in the proof. However, as for the first term in the right side of the inequality, How do we know that $alpha(b)$ and $-alpha(a)$ terms are in $sum_i in A Delta alpha_i$? $A$ might not contain $1$ and $n$ so that $Deltaalpha(x_n)=alpha(b) - alpha(x_n-1)$ and $Deltaalpha(x_1)=alpha(x_1)-alpha(a)$ are not a part of $sum_i in A Delta alpha_i$.
Thank you in advance!
real-analysis general-topology analysis riemann-integration stieltjes-integral
edited 47 mins ago
Henno Brandsma
99.5k344107
asked 5 hours ago
Hunnam
615
The theorem and the proof above is from Rudin.
I have a question about the last part of the proof.
For the inequality $sum_i in A(M_i^*-m^*_i)Deltaalpha_i+sum_i in B(M_i^*-m^*_i)Deltaalpha_i le epsilon[alpha(b) - alpha(a)]+2Kdelta$, I understand that the second term in the right side is made so because $M_i^*-m^*_ile2K$ for $iin B$ and $sum_i in BDeltaalpha_i<delta$ as proved in the proof. However, as for the first term in the right side of the inequality, How do we know that $alpha(b)$ and $-alpha(a)$ terms are in $sum_i in A Delta alpha_i$? $A$ might not contain $1$ and $n$ so that $Deltaalpha(x_n)=alpha(b) - alpha(x_n-1)$ and $Deltaalpha(x_1)=alpha(x_1)-alpha(a)$ are not a part of $sum_i in A Delta alpha_i$.
Thank you in advance!
real-analysis general-topology analysis riemann-integration stieltjes-integral
real-analysis general-topology analysis riemann-integration stieltjes-integral
edited 47 mins ago
Henno Brandsma
99.5k344107
asked 5 hours ago
Hunnam
615
edited 47 mins ago
Henno Brandsma
99.5k344107
asked 5 hours ago
Hunnam
615
edited 47 mins ago
Henno Brandsma
99.5k344107
edited 47 mins ago
Henno Brandsma
99.5k344107
edited 47 mins ago
Henno Brandsma
99.5k344107
99.5k344107
asked 5 hours ago
Hunnam
615
asked 5 hours ago
Hunnam
615
asked 5 hours ago
Hunnam
615
615
What is $alpha$? Is it an increasing function?
– Lord Shark the Unknown
4 hours ago
Yes. According to Rudin, $alpha$ is a monotonically increasing function on $[a, b]$ which is bounded.
– Hunnam
4 hours ago
Note that $A subseteq 1,2,dots,n$.
– xbh
4 hours ago
add a comment |
What is $alpha$? Is it an increasing function?
– Lord Shark the Unknown
4 hours ago
Yes. According to Rudin, $alpha$ is a monotonically increasing function on $[a, b]$ which is bounded.
– Hunnam
4 hours ago
Note that $A subseteq 1,2,dots,n$.
– xbh
4 hours ago
What is $alpha$? Is it an increasing function?
– Lord Shark the Unknown
4 hours ago
What is $alpha$? Is it an increasing function?
– Lord Shark the Unknown
4 hours ago
Yes. According to Rudin, $alpha$ is a monotonically increasing function on $[a, b]$ which is bounded.
– Hunnam
4 hours ago
Yes. According to Rudin, $alpha$ is a monotonically increasing function on $[a, b]$ which is bounded.
– Hunnam
4 hours ago
Note that $A subseteq 1,2,dots,n$.
– xbh
4 hours ago
Note that $A subseteq 1,2,dots,n$.
– xbh
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Each $M_i^*-m_i^*ge0$ and $Deltaalpha_ige0$ as $alpha_igealpha_i-1$.
Also $M_i^*-m_i^*leepsilon$ for $iin A$. Therefore
$$sum_iin A(M_i^*-m_i^*)Deltaalpha_ileepsilonsum_iin A
Deltaalpha_i.$$
But
$$sum_iin ADeltaalpha_ilesum_i=1^nDeltaalpha_i=alpha(b)-alpha(a)$$
as $Deltaalpha_ige0$ for $inotin A$. Putting this together
gives the given bound.
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
add a comment |
up vote
2
down vote
Since $M_i - m_i geqslant 0$ (supremum minus infimum) and $alpha$ is increasing we have
$$sum_i in A(M_i-m_i) Delta alpha_i leqslant sum_i =1^n(M_i-m_i) Delta alpha_i leqslant epsilon[alpha(b) - alpha(a)]$$
answered 4 hours ago
RRL
46.2k42365
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Each $M_i^*-m_i^*ge0$ and $Deltaalpha_ige0$ as $alpha_igealpha_i-1$.
Also $M_i^*-m_i^*leepsilon$ for $iin A$. Therefore
$$sum_iin A(M_i^*-m_i^*)Deltaalpha_ileepsilonsum_iin A
Deltaalpha_i.$$
But
$$sum_iin ADeltaalpha_ilesum_i=1^nDeltaalpha_i=alpha(b)-alpha(a)$$
as $Deltaalpha_ige0$ for $inotin A$. Putting this together
gives the given bound.
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
add a comment |
up vote
3
down vote
accepted
Each $M_i^*-m_i^*ge0$ and $Deltaalpha_ige0$ as $alpha_igealpha_i-1$.
Also $M_i^*-m_i^*leepsilon$ for $iin A$. Therefore
$$sum_iin A(M_i^*-m_i^*)Deltaalpha_ileepsilonsum_iin A
Deltaalpha_i.$$
But
$$sum_iin ADeltaalpha_ilesum_i=1^nDeltaalpha_i=alpha(b)-alpha(a)$$
as $Deltaalpha_ige0$ for $inotin A$. Putting this together
gives the given bound.
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Each $M_i^*-m_i^*ge0$ and $Deltaalpha_ige0$ as $alpha_igealpha_i-1$.
Also $M_i^*-m_i^*leepsilon$ for $iin A$. Therefore
$$sum_iin A(M_i^*-m_i^*)Deltaalpha_ileepsilonsum_iin A
Deltaalpha_i.$$
But
$$sum_iin ADeltaalpha_ilesum_i=1^nDeltaalpha_i=alpha(b)-alpha(a)$$
as $Deltaalpha_ige0$ for $inotin A$. Putting this together
gives the given bound.
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
Each $M_i^*-m_i^*ge0$ and $Deltaalpha_ige0$ as $alpha_igealpha_i-1$.
Also $M_i^*-m_i^*leepsilon$ for $iin A$. Therefore
$$sum_iin A(M_i^*-m_i^*)Deltaalpha_ileepsilonsum_iin A
Deltaalpha_i.$$
But
$$sum_iin ADeltaalpha_ilesum_i=1^nDeltaalpha_i=alpha(b)-alpha(a)$$
as $Deltaalpha_ige0$ for $inotin A$. Putting this together
gives the given bound.
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
answered 4 hours ago
Lord Shark the Unknown
95.8k957125
95.8k957125
add a comment |
add a comment |
up vote
2
down vote
Since $M_i - m_i geqslant 0$ (supremum minus infimum) and $alpha$ is increasing we have
$$sum_i in A(M_i-m_i) Delta alpha_i leqslant sum_i =1^n(M_i-m_i) Delta alpha_i leqslant epsilon[alpha(b) - alpha(a)]$$
answered 4 hours ago
RRL
46.2k42365
add a comment |
up vote
2
down vote
Since $M_i - m_i geqslant 0$ (supremum minus infimum) and $alpha$ is increasing we have
$$sum_i in A(M_i-m_i) Delta alpha_i leqslant sum_i =1^n(M_i-m_i) Delta alpha_i leqslant epsilon[alpha(b) - alpha(a)]$$
answered 4 hours ago
RRL
46.2k42365
add a comment |
up vote
2
down vote
up vote
2
down vote
Since $M_i - m_i geqslant 0$ (supremum minus infimum) and $alpha$ is increasing we have
$$sum_i in A(M_i-m_i) Delta alpha_i leqslant sum_i =1^n(M_i-m_i) Delta alpha_i leqslant epsilon[alpha(b) - alpha(a)]$$
answered 4 hours ago
RRL
46.2k42365
Since $M_i - m_i geqslant 0$ (supremum minus infimum) and $alpha$ is increasing we have
$$sum_i in A(M_i-m_i) Delta alpha_i leqslant sum_i =1^n(M_i-m_i) Delta alpha_i leqslant epsilon[alpha(b) - alpha(a)]$$
answered 4 hours ago
RRL
46.2k42365
answered 4 hours ago
RRL
46.2k42365
answered 4 hours ago
RRL
46.2k42365
answered 4 hours ago
RRL
46.2k42365
46.2k42365
add a comment |
add a comment |
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What is $alpha$? Is it an increasing function?
– Lord Shark the Unknown
4 hours ago
Yes. According to Rudin, $alpha$ is a monotonically increasing function on $[a, b]$ which is bounded.
– Hunnam
4 hours ago
Note that $A subseteq 1,2,dots,n$.
– xbh
4 hours ago