Convergence of Riemann-Stieltjes sums
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Let $f$ be Riemann-Stieltjes integrable with respect $G$ (increasing function). My definition for Riemann-Stieltjes integration is: for every $epsilon$ there is a partition $mathcalP_epsilon$ such that when $mathcalP_epsilon subset mathcalP$ then $left|S(mathcalP,f,G, t_i) - int_a^bf dG right| < epsilon$ for any set of tags $t_i$.
For the Riemann integral and it is true that the integral is the limit of Riemann sums: $lim_mathcalPS(mathcalP,f)= int_a^bf(x)dx$.
Is this true for the Riemann-Stieltjes integral?
I know the integral may not exist when $f$ and $G$ are discontinuous at the same point, and I think I would need to add the condition that prevents this as well as $f$ is integrable and $G$ is increasing.
riemann-integration riemann-sum stieltjes-integral
asked Aug 25 at 0:07
WoodWorker
423210
add a comment |Â
up vote
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down vote
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Let $f$ be Riemann-Stieltjes integrable with respect $G$ (increasing function). My definition for Riemann-Stieltjes integration is: for every $epsilon$ there is a partition $mathcalP_epsilon$ such that when $mathcalP_epsilon subset mathcalP$ then $left|S(mathcalP,f,G, t_i) - int_a^bf dG right| < epsilon$ for any set of tags $t_i$.
For the Riemann integral and it is true that the integral is the limit of Riemann sums: $lim_mathcalPS(mathcalP,f)= int_a^bf(x)dx$.
Is this true for the Riemann-Stieltjes integral?
I know the integral may not exist when $f$ and $G$ are discontinuous at the same point, and I think I would need to add the condition that prevents this as well as $f$ is integrable and $G$ is increasing.
riemann-integration riemann-sum stieltjes-integral
asked Aug 25 at 0:07
WoodWorker
423210
You have $g$ instead of $G$ in one place (typo?). Your statement (including $G$ and $f$ not having discontinuity at the same point) is correct.
– herb steinberg
Aug 25 at 0:16
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f$ be Riemann-Stieltjes integrable with respect $G$ (increasing function). My definition for Riemann-Stieltjes integration is: for every $epsilon$ there is a partition $mathcalP_epsilon$ such that when $mathcalP_epsilon subset mathcalP$ then $left|S(mathcalP,f,G, t_i) - int_a^bf dG right| < epsilon$ for any set of tags $t_i$.
For the Riemann integral and it is true that the integral is the limit of Riemann sums: $lim_mathcalPS(mathcalP,f)= int_a^bf(x)dx$.
Is this true for the Riemann-Stieltjes integral?
I know the integral may not exist when $f$ and $G$ are discontinuous at the same point, and I think I would need to add the condition that prevents this as well as $f$ is integrable and $G$ is increasing.
riemann-integration riemann-sum stieltjes-integral
asked Aug 25 at 0:07
WoodWorker
423210
Let $f$ be Riemann-Stieltjes integrable with respect $G$ (increasing function). My definition for Riemann-Stieltjes integration is: for every $epsilon$ there is a partition $mathcalP_epsilon$ such that when $mathcalP_epsilon subset mathcalP$ then $left|S(mathcalP,f,G, t_i) - int_a^bf dG right| < epsilon$ for any set of tags $t_i$.
For the Riemann integral and it is true that the integral is the limit of Riemann sums: $lim_mathcalPS(mathcalP,f)= int_a^bf(x)dx$.
Is this true for the Riemann-Stieltjes integral?
I know the integral may not exist when $f$ and $G$ are discontinuous at the same point, and I think I would need to add the condition that prevents this as well as $f$ is integrable and $G$ is increasing.
riemann-integration riemann-sum stieltjes-integral
asked Aug 25 at 0:07
WoodWorker
423210
edited Aug 25 at 0:22
asked Aug 25 at 0:07
WoodWorker
423210
asked Aug 25 at 0:07
WoodWorker
423210
asked Aug 25 at 0:07
WoodWorker
423210
423210
You have $g$ instead of $G$ in one place (typo?). Your statement (including $G$ and $f$ not having discontinuity at the same point) is correct.
– herb steinberg
Aug 25 at 0:16
add a comment |Â
You have $g$ instead of $G$ in one place (typo?). Your statement (including $G$ and $f$ not having discontinuity at the same point) is correct.
– herb steinberg
Aug 25 at 0:16
You have $g$ instead of $G$ in one place (typo?). Your statement (including $G$ and $f$ not having discontinuity at the same point) is correct.
– herb steinberg
Aug 25 at 0:16
You have $g$ instead of $G$ in one place (typo?). Your statement (including $G$ and $f$ not having discontinuity at the same point) is correct.
– herb steinberg
Aug 25 at 0:16
add a comment |Â
1 Answer
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1
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When the Riemann-Stieltjes integral exists, there is no guarantee that it is the limit of sums as the partition norm tends to $0$ without stronger conditions than what you assume. In this way, it is unlike the Riemann integral.
Also, your statement that the integral will fail to exist when $f$ and $G$ have common points of discontinuity is inaccurate. The integral does not exist if the functions are both discontinuous from the right or both discontinuous from the left.
Consider the following counterexample where the integral exists but the limit of sums does not. Notice that both integrand and integrator are discontinuous at $x = 1/2$, although not both from the left or right.
$$f(x) = begincases0, quad 0 leqslant x < 1/2
\1, quad1/2 leqslant x leqslant 1 endcases\ G(x) = begincases0, quad 0 leqslant x leqslant 1/2
\1, quad1/2 < x leqslant 1 endcases$$
With partition $P = (0,1/2,1)$ we have $U(P,f,G) = L(P,f,G) = 1$. Since upper and lower Darboux integrals satisfy
$$1 = L(P,f,G) leqslant underlineint_0^1 f , dG leqslant overlineint_0^1 f , dG leqslant U(P,f,G) = 1,$$
the integral exists with $displaystyleint_0^1 f ,dG = 1$.
However, for a sequence of partitions $P_n = left(0,frac12n,ldots, frac12 - frac12n , frac12 + frac12n, ldots ,1right)$, we have $|P_n| = 1/n to 0$ where tags can be chosen such that $S(P_n,f,G)$ converges either to $0$ or $1$ and there is no unique limit.
answered Aug 25 at 0:38
RRL
44.2k42362
Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works?
– WoodWorker
Aug 27 at 16:44
You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous.
– RRL
Aug 27 at 17:53
1
Also this answer provides a more detailed proof.
– RRL
Aug 27 at 19:56
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
When the Riemann-Stieltjes integral exists, there is no guarantee that it is the limit of sums as the partition norm tends to $0$ without stronger conditions than what you assume. In this way, it is unlike the Riemann integral.
Also, your statement that the integral will fail to exist when $f$ and $G$ have common points of discontinuity is inaccurate. The integral does not exist if the functions are both discontinuous from the right or both discontinuous from the left.
Consider the following counterexample where the integral exists but the limit of sums does not. Notice that both integrand and integrator are discontinuous at $x = 1/2$, although not both from the left or right.
$$f(x) = begincases0, quad 0 leqslant x < 1/2
\1, quad1/2 leqslant x leqslant 1 endcases\ G(x) = begincases0, quad 0 leqslant x leqslant 1/2
\1, quad1/2 < x leqslant 1 endcases$$
With partition $P = (0,1/2,1)$ we have $U(P,f,G) = L(P,f,G) = 1$. Since upper and lower Darboux integrals satisfy
$$1 = L(P,f,G) leqslant underlineint_0^1 f , dG leqslant overlineint_0^1 f , dG leqslant U(P,f,G) = 1,$$
the integral exists with $displaystyleint_0^1 f ,dG = 1$.
However, for a sequence of partitions $P_n = left(0,frac12n,ldots, frac12 - frac12n , frac12 + frac12n, ldots ,1right)$, we have $|P_n| = 1/n to 0$ where tags can be chosen such that $S(P_n,f,G)$ converges either to $0$ or $1$ and there is no unique limit.
answered Aug 25 at 0:38
RRL
44.2k42362
Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works?
– WoodWorker
Aug 27 at 16:44
You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous.
– RRL
Aug 27 at 17:53
1
Also this answer provides a more detailed proof.
– RRL
Aug 27 at 19:56
add a comment |Â
up vote
1
down vote
accepted
When the Riemann-Stieltjes integral exists, there is no guarantee that it is the limit of sums as the partition norm tends to $0$ without stronger conditions than what you assume. In this way, it is unlike the Riemann integral.
Also, your statement that the integral will fail to exist when $f$ and $G$ have common points of discontinuity is inaccurate. The integral does not exist if the functions are both discontinuous from the right or both discontinuous from the left.
Consider the following counterexample where the integral exists but the limit of sums does not. Notice that both integrand and integrator are discontinuous at $x = 1/2$, although not both from the left or right.
$$f(x) = begincases0, quad 0 leqslant x < 1/2
\1, quad1/2 leqslant x leqslant 1 endcases\ G(x) = begincases0, quad 0 leqslant x leqslant 1/2
\1, quad1/2 < x leqslant 1 endcases$$
With partition $P = (0,1/2,1)$ we have $U(P,f,G) = L(P,f,G) = 1$. Since upper and lower Darboux integrals satisfy
$$1 = L(P,f,G) leqslant underlineint_0^1 f , dG leqslant overlineint_0^1 f , dG leqslant U(P,f,G) = 1,$$
the integral exists with $displaystyleint_0^1 f ,dG = 1$.
However, for a sequence of partitions $P_n = left(0,frac12n,ldots, frac12 - frac12n , frac12 + frac12n, ldots ,1right)$, we have $|P_n| = 1/n to 0$ where tags can be chosen such that $S(P_n,f,G)$ converges either to $0$ or $1$ and there is no unique limit.
answered Aug 25 at 0:38
RRL
44.2k42362
Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works?
– WoodWorker
Aug 27 at 16:44
You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous.
– RRL
Aug 27 at 17:53
1
Also this answer provides a more detailed proof.
– RRL
Aug 27 at 19:56
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
When the Riemann-Stieltjes integral exists, there is no guarantee that it is the limit of sums as the partition norm tends to $0$ without stronger conditions than what you assume. In this way, it is unlike the Riemann integral.
Also, your statement that the integral will fail to exist when $f$ and $G$ have common points of discontinuity is inaccurate. The integral does not exist if the functions are both discontinuous from the right or both discontinuous from the left.
Consider the following counterexample where the integral exists but the limit of sums does not. Notice that both integrand and integrator are discontinuous at $x = 1/2$, although not both from the left or right.
$$f(x) = begincases0, quad 0 leqslant x < 1/2
\1, quad1/2 leqslant x leqslant 1 endcases\ G(x) = begincases0, quad 0 leqslant x leqslant 1/2
\1, quad1/2 < x leqslant 1 endcases$$
With partition $P = (0,1/2,1)$ we have $U(P,f,G) = L(P,f,G) = 1$. Since upper and lower Darboux integrals satisfy
$$1 = L(P,f,G) leqslant underlineint_0^1 f , dG leqslant overlineint_0^1 f , dG leqslant U(P,f,G) = 1,$$
the integral exists with $displaystyleint_0^1 f ,dG = 1$.
However, for a sequence of partitions $P_n = left(0,frac12n,ldots, frac12 - frac12n , frac12 + frac12n, ldots ,1right)$, we have $|P_n| = 1/n to 0$ where tags can be chosen such that $S(P_n,f,G)$ converges either to $0$ or $1$ and there is no unique limit.
answered Aug 25 at 0:38
RRL
44.2k42362
When the Riemann-Stieltjes integral exists, there is no guarantee that it is the limit of sums as the partition norm tends to $0$ without stronger conditions than what you assume. In this way, it is unlike the Riemann integral.
Also, your statement that the integral will fail to exist when $f$ and $G$ have common points of discontinuity is inaccurate. The integral does not exist if the functions are both discontinuous from the right or both discontinuous from the left.
Consider the following counterexample where the integral exists but the limit of sums does not. Notice that both integrand and integrator are discontinuous at $x = 1/2$, although not both from the left or right.
$$f(x) = begincases0, quad 0 leqslant x < 1/2
\1, quad1/2 leqslant x leqslant 1 endcases\ G(x) = begincases0, quad 0 leqslant x leqslant 1/2
\1, quad1/2 < x leqslant 1 endcases$$
With partition $P = (0,1/2,1)$ we have $U(P,f,G) = L(P,f,G) = 1$. Since upper and lower Darboux integrals satisfy
$$1 = L(P,f,G) leqslant underlineint_0^1 f , dG leqslant overlineint_0^1 f , dG leqslant U(P,f,G) = 1,$$
the integral exists with $displaystyleint_0^1 f ,dG = 1$.
However, for a sequence of partitions $P_n = left(0,frac12n,ldots, frac12 - frac12n , frac12 + frac12n, ldots ,1right)$, we have $|P_n| = 1/n to 0$ where tags can be chosen such that $S(P_n,f,G)$ converges either to $0$ or $1$ and there is no unique limit.
answered Aug 25 at 0:38
RRL
44.2k42362
answered Aug 25 at 0:38
RRL
44.2k42362
answered Aug 25 at 0:38
RRL
44.2k42362
answered Aug 25 at 0:38
RRL
44.2k42362
44.2k42362
Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works?
– WoodWorker
Aug 27 at 16:44
You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous.
– RRL
Aug 27 at 17:53
1
Also this answer provides a more detailed proof.
– RRL
Aug 27 at 19:56
add a comment |Â
Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works?
– WoodWorker
Aug 27 at 16:44
You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous.
– RRL
Aug 27 at 17:53
1
Also this answer provides a more detailed proof.
– RRL
Aug 27 at 19:56
Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works?
– WoodWorker
Aug 27 at 16:44
Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works?
– WoodWorker
Aug 27 at 16:44
You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous.
– RRL
Aug 27 at 17:53
You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous.
– RRL
Aug 27 at 17:53
1
1
Also this answer provides a more detailed proof.
– RRL
Aug 27 at 19:56
Also this answer provides a more detailed proof.
– RRL
Aug 27 at 19:56
add a comment |Â
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You have $g$ instead of $G$ in one place (typo?). Your statement (including $G$ and $f$ not having discontinuity at the same point) is correct.
– herb steinberg
Aug 25 at 0:16