net convergence implies bounded?












1












$begingroup$


I just saw the following theorem:




Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.




in this answer.



but I'm confused by this step in the proof:




Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$




Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?






real-analysis general-topology convergence riemann-integration







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$endgroup$












  • $begingroup$
    In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
    $endgroup$
    – Daniel Fischer♦
    Aug 31 '17 at 15:21
















1












$begingroup$


I just saw the following theorem:




Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.




in this answer.



but I'm confused by this step in the proof:




Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$




Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?






real-analysis general-topology convergence riemann-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
    $endgroup$
    – Daniel Fischer♦
    Aug 31 '17 at 15:21














1












1








1


1



$begingroup$


I just saw the following theorem:




Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.




in this answer.



but I'm confused by this step in the proof:




Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$




Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?






real-analysis general-topology convergence riemann-integration







share|cite|improve this question











$endgroup$




I just saw the following theorem:




Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.




in this answer.



but I'm confused by this step in the proof:




Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$




Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?






real-analysis general-topology convergence riemann-integration




real-analysis general-topology convergence riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 31 '17 at 13:50







Lookout

















asked Aug 30 '17 at 3:33









LookoutLookout

897818




897818












  • $begingroup$
    In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
    $endgroup$
    – Daniel Fischer♦
    Aug 31 '17 at 15:21


















  • $begingroup$
    In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
    $endgroup$
    – Daniel Fischer♦
    Aug 31 '17 at 15:21
















$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21




$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21










1 Answer
1






active

oldest

votes


















0












$begingroup$

You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
    $endgroup$
    – Lookout
    Aug 30 '17 at 7:23











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
    $endgroup$
    – Lookout
    Aug 30 '17 at 7:23
















0












$begingroup$

You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
    $endgroup$
    – Lookout
    Aug 30 '17 at 7:23














0












0








0





$begingroup$

You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.






share|cite|improve this answer









$endgroup$



You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 30 '17 at 6:32









Kavi Rama MurthyKavi Rama Murthy

64.5k42765




64.5k42765












  • $begingroup$
    But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
    $endgroup$
    – Lookout
    Aug 30 '17 at 7:23


















  • $begingroup$
    But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
    $endgroup$
    – Lookout
    Aug 30 '17 at 7:23
















$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23




$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23


















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