Sufficient condition on integrator that Riemann-Stieltjes integrable function on $[0,1]$ is Lebesgue integrable.

Asked Aug 12 '19 at 11:28

Active Aug 13 '19 at 06:43

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Let $\alpha$ be nondecreasing on $[0,1]$. Then Riemann-Stieltjes integrable function is well defined with respect to the integrator $\alpha$.

I was wondering if the following assertion is true:

(1) If $f$ is Riemann-Stieltjes integrable on $[0,1],$ then $f$ is Lebesgue integrable on $[0,1]$ with respect to the measure induced by $\alpha$.

(2) Do we need additional hypotheses on $\alpha$ for guaranteeing the above assertion? For example, $\alpha$ is right-continuous.

I would be grateful if you give any comment for my question.

edited Jun 12 '20 at 10:38

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asked Aug 12 '19 at 11:28

04170706

2

@G. Chiusole Certainly not a duplicate of the question you linked to. – uniquesolution Aug 12 '19 at 11:44
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Monotone functions on the open interval $(0,1)$ need not be bounded, let alone be of bounded variation, so in general the Riemann-Steltjes integral will not exist. In order for the question to make sense, you may want to consider functions on the closed interval $[0,1]$. – uniquesolution Aug 12 '19 at 11:46
@uniquesolution Thanks a lot! I corrected it. – 04170706 Aug 12 '19 at 11:48
@uniquesolution How so? It boils down to the question "Does Riemann integrable imply Lebesgue integrable" – G. Chiusole Aug 12 '19 at 11:52
@G.Chiusole Thanks for your comment. For $\alpha(x)=x$ on [0,1], the assertion is true. Note that in this case, $\alpha$ is continuous. – 04170706 Aug 12 '19 at 11:57
@04170706 ah yes, my bad. – G. Chiusole Aug 12 '19 at 16:25
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Look at "WHEEDEN, Richard – Measure and Integral 2nd ed". – Lázaro Albuquerque Aug 12 '19 at 16:29
@lzralbu Thanks a lot. I will check it later, since I don’t have the book now. – 04170706 Aug 13 '19 at 01:05

Sufficient condition on integrator that Riemann-Stieltjes integrable function on $[0,1]$ is Lebesgue integrable.

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