Riemann integrable

**BBrown** · May 7th 2010, 10:54 AM

Anyone know more about reimann-stieltjes integrable functions? I'm researching them and i need to show if $g'$ is integrable with $f$ on $[a,b]$ continuous, then

$\int_a^b f(x)dg(x)$ exists and $= \int^b_a f(x)g'(x)$

**tonio** · May 7th 2010, 12:44 PM

Originally Posted by BBrown

Anyone know more about reimann-stieltjes integrable functions? I'm researching them and i need to show if $g'$ is integrable with $f$ on $[a,b]$ continuous, then

$\int_a^b f(x)dg(x)$ exists and $= \int^b_a f(x)g'(x)$

Look at each summand defining $\int_a^b f(x)dg(x)$ : it is of the form $f(c_i)(g(x_{i+1})-g(x_i))$ , where $[x_i,\,x_{i+1}]$ is a subinterval of $[a,\,b]$ and $c_i\in [x_i,\,x_{i+1}]$ .

Well, now apply the mean value theorem to the function $g\;:\;g(x_{i+1})-g(x_i)=g'(d_i)(x_{i+1}-x_i)\,,\,\,d_i\in (x_i,\,x_{i+1})$ , substitute this in every summand of

the sum defining $\int_a^b f(x)dg(x)$ and get the Riemann sum of exactly what you need since we're given $g'(x)$ is Riemann integrable...

Tonio

**Plato** · May 7th 2010, 01:22 PM

Originally Posted by tonio

Look at each summand defining $\int_a^b f(x)dg(x)$ : it is of the form $f(c_i)(g(x_{i+1})-g(x_i))$ , where $[x_i,\,x_{i+1}]$ is a subinterval of $[a,\,b]$ and $c_i\in [x_i,\,x_{i+1}]$ .
Well, now apply the mean value theorem to the function $g\;:\;g(x_{i+1})-g(x_i)=g'(d_i)(x_{i+1}-x_i)\,,\,\,d_i\in (x_i,\,x_{i+1})$ , substitute this in every summand of
the sum defining $\int_a^b f(x)dg(x)$ and get the Riemann sum of exactly what you need since we're given $g'(x)$ is Riemann integrable...

All of the above is true provided we know that $\int_a^b f(x)dg(x)$ exists.
Do we know it exists?
Because $g'$ is Riemann integrable implies that it is bounded on $[a,b].$
So $g$ has the Lipschitz Condition.
Thus $g$ is of bounded variation.
Thus that integral does exist.

**tonio** · May 7th 2010, 01:41 PM

Originally Posted by Plato

All of the above is true provided we know that $\int_a^b f(x)dg(x)$ exists.
Do we know it exists?
We need to more about $f$ and/or $g$ .

Well, f is cont. and thus Riemann integ., and we're given g' is (Riemann) int., so fg' is Riemann int., and my first post proves the Riemann-Stieltjes int. on the left equals the Riemann int. on the right, so from the existence of the RHS and the equality between both sides it follows the existence of the LHS.

Tonio

**BBrown** · May 7th 2010, 03:34 PM

Thank you so much... I should have known: when in doubt, apply the mean value thm.

Thanks for the guidance I truly appreciate it.

**Plato** · May 7th 2010, 03:52 PM

Originally Posted by tonio

Well, f is cont. and thus Riemann integ., and we're given g' is (Riemann) int., so fg' is Riemann int., and my first post proves the Riemann-Stieltjes int. on the left equals the Riemann int. on the right, so from the existence of the RHS and the equality between both sides it follows the existence of the LHS.

Frankly I do not find that argument persuasive.
I am not sure that one approximating sum can prove the existence of the integral.
Here is my point: A necessary and sufficient condition that $\int_a^b {fdg}$ exist for every function $f$ continuous on $[a,b]$ is that $g$ be of bounded variation on $[a,b]$

**tonio** · May 7th 2010, 11:18 PM

Originally Posted by Plato

Frankly I do not find that argument persuasive.
I am not sure that one approximating sum can prove the existence of the integral.
Here is my point: A necessary and sufficient condition that $\int_a^b {fdg}$ exist for every function $f$ continuous on $[a,b]$ is that $g$ be of bounded variation on $[a,b]$

I think my argument is sound, and there's no approximation in it: it shows that the sum whose limit defines the Riemann-Stieltjes integral on the LHS equals the sum whose limit limit defines the Riemann integral on the RHS, and when we pass to the limit in both sides (when the number of partition points goes to infinity and the partition's length goes to zero), this last sum's limit exists since fg' is a Riemann integrable function, as already said, from which also follows the existence of the LHS's linit.

The bounded variation condition on g is fine to know beforehand that the LHS integral exists (together with f being continuous or simply Riemann integrable), but I think we can also deduce its existence by the above argument.

Tonio

**BBrown** · May 8th 2010, 10:30 AM

Either of you looked at AidenMatthew's post about integrability? He helped me out before and I'd like to help him but I think I'm making the problem easier than it is. Ideas?

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