Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x)

mm1239

New member
Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x)

Is this True/False- and demonstrate with a short proof or counterexample:

Let f(x) be the step function f(x) = 0 if x<0, f(x) = 1 if x>= (greater than or equal to) 0. Every continuous function on [-1,1] is in Riemann(x).

Thanks in advance for the help!

Evgeny.Makarov

Well-known member
MHB Math Scholar
Re: Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x

Let f(x) be the step function f(x) = 0 if x<0, f(x) = 1 if x>= (greater than or equal to) 0. Every continuous function on [-1,1] is in Riemann(x).
What is Riemann(x)? And what do you think yourself about the question?

mm1239

New member
Re: Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x

Sorry I think the question should actually end with "is in Riemann wrt f"

I think the answer is true since the function is clearly bounded and we can show f to be integrable. Is that correct?

What is Riemann(x)? And what do you think yourself about the question?

Evgeny.Makarov

Well-known member
MHB Math Scholar
Re: Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x

Sorry I think the question should actually end with "is in Riemann wrt f"
Sorry, but it is still not clear to me what it means for a function to be in Riemann wrt another function.

mm1239

New member
Re: Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x

Yes, I'm not sure either. The exact question is:

"Is the following question true or false? Motivate your answer with a short proof or counterexample:

Let theta(x) be the step function theta(x) = 0 if x<0 and theta(x) = 1 if x>= 0. Every continuous function on [-1,1] is in R(theta)."

There's also a follow-up question that asks whether "every left continuous function on [-1,1] is in R(theta)."

Sorry, but it is still not clear to me what it means for a function to be in Riemann wrt another function.

Evgeny.Makarov

Well-known member
MHB Math Scholar
Re: Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x

You have to find out the definition of R(θ) in your textbook or lecture notes. The only guess I have is that it may denote the set of functions $$f$$ whose Riemann–Stieltjes integral $$\int f(x)\,d\theta(x)$$ exists. Maybe someone has a better guess.

mm1239

New member
Re: Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x

Just looked it up- that's it! Still not sure how to prove this or the follow-up question though- any idea?

You have to find out the definition of R(θ) in your textbook or lecture notes. The only guess I have is that it may denote the set of functions $$f$$ whose Riemann–Stieltjes integral $$\int f(x)\,d\theta(x)$$ exists. Maybe someone has a better guess.

Plato

Well-known member
MHB Math Helper
Re: Let f(x) be the step f'n f(x)=0 if x<0, f(x)=1 if x>=0. Every cts. f'n on [-1,1] is in Riemann{x

Just looked it up- that's it! Still not sure how to prove this or the follow-up question though- any idea?

This is a completely standard theorem is R-S integrals.
If $f$ is continuous on $[a,b]$ and $g$ is monotone non-decreasing there then $f$ is $g\text{-integerable}$ on $[a,b]$.

In this case it almost trivial. Any division of $[-1,1]$ has a refinement that includes 0 as an endpoint.
On any sub interval $[x_j,x_k]$ in the refinement if $x_k\ne 0$ then $\theta(x_k)-\theta(x_j)=0$.

Also, you know that $f$ is uniformity continuous on $[-1,1]$.