Riemann-Stieltjes Problem

[jdcasey9]

Homework Statement

Give an example where f² $$\in$$ R_$$\alpha$$, but f$$\notin$$ R_$$\alpha$$[a,b].

Homework Equations

Let $$\alpha$$: [a,b]$$\rightarrow$$ R(Real numbers) be increasing. A bounded function f: [a,b] $$\rightarrow$$ R(real numbers) is in R_$$\alpha$$[a,b] if and only if, given $$\epsilon$$ $$\succ$$0, there exists a partition P of [a,b] such that U(f,P) - L(f,P) $$\prec$$$$\epsilon$$.

The Attempt at a Solution

Would f(x) = (1/x)(sinx) work? Because (f(x))² converges and f(x) diverges, right?

 

[mighty2000]

There are actually many examples where f2 \in R\alpha, but f\notin R\alpha[a,b]. One such example is the function f(x) = 1/x on the interval [0,1]. This function is not Riemann integrable on [0,1] because it is unbounded at x = 0. However, if we square this function, we get f2(x) = 1/x2, which is Riemann integrable on [0,1] since it is bounded on this interval. Therefore, f2 \in R\alpha, but f\notin R\alpha[0,1].

Another example is the function f(x) = x on the interval [0,1]. This function is Riemann integrable on [0,1], but if we take the square root of f(x), we get f(x) = √x, which is not Riemann integrable on [0,1] since it is unbounded at x = 0. Therefore, f2 \in R\alpha, but f\notin R\alpha[0,1].

In general, any function that is not Riemann integrable on [a,b] but becomes Riemann integrable when squared will satisfy the condition f2 \in R\alpha, but f\notin R\alpha[a,b]. This is because the Riemann integrability of a function is dependent on its boundedness, and squaring a function can change its boundedness. I hope this helps to clarify the concept.