Determine if Riemann-Stieltjes integrable and calculate the integral if so

Question

Here's the detailed problem statement.

For each of the following functions $f$ and monotonically increasing function $\alpha$, determine if $f\in \mathcal{R}_{\alpha}[-1,1]$, and if so calculate the integral using the definition of RS integral (you may apply Theorem 6.6 of Baby Rudin if you wish)

(a) $f(x)= \begin{cases} -1 \ \ (x\leq 0) \\ 1 \ \ \ \ \ (x>0) \end{cases}$

$\ \ \ \ \ \ \alpha(x)= \begin{cases} 0 \ \ \ \ \ (x<0) \\ 1 \ \ \ \ \ (x\geq 0) \end{cases}$

I think this problem is all about finding a nice partition $P$ so that $U(P,f,\alpha)=L(P,f,\alpha)$ thanks to Theorem 6.6. Now I have no idea how to find it for (a). I drew the graph myself and found the problem, that is, how to deal with $x=0$.

Answer 1

As I wrote in comment, lets consider partition where $\exists k$ for which $x_{k-1}<0<x_k$. For such partition from sum $\sum\limits_{i=1}^{n}f(\xi_i)\big(\alpha(x_{i})-\alpha(x_{i-1})\big)$ we will have only one non zero summand $f(\xi_k)\big(\alpha(x_{k})-\alpha(x_{k-1})\big)=f(\xi_k)$. Now taking $\xi_k<0$ and $\xi_k>0$ we will have whole sum $-1$ and $1$ respectively, so limit from formal definition doesn't exist.

Answer 2

There are two definitions of the Riemann-Stieltjes integral. For this problem, the integral exists with value $-1$ under one definition, but fails to exist under the other.

Given $f,\alpha:[a,b] \to \mathbb{R}$ and a partition $P = (x_0,x_1, \ldots,x_n)$ of the interval $[a,b]$, the Riemann-Stieltjes sum corresponding to a choice of intermediate points $c_j \in [x_{j-1},x_j]$ is defined as

$$S(P,f,\alpha) = \sum_{j=1}^n f(c_j)[\alpha(x_j)- \alpha(x_{j-1})],$$

and, with $M_j := \sup\{(f(x): x \in [x_{j-1},x_j]\}$ and $m_j := \sup\{(f(x): x \in [x_{j-1},x_j]\}$, the upper and lower Darboux sums are defined as

$$U(P,f,\alpha) = \sum_{j=1}^nM_j[\alpha(x_j)- \alpha(x_{j-1})],\quad L(P,f,\alpha) = \sum_{j=1}^nm_j[\alpha(x_j)- \alpha(x_{j-1})]$$

Two definitions of the Riemann-Stieltjes integral are:

D1: We say that $f$ is Riemann-Stieltjes with respect to $\alpha$ if there exists a real number $I$ with the property that for every $\epsilon > 0$ there exists a partition $P_\epsilon$ such that for every refinement $P \supset P_\epsilon$ we have $|S(P,f,\alpha)- I| < \epsilon$ for any choice of intermediate points .

D2: We say that $f$ is Riemann-Stieltjes with respect to $\alpha$ if there exists a real number $I$ with the property that for every $\epsilon > 0$ there exists $\delta > 0$ such that for any partition with mesh $\|P\| = \max_{1 \leqslant j \leqslant n}(x_j-x_{j-1})< \epsilon$, we have $|S(P,f,\alpha)- I| < \epsilon$ for any choice of intermediate points .

Unlike the Riemann integral, where $\alpha(x) = x$, the two definitions are not always equivalent.

When $\alpha$ is nondecreasing (as in this problem), we can prove integrability under definition D1 and evaluate the integral using Darboux sums. For bounded functions, there always exist upper and lower integrals satisfying

$$\sup_{P} L(P,f,\alpha)= \underline{\int_a}^b f\, d\alpha \leqslant \overline{\int_a}^b f \, d\alpha = \inf_{P} U(P,f,\alpha)$$

Then the Riemann-Stieltjes integral exists if and only if the upper and lower integrals are equal, with

$$\int_a^b f \, d\alpha = \underline{\int_a}^b f\, d\alpha = \overline{\int_a}^b f \, d\alpha $$

The Riemann-Stieltjes integral for this problem exists under definition D1.

Taking the partition $P'$ with points $\{-1,0,1\}$, it is easy to show that $L(P',f,\alpha) = U(P',f,\alpha) = -1$, and, hence,

$$-1 = L(P',f,\alpha) \leqslant \underline{\int_a}^b f\, d\alpha \leqslant \overline{\int_a}^b f \, d\alpha \leqslant U(P'.f.\alpha) = -1,$$

which implies that the upper and lower integrals, and therefore the Riemann-Stieltjes integral, are equal to $-1$.

The Riemann-Stieltjes integral for this problem does not exist under definition D2.

Consider the sequence of partitions $P_n = (-1,-1+\frac{1}{2n},\ldots,-\frac{1}{2n},\frac{1}{2n},\ldots, 1- \frac{1}{2n},1)$. The partiton mesh is $\|P_n\| = \frac{1}{n}$ which tends to $0$ as $n \to \infty$. However, with intermediate point $c \in [-\frac{1}{2n},\frac{1}{2n}]$ and any selection for the other points we have

$$S(P_n,f,\alpha) = c \left[\alpha\left(\frac{1}{2n}\right) - \alpha \left(-\frac{1}{2n}\right)\right]= c$$

Hence, for any $n$ and arbitrarily small $\|P_n\|$ there is always an intermediate point $c$ such that $S(P_n,f,\alpha) = 1$ and one such that $S(P_n,f,\alpha) = -1$.