I saw the following exercise in Bartle’s Elements of Real Analysis:
If f is a function from the interval [a,b] to the reals such that f is discontinuous at some point of the interval, then there exists a monotone increasing function g such that f isn’t g-integrable.
I understand that the g must be continuous at the point of discontinuity of the f for the f to be g-integrable. However, I don’t know how the proof should follow.
On
Hints: for convenience, take $a=0$ and $b=1$ and suppose $f$ has a discontinuity at $x=1/2.$
$\text{Set}\ g(x)=\left\{ \begin{array}{lr} 0 & x<1/2\\ \frac{1}{2} & x=1/2\\ 1 & x>1/2 \end{array} \right\},$ note that $\text{osc}\ f=\epsilon>0$ at $x=1/2$
and prove that any interval containing $1/2$ satisfies $U(f,g)-L(f,g)>\epsilon.$
We have non-existence of the Riemann-Stieltjes integral if $f$ and $g$ are either both discontinuous from the right or both discontinuous from the left. If $f$ is discontinuous (from both the right and left) at $\xi \in (a,b)$ then we can choose any monotone increasing function $g$ that is either right- or left-continuous at $\xi$ and the integral will fail to exist.
For proof, suppose (without loss of generality) that $f$ and $g$ are discontinuous from the right at $\xi \in (a,b).$ Consider any partition $P = (x_0,x_1, \ldots, x_{i-1},\xi, x_i, \ldots, x_n)$ with $\xi$ as a partition point and $x_i - \xi = \delta_i$.
Since $f$ and $g$ are right-discontinuous, there exists $\epsilon > 0$ such that for every $\delta > 0$ (including $\delta_i$), there are points $y_1, y_2 \in (\xi, \xi + \delta)$ such that $|f(y_1) - f(\xi)| \geqslant \epsilon$ and $|g(y_2) - g(\xi)| \geqslant \epsilon$.
Consider the difference between upper and lower Riemann-Stieltjes sums. We have
$$U(P,f,g) - L(P,f,g) = [\sup_{x \in [\xi,x_i]} f(x) - \inf_{x \in [\xi,x_i]} f(x)](g(x_i) - g(\xi)) + \text{other nonnegative terms} \\ \geqslant [\sup_{x \in [\xi,x_i]} f(x) - \inf_{x \in [\xi,x_i]} f(x)](g(x_i) - g(\xi)) $$
Since $g(x_i) - g(\xi) \geqslant g(y_2) - g(\xi) \geqslant \epsilon$ and $\sup_{x \in [\xi,x_i]} f(x) - \inf_{x \in [\xi,x_i]} f(x) \geqslant \epsilon$ it follows that
$$U(P,f,g) - L(P,f,g) \geqslant \epsilon^2$$
Consequently, the Riemann criterion for integrability is not satisfied. If $f$ were Riemann-Stieltjes integrable with respect to $g$ then there must exist a partition $P'$ such that $U(P',f,g) - L(P',f,g) < \epsilon^2$. Adding the point $\xi$ to $P'$ -- if it is not already present -- we obtain a refined partition $P$ as above where $U(P,f,g) \leqslant U(P',f,g)$ and $L(P,f,g) \geqslant L(P',f,g)$. We then have
$$U(P,f,g) - L(P,f,g) \leqslant U(P',f,g) - L(P',f,g) < \epsilon^2,$$
a contradiction. Therefore the Riemann-Stieltjes integral of $f$ with respect to $g$ does not exist.