In many theorems about the Riemann-Stieltjes integral they required the hypothesis of $f$ to be bounded to then conclude that $f$ is Riemann-Stieltjes integrable.
For example, suppose that $f$ is bounded in $I = [a,b]$, $f$ has only finitely many points of discontinuity in $I$, and that the monotonically increasing function $\alpha$ is continuous at each point of discontinuity of $f$, then $f$ is Riemann-Stieltjes integrable.
What if we remove the bounded hypothesis?
Could there exist an unbounded function $f$ in a given interval $[a,b]$ such that $\int_a^bf\,d\alpha$ exist?
Maybe a counterexample?
Best Answer
A function $f$ cannot be both unbounded and Riemann-Stieltjes integrable.
This can be shown by producing an $\epsilon > 0$ such that for any real number $A$ and any $\delta > 0$ there is a tagged partition $P$ with $\|P\| < \delta$ and with a Riemann-Stieltjes sum satisfying
$$|S(P,f,\alpha) - A| > \epsilon$$
Given any partition $P$, since $f$ is unbounded, it must be unbounded on at least one subinterval $[x_{j-1},x_j]$ of P. Using the reverse triangle inequality we have
$$|S(P,f,\alpha) - A| = \left|f(t_j)(\alpha(x_j) - \alpha(x_{j-1})) + \sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right| \\ \geqslant |f(t_j)|(\alpha(x_j) - \alpha(x_{j-1})) - \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|$$
Since $f$ is unbounded on $[x_{j-1},x_j]$, choose a partition tag $t_j$ such that
$$|f(t_j)| > \frac{\epsilon + \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|}{\alpha(x_j) - \alpha(x_{j-1})},$$
and it follows that no matter how fine the partition $P$ we have
$$|S(P,f, \alpha) - A| > \epsilon.$$
Thus, when $f$ is unbounded, it is impossible to find $A$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(P,f,\alpha) - A| < \epsilon$ holds. We can always select the tags so that the inequality is violated.