Riemann-Stieltjes integral of unbounded function

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.

Remember that the Riemann/Darboux integral requires the function to be bounded, or at least one of the upper and lower sums for a given partition will always diverge. We see the same situation in the Darboux formulation of Riemann–Stieltjes integrability.

Of course, one can formulate an improper Riemann–Stieltjes integral in exactly the same way as the improper Riemann integral: see, e.g., Burkill & Burkill, § 6.3.